MATH 117 - ELEMENTS OF
MATHEMATICS - Jacobian matrix
Question Bank - Set 4
Liberty University
Question 1
Question
Find the Jacobian matrix of the following vector-valued function:
f(x, y, z) =
xyz
x2+y2+z2
exy
Solution
To find the Jacobian matrix of a vector-valued function, we need to compute
the partial derivatives of each component function with respect to each variable.
Step 1: Compute the partial derivatives of each component function with
respect to x.
∂f1
∂x =yz, ∂f2
∂x = 2x, ∂f3
∂x =yexy
Step 2: Compute the partial derivatives of each component function with
respect to y.
∂f1
∂y =xz, ∂f2
∂y = 2y, ∂f3
∂y =xexy
Step 3: Compute the partial derivatives of each component function with
respect to z.
∂f1
∂z =xy, ∂f2
∂z = 2z, ∂f3
∂z = 0
Step 4: Assemble the Jacobian matrix J.
J=
yz xz xy
2x2y0
yexy xexy 0
Question 2
Question
Let f(x, y) = 3x2y
x+y2. Find the Jacobian matrix of f.
Solution
We want to find the Jacobian matrix of the function f(x, y) = 3x2y
x+y2.
Step 1: Calculate the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (3x2y)
∂
∂x (x+y2)
=6xy
1
∂f
∂y ="∂
∂y (3x2y)
∂
∂y (x+y2)#
=3x2
2y
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
where f1and f2are the components of f.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
Jf=6xy 1
3x22y
Therefore, the Jacobian matrix of fis
Jf=6xy 1
3x22y
Question 3
Question
Let f(x, y, z) = (x2+y, 2y+z, 3z+x) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,2,3).
2
Solution
To find the Jacobian matrix of fat the point (1,2,3), we must first define the
function fand then compute its partial derivatives with respect to each variable
at that point.
Step 1: Define the function f
f(x, y, z)=(x2+y, 2y+z, 3z+x)
Step 2: Compute partial derivatives of fwith respect to each
variable The Jacobian matrix of fat (1,2,3) is given by:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Where ∂f1
∂x is the partial derivative of the first component of fwith respect
to x, and so on.
Step 3: Compute the partial derivatives
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2,∂f2
∂z = 1
∂f3
∂x = 1,∂f3
∂y = 0,∂f3
∂z = 3
Step 4: Evaluate the Jacobian matrix at (1,2,3) Substitute x= 1,
y= 2, and z= 3 into the partial derivatives to find the Jacobian matrix at
(1,2,3):
Jf(1,2,3) =
210
021
103
Question 4
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2) be a vector-valued function. Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to the variables x,y, and z. The Jacobian
matrix will then be the matrix of these partial derivatives.
3
Step 1: Compute the partial derivatives of each component function.
For the first component function, f1(x, y, z) = x2+y2, we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
For the second component function, f2(x, y, z) = y2+z2, we have:
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
For the third component function, f3(x, y, z) = z2+x2, we have:
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix.
The Jacobian matrix of fis given by:
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y0
0 2y2z
2x0 2z
Question 5
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, ex+y). Find the Jacobian
matrix of fat the point (1,0,1).
Solution
To find the Jacobian matrix of fat the point (1,0,1), we need to compute the
partial derivatives of fwith respect to each variable at that point.
Step 1: Compute the partial derivatives of fLet F(x, y, z) = (u, v) =
f(x, y, z). Then, the Jacobian matrix of fis given by:
Jf= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
To find the partial derivatives, we have:
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
∂v
∂x =ex+y,∂v
∂y =ex+y,∂v
∂z = 0
4
Step 2: Evaluate the partial derivatives at the point (1,0,1) Substi-
tute x= 1, y= 0, and z= 1 into the partial derivatives:
∂u
∂x = 2(1) = 2,∂u
∂y = 1,∂u
∂z = 0
∂v
∂x =e1+0 =e, ∂v
∂y =e1+0 =e, ∂v
∂z = 0
Step 3: Construct the Jacobian matrix at the point (1,0,1) The
Jacobian matrix of fat the point (1,0,1) is:
Jf(1,0,1) = 210
e e 0
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, yz). Calculate the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet u=x2yand v=yz.
Then, the components of fare uand v. We will compute the partial derivatives
of uand vwith respect to x,y, and z.
∂u
∂x =∂(x2y)
∂x = 2xy
∂u
∂y =∂(x2y)
∂y =x2
∂u
∂z = 0
∂v
∂x = 0
∂v
∂y =∂(yz)
∂y =z
∂v
∂z =∂(yz)
∂z =y
5
Step 2: Assemble the Jacobian matrix The Jacobian matrix of fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the computed partial derivatives into the matrix:
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of fis:
Jf=2xy x20
0z y
Question 7
Question
Let f:R3→R2be a differentiable function given by f(x, y, z) = (xy +z, x2+
yz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f:R3→R2, we need to compute the partial
derivatives of each component function.
Step 1: Compute the partial derivatives of f: Let u=xy+zand v=x2+yz.
We have: ∂u
∂x =y, ∂u
∂y =x, ∂u
∂z = 1
and ∂v
∂x = 2x, ∂v
∂y =z, ∂v
∂z =y
Step 2: Assemble the Jacobian matrix: The Jacobian matrix Jfof fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the partial derivatives into the Jacobian matrix:
Jf=y x 1
2x z y
Hence, the Jacobian matrix of fis:
Jf=y x 1
2x z y
6
Question 8
Question
Consider the following system of equations:
(x2+ 2y2= 6
xy = 3
Find the Jacobian matrix of the system at the point (x, y) = (1,3).
Solution
To find the Jacobian matrix of the system at the point (x, y) = (1,3), we first
need to find the partial derivatives of the two equations.
Step 1: Find the partial derivatives
For the first equation x2+ 2y2= 6, we have:
∂
∂x (x2+ 2y2)=2x
∂
∂y (x2+ 2y2)=4y
For the second equation xy = 3, we have:
∂
∂x (xy) = y
∂
∂y (xy) = x
Step 2: Evaluate the partial derivatives at (x, y) = (1,3)
Evaluating at (x, y) = (1,3), we get:
∂
∂x (x2+ 2y2)(1,3)
= 2 ×1=2
∂
∂y (x2+ 2y2)(1,3)
= 4 ×3 = 12
∂
∂x (xy)(1,3)
= 3
∂
∂y (xy)(1,3)
= 1
7
Step 3: Assemble the Jacobian matrix The Jacobian matrix is formed
by arranging the partial derivatives in the following order:
J="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute the evaluated partial derivatives into the Jacobian matrix:
J=2 12
3 1
Therefore, the Jacobian matrix of the system at the point (x, y) = (1,3) is
2 12
3 1 .
Question 9
Question
Let f(x, y, z) =
x2y
ysin(z)
xyz
. Compute the Jacobian matrix of fwith respect to
x,y, and z.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂(x2y)
∂x = 2xy,
∂(x2y)
∂y =x2,
∂(ysin(z))
∂y = sin(z),
∂(ysin(z))
∂z =ycos(z),
∂(xyz)
∂x =yz,
∂(xyz)
∂y =xz,
∂(xyz)
∂z =xy.
Step 2: Assemble the partial derivatives into the Jacobian matrix of f.
Jacobian matrix of f=
2xy x20
0 sin(z)ycos(z)
yz xz xy
.
8
Question 10
Question
Let f(x, y, z) =
exyz
sin(xy) + z2
xyz
. Find the Jacobian matrix of f.
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
yzexyz
ycos(xy)
yz
,
∂f
∂y =
xzexyz
xcos(xy)
xz
,
∂f
∂z =
yxexyz
0
xy
.
Step 2: Construct the Jacobian matrix by arranging the partial derivatives.
Jacobian matrix J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Therefore, the Jacobian matrix of fis given by:
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Question 11
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2y, eyz , xyz).
Find the Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable (x, y, z).
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for f1(x, y, z) = x2y:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for f2(x, y, z) = eyz:
∂f2
∂x = 0,∂f2
∂y =zeyz,∂f2
∂z =yeyz
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for f3(x, y, z) = xyz:
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Construct the Jacobian matrix of fusing the partial derivatives
computed above:
Jacobian matrix =
2xy x20
0zeyz yeyz
yz xz xy
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, xz). Calculate the
Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x2y),∂
∂x (xz)= (2xy, z)
∂f
∂y =∂
∂y (x2y),∂
∂y (xz)= (x2,0)
∂f
∂z =∂
∂z (x2y),∂
∂z (xz)= (0, x)
10
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1)(2),3) = (2,3)
∂f
∂y (1,2,3)
= (12,0) = (1,0)
∂f
∂z (1,2,3)
= (0,1) = (0,1)
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
The Jacobian matrix of fat the point (1,2,3) is:
2 3
1 0
0 1
Question 13
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+z, yz).
Find the Jacobian matrix of fat the point (1,−2,3).
Solution
Step 1: Compute the partial derivatives of the function f.The Jacobian matrix
of a function is the matrix of all its first-order partial derivatives. For a function
f:Rn→Rm, the Jacobian matrix is an m×nmatrix.
Given f(x, y, z)=(x2+z, yz), the partial derivatives of fare:
∂f
∂x = (2x, 0) and ∂f
∂y = (0, z) and ∂f
∂z = (1, y)
Step 2: Evaluate the partial derivatives at the point (1,−2,3).Now, evaluate
these partial derivatives at the given point (1,−2,3):
∂f
∂x (1,−2,3) = (2(1),0) = (2,0)
∂f
∂y (1,−2,3) = (0,3) = (0,3)
∂f
∂z (1,−2,3) = (1,−2) = (1,−2)
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat the
point (1,−2,3) is:
Jf(1,−2,3) =
2 0
0 3
1−2
11
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2y, yz). Compute the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2y.
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jfusing the computed partial deriva-
tives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Question 15
Question
Let f:R3→R2be given by f(x, y, z) = (x2, yz). Determine the Jacobian
matrix of f.
Solution
To determine the Jacobian matrix of f, we need to compute the partial deriva-
tives of each component function.
12
Step 1: Compute ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2)=2x,
∂f1
∂y = 0,
∂f1
∂z = 0.
Step 2: Compute ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,
∂f2
∂y =z,
∂f2
∂z =y.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y.
Therefore, the Jacobian matrix of fis 2x0 0
0z y.
Question 16
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xyz). Compute the Jacobian
matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x, y, z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
13
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix Jfof fusing the computed partial
derivatives:
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis 2x z y
yz xz xy.
Question 17
Question
Let f(x, y, z) = (x+y+z, xy, xyz). Compute the Jacobian matrix Jfof fat
the point (1,2,3).
Solution
To compute the Jacobian matrix Jfof the function f(x, y, z)=(x+y+
z, xy, xyz) at the point (1,2,3), we need to find the partial derivatives of each
component of fwith respect to x,y, and zat the given point.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x+y+z) = 1,
∂f1
∂y =∂
∂y (x+y+z) = 1,
∂f1
∂z =∂
∂z (x+y+z)=1.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy) = y= 2,
∂f2
∂y =∂
∂y (xy) = x= 1,
∂f2
∂z = 0.
14
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz) = yz = 6,
∂f3
∂y =∂
∂y (xyz) = xz = 3,
∂f3
∂z =∂
∂z (xyz) = xy = 2.
Step 4: Assemble the Jacobian matrix Jf.
Jf=
111
210
632
Question 18
Question
Let f(x, y, z) = (x2+y2, xy +z, xz −y). Find the Jacobian matrix of fat the
point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 1
∂f3
∂x =z, ∂f3
∂y =−1,∂f3
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 2(2) = 4,∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 1,∂f2
∂z (1,2,3) = 1
∂f3
∂x (1,2,3) = 3,∂f3
∂y (1,2,3) = −1,∂f3
∂z (1,2,3) = 1
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
Jf(1,2,3) =
240
211
3−1 1
15
Question 19
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y −
z). Determine the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x z
∂f
∂y =z1
∂f
∂z =y−1
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2 3
∂f
∂y (1,2,3) = 3 1
∂f
∂z (1,2,3) = 2−1
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2 3 2
3 1 −1
Question 20
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xz). Calculate the Jacobian
matrix of fat the point (1,2,3).
Solution
To calculate the Jacobian matrix of fat the point (1,2,3), we first need to find
the partial derivatives of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y.
16
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x.
Step 3: Evaluate the partial derivatives at (1,2,3).
Plugging in (x, y, z) = (1,2,3), we have:
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2,
∂f2
∂x = 3,∂f2
∂y = 0,∂f2
∂z = 1.
Step 4: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=232
301.
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2+y, y2−z). Compute
the Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: Calculate the partial derivatives of fwith respect to each variable.
f1(x, y, z) = ∂
∂x (x2+y) = 2x,
f2(x, y, z) = ∂
∂y (x2+y) = 1,
f3(x, y, z) = ∂
∂z (x2+y)=0,
f4(x, y, z) = ∂
∂x (y2−z)=0,
f5(x, y, z) = ∂
∂y (y2−z)=2y,
f6(x, y, z) = ∂
∂z (y2−z) = −1.
17
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
f1(1,2,−1) = 2(1) = 2,
f2(1,2,−1) = 1,
f3(1,2,−1) = 0,
f4(1,2,−1) = 0,
f5(1,2,−1) = 2(2) = 4,
f6(1,2,−1) = −1.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in a 2 ×3 matrix.
Jf(1,2,−1) = 2 1 0
0 4 −1.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, z3). Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f1
∂x = 2xand ∂f2
∂x = 0
Step 2: Compute the partial derivatives of fwith respect to y:
∂f1
∂y =zand ∂f2
∂y = 0
Step 3: Compute the partial derivatives of fwith respect to z:
∂f1
∂z =yand ∂f2
∂z = 3z2
Step 4: Assemble the partial derivatives into the Jacobian matrix J(f):
J(f) = 2x z
y3z2
Therefore, the Jacobian matrix of the function f(x, y, z) = (x2+yz, z3) is
given by
J(f) = 2x z
y3z2
.
18
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Given f(x, y, z) = x2y+y2z+z2x, the Jacobian matrix of fis given by:
Jf=h∂f
∂x
∂f
∂y
∂f
∂z i
Step 1: Find ∂f
∂x .
∂f
∂x = 2xy +z2
Step 2: Find ∂f
∂y .
∂f
∂y =x2+ 2yz
Step 3: Find ∂f
∂z .
∂f
∂z =y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3). Substitute
x= 1, y= 2, z= 3 into the partial derivatives calculated in steps 1-3:
∂f
∂x = 2(1)(2) + (3)2= 4 + 9 = 13
∂f
∂y = (1)2+ 2(2)(3) = 1 + 12 = 13
∂f
∂z = (2)2+ 2(3)(1) = 4 + 6 = 10
Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 13 13 10
Question 24
Question
Let f(x, y, z) =
x2+yz
2xz
3xy
. Find the Jacobian matrix of fat the point (1,2,3).
19
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
0
3y
,∂f
∂y =
z
0
3x
,∂f
∂z =
y
2x
0
.
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x =
2(1)
0
3(2)
=
2
0
6
,
∂f
∂y =
3
0
3(1)
=
3
0
3
,
∂f
∂z =
2
2(1)
0
=
2
2
0
.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
206
303
220
.
Question 25
Question
Let f(x, y, z)=(x2+y2+z2, xyz, x +y+z). Calculate the Jacobian matrix of
f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f
∂x
For the first component: x2+y2+z2, we have:
∂(x2+y2+z2)
∂x = 2x
For the second component: xyz, we have:
∂(xyz)
∂x =yz
20
For the third component: x+y+z, we have:
∂(x+y+z)
∂x = 1
Therefore, the first row of the Jacobian matrix is 2x yz 1.
Step 2: Compute ∂f
∂y
Following the same process, we find:
∂f
∂y =2y xz 1
Step 3: Compute ∂f
∂z
Similarly, we find:
∂f
∂z =2z xy 1
Step 4: Assemble the Jacobian matrix
Thus, the Jacobian matrix of fis:
Jf=
2x yz 1
2y xz 1
2z xy 1
Question 26
Question
Let f(x, y)=(xy2, x2y, x +y) be a vector-valued function. Find the Jacobian
matrix of fat the point (2,1).
Solution
To find the Jacobian matrix of a vector-valued function, we find the partial
derivatives of each component function with respect to each variable. The Ja-
cobian matrix of fat the point (2,1) will be a 3 ×2 matrix.
Step 1: Find the partial derivatives of each component function.
Let’s denote u=xand v=y. Then,
∂f1
∂u =y2,∂f1
∂v = 2xy
∂f2
∂u = 2xy, ∂f2
∂v =x2
∂f3
∂u = 1,∂f3
∂v = 1
Step 2: Evaluate the partial derivatives at the point (2,1).At the
point (2,1), we have:
∂f1
∂u = 1,∂f1
∂v = 4
21
∂f2
∂u = 4,∂f2
∂v = 4
∂f3
∂u = 1,∂f3
∂v = 1
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat
(2,1) is:
Jf(2,1) =
1 4
4 4
1 1
.
Question 27
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+y2, yz).
Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z
Let f1(x, y, z) = x2+y2. Then,
∂f1
∂x = 2x, ∂f1
∂y = 2y, and ∂f1
∂z = 0.
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z
Let f2(x, y, z) = yz. Then,
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Put it all together to form the Jacobian matrix of f
The Jacobian matrix of fis given by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y.
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+y, xyz). Find the
Jacobian matrix of f.
22
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Assemble the Jacobian matrix:
Jf(x, y, z) = 2x1 0
yz xz xy
Question 29
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, z2−x). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the component functions of f.
Step 2: To find the Jacobian matrix of f, we first compute the partial
derivatives of fwith respect to x,y, and z:
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2z
Step 3: Now evaluate these partial derivatives at the point (1,2,3) to find
the Jacobian matrix at that point:
At (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2(3) = 6
Step 4: Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 2 3 2
−106
23
Question 30
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Compute the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y, yz), we have:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with f(x, y, z)=(x2+y, yz), we have:
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
The Jacobian matrix Jfis formed by arranging the partial derivatives in a
matrix. Therefore,
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0z y
So, the Jacobian matrix of fis 2x1 0
0z y.
24
Question 2
Question
Let f(x, y) = 3x2y
x+y2. Find the Jacobian matrix of f.
Solution
We want to find the Jacobian matrix of the function f(x, y) = 3x2y
x+y2.
Step 1: Calculate the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (3x2y)
∂
∂x (x+y2)
=6xy
1
∂f
∂y ="∂
∂y (3x2y)
∂
∂y (x+y2)#
=3x2
2y
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
where f1and f2are the components of f.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
Jf=6xy 1
3x22y
Therefore, the Jacobian matrix of fis
Jf=6xy 1
3x22y
Question 3
Question
Let f(x, y, z) = (x2+y, 2y+z, 3z+x) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,2,3).
2
Solution
To find the Jacobian matrix of fat the point (1,2,3), we must first define the
function fand then compute its partial derivatives with respect to each variable
at that point.
Step 1: Define the function f
f(x, y, z)=(x2+y, 2y+z, 3z+x)
Step 2: Compute partial derivatives of fwith respect to each
variable The Jacobian matrix of fat (1,2,3) is given by:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Where ∂f1
∂x is the partial derivative of the first component of fwith respect
to x, and so on.
Step 3: Compute the partial derivatives
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2,∂f2
∂z = 1
∂f3
∂x = 1,∂f3
∂y = 0,∂f3
∂z = 3
Step 4: Evaluate the Jacobian matrix at (1,2,3) Substitute x= 1,
y= 2, and z= 3 into the partial derivatives to find the Jacobian matrix at
(1,2,3):
Jf(1,2,3) =
210
021
103
Question 4
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2) be a vector-valued function. Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to the variables x,y, and z. The Jacobian
matrix will then be the matrix of these partial derivatives.
3
Step 1: Compute the partial derivatives of each component function.
For the first component function, f1(x, y, z) = x2+y2, we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
For the second component function, f2(x, y, z) = y2+z2, we have:
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
For the third component function, f3(x, y, z) = z2+x2, we have:
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix.
The Jacobian matrix of fis given by:
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y0
0 2y2z
2x0 2z
Question 5
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, ex+y). Find the Jacobian
matrix of fat the point (1,0,1).
Solution
To find the Jacobian matrix of fat the point (1,0,1), we need to compute the
partial derivatives of fwith respect to each variable at that point.
Step 1: Compute the partial derivatives of fLet F(x, y, z) = (u, v) =
f(x, y, z). Then, the Jacobian matrix of fis given by:
Jf= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
To find the partial derivatives, we have:
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
∂v
∂x =ex+y,∂v
∂y =ex+y,∂v
∂z = 0
4
Step 2: Evaluate the partial derivatives at the point (1,0,1) Substi-
tute x= 1, y= 0, and z= 1 into the partial derivatives:
∂u
∂x = 2(1) = 2,∂u
∂y = 1,∂u
∂z = 0
∂v
∂x =e1+0 =e, ∂v
∂y =e1+0 =e, ∂v
∂z = 0
Step 3: Construct the Jacobian matrix at the point (1,0,1) The
Jacobian matrix of fat the point (1,0,1) is:
Jf(1,0,1) = 210
e e 0
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, yz). Calculate the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet u=x2yand v=yz.
Then, the components of fare uand v. We will compute the partial derivatives
of uand vwith respect to x,y, and z.
∂u
∂x =∂(x2y)
∂x = 2xy
∂u
∂y =∂(x2y)
∂y =x2
∂u
∂z = 0
∂v
∂x = 0
∂v
∂y =∂(yz)
∂y =z
∂v
∂z =∂(yz)
∂z =y
5
Step 2: Assemble the Jacobian matrix The Jacobian matrix of fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the computed partial derivatives into the matrix:
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of fis:
Jf=2xy x20
0z y
Question 7
Question
Let f:R3→R2be a differentiable function given by f(x, y, z) = (xy +z, x2+
yz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f:R3→R2, we need to compute the partial
derivatives of each component function.
Step 1: Compute the partial derivatives of f: Let u=xy+zand v=x2+yz.
We have: ∂u
∂x =y, ∂u
∂y =x, ∂u
∂z = 1
and ∂v
∂x = 2x, ∂v
∂y =z, ∂v
∂z =y
Step 2: Assemble the Jacobian matrix: The Jacobian matrix Jfof fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the partial derivatives into the Jacobian matrix:
Jf=y x 1
2x z y
Hence, the Jacobian matrix of fis:
Jf=y x 1
2x z y
6
Question 8
Question
Consider the following system of equations:
(x2+ 2y2= 6
xy = 3
Find the Jacobian matrix of the system at the point (x, y) = (1,3).
Solution
To find the Jacobian matrix of the system at the point (x, y) = (1,3), we first
need to find the partial derivatives of the two equations.
Step 1: Find the partial derivatives
For the first equation x2+ 2y2= 6, we have:
∂
∂x (x2+ 2y2)=2x
∂
∂y (x2+ 2y2)=4y
For the second equation xy = 3, we have:
∂
∂x (xy) = y
∂
∂y (xy) = x
Step 2: Evaluate the partial derivatives at (x, y) = (1,3)
Evaluating at (x, y) = (1,3), we get:
∂
∂x (x2+ 2y2)(1,3)
= 2 ×1=2
∂
∂y (x2+ 2y2)(1,3)
= 4 ×3 = 12
∂
∂x (xy)(1,3)
= 3
∂
∂y (xy)(1,3)
= 1
7
Step 3: Assemble the Jacobian matrix The Jacobian matrix is formed
by arranging the partial derivatives in the following order:
J="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute the evaluated partial derivatives into the Jacobian matrix:
J=2 12
3 1
Therefore, the Jacobian matrix of the system at the point (x, y) = (1,3) is
2 12
3 1 .
Question 9
Question
Let f(x, y, z) =
x2y
ysin(z)
xyz
. Compute the Jacobian matrix of fwith respect to
x,y, and z.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂(x2y)
∂x = 2xy,
∂(x2y)
∂y =x2,
∂(ysin(z))
∂y = sin(z),
∂(ysin(z))
∂z =ycos(z),
∂(xyz)
∂x =yz,
∂(xyz)
∂y =xz,
∂(xyz)
∂z =xy.
Step 2: Assemble the partial derivatives into the Jacobian matrix of f.
Jacobian matrix of f=
2xy x20
0 sin(z)ycos(z)
yz xz xy
.
8
Question 10
Question
Let f(x, y, z) =
exyz
sin(xy) + z2
xyz
. Find the Jacobian matrix of f.
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
yzexyz
ycos(xy)
yz
,
∂f
∂y =
xzexyz
xcos(xy)
xz
,
∂f
∂z =
yxexyz
0
xy
.
Step 2: Construct the Jacobian matrix by arranging the partial derivatives.
Jacobian matrix J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Therefore, the Jacobian matrix of fis given by:
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Question 11
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2y, eyz , xyz).
Find the Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable (x, y, z).
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for f1(x, y, z) = x2y:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for f2(x, y, z) = eyz:
∂f2
∂x = 0,∂f2
∂y =zeyz,∂f2
∂z =yeyz
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for f3(x, y, z) = xyz:
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Construct the Jacobian matrix of fusing the partial derivatives
computed above:
Jacobian matrix =
2xy x20
0zeyz yeyz
yz xz xy
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, xz). Calculate the
Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x2y),∂
∂x (xz)= (2xy, z)
∂f
∂y =∂
∂y (x2y),∂
∂y (xz)= (x2,0)
∂f
∂z =∂
∂z (x2y),∂
∂z (xz)= (0, x)
10
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1)(2),3) = (2,3)
∂f
∂y (1,2,3)
= (12,0) = (1,0)
∂f
∂z (1,2,3)
= (0,1) = (0,1)
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
The Jacobian matrix of fat the point (1,2,3) is:
2 3
1 0
0 1
Question 13
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+z, yz).
Find the Jacobian matrix of fat the point (1,−2,3).
Solution
Step 1: Compute the partial derivatives of the function f.The Jacobian matrix
of a function is the matrix of all its first-order partial derivatives. For a function
f:Rn→Rm, the Jacobian matrix is an m×nmatrix.
Given f(x, y, z)=(x2+z, yz), the partial derivatives of fare:
∂f
∂x = (2x, 0) and ∂f
∂y = (0, z) and ∂f
∂z = (1, y)
Step 2: Evaluate the partial derivatives at the point (1,−2,3).Now, evaluate
these partial derivatives at the given point (1,−2,3):
∂f
∂x (1,−2,3) = (2(1),0) = (2,0)
∂f
∂y (1,−2,3) = (0,3) = (0,3)
∂f
∂z (1,−2,3) = (1,−2) = (1,−2)
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat the
point (1,−2,3) is:
Jf(1,−2,3) =
2 0
0 3
1−2
11
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2y, yz). Compute the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2y.
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jfusing the computed partial deriva-
tives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Question 15
Question
Let f:R3→R2be given by f(x, y, z) = (x2, yz). Determine the Jacobian
matrix of f.
Solution
To determine the Jacobian matrix of f, we need to compute the partial deriva-
tives of each component function.
12
Step 1: Compute ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2)=2x,
∂f1
∂y = 0,
∂f1
∂z = 0.
Step 2: Compute ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,
∂f2
∂y =z,
∂f2
∂z =y.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y.
Therefore, the Jacobian matrix of fis 2x0 0
0z y.
Question 16
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xyz). Compute the Jacobian
matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x, y, z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
13
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix Jfof fusing the computed partial
derivatives:
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis 2x z y
yz xz xy.
Question 17
Question
Let f(x, y, z) = (x+y+z, xy, xyz). Compute the Jacobian matrix Jfof fat
the point (1,2,3).
Solution
To compute the Jacobian matrix Jfof the function f(x, y, z)=(x+y+
z, xy, xyz) at the point (1,2,3), we need to find the partial derivatives of each
component of fwith respect to x,y, and zat the given point.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x+y+z) = 1,
∂f1
∂y =∂
∂y (x+y+z) = 1,
∂f1
∂z =∂
∂z (x+y+z)=1.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy) = y= 2,
∂f2
∂y =∂
∂y (xy) = x= 1,
∂f2
∂z = 0.
14
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz) = yz = 6,
∂f3
∂y =∂
∂y (xyz) = xz = 3,
∂f3
∂z =∂
∂z (xyz) = xy = 2.
Step 4: Assemble the Jacobian matrix Jf.
Jf=
111
210
632
Question 18
Question
Let f(x, y, z) = (x2+y2, xy +z, xz −y). Find the Jacobian matrix of fat the
point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 1
∂f3
∂x =z, ∂f3
∂y =−1,∂f3
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 2(2) = 4,∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 1,∂f2
∂z (1,2,3) = 1
∂f3
∂x (1,2,3) = 3,∂f3
∂y (1,2,3) = −1,∂f3
∂z (1,2,3) = 1
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
Jf(1,2,3) =
240
211
3−1 1
15
Question 19
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y −
z). Determine the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x z
∂f
∂y =z1
∂f
∂z =y−1
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2 3
∂f
∂y (1,2,3) = 3 1
∂f
∂z (1,2,3) = 2−1
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2 3 2
3 1 −1
Question 20
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xz). Calculate the Jacobian
matrix of fat the point (1,2,3).
Solution
To calculate the Jacobian matrix of fat the point (1,2,3), we first need to find
the partial derivatives of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y.
16
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x.
Step 3: Evaluate the partial derivatives at (1,2,3).
Plugging in (x, y, z) = (1,2,3), we have:
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2,
∂f2
∂x = 3,∂f2
∂y = 0,∂f2
∂z = 1.
Step 4: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=232
301.
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2+y, y2−z). Compute
the Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: Calculate the partial derivatives of fwith respect to each variable.
f1(x, y, z) = ∂
∂x (x2+y) = 2x,
f2(x, y, z) = ∂
∂y (x2+y) = 1,
f3(x, y, z) = ∂
∂z (x2+y)=0,
f4(x, y, z) = ∂
∂x (y2−z)=0,
f5(x, y, z) = ∂
∂y (y2−z)=2y,
f6(x, y, z) = ∂
∂z (y2−z) = −1.
17
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
f1(1,2,−1) = 2(1) = 2,
f2(1,2,−1) = 1,
f3(1,2,−1) = 0,
f4(1,2,−1) = 0,
f5(1,2,−1) = 2(2) = 4,
f6(1,2,−1) = −1.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in a 2 ×3 matrix.
Jf(1,2,−1) = 2 1 0
0 4 −1.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, z3). Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f1
∂x = 2xand ∂f2
∂x = 0
Step 2: Compute the partial derivatives of fwith respect to y:
∂f1
∂y =zand ∂f2
∂y = 0
Step 3: Compute the partial derivatives of fwith respect to z:
∂f1
∂z =yand ∂f2
∂z = 3z2
Step 4: Assemble the partial derivatives into the Jacobian matrix J(f):
J(f) = 2x z
y3z2
Therefore, the Jacobian matrix of the function f(x, y, z) = (x2+yz, z3) is
given by
J(f) = 2x z
y3z2
.
18
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Given f(x, y, z) = x2y+y2z+z2x, the Jacobian matrix of fis given by:
Jf=h∂f
∂x
∂f
∂y
∂f
∂z i
Step 1: Find ∂f
∂x .
∂f
∂x = 2xy +z2
Step 2: Find ∂f
∂y .
∂f
∂y =x2+ 2yz
Step 3: Find ∂f
∂z .
∂f
∂z =y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3). Substitute
x= 1, y= 2, z= 3 into the partial derivatives calculated in steps 1-3:
∂f
∂x = 2(1)(2) + (3)2= 4 + 9 = 13
∂f
∂y = (1)2+ 2(2)(3) = 1 + 12 = 13
∂f
∂z = (2)2+ 2(3)(1) = 4 + 6 = 10
Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 13 13 10
Question 24
Question
Let f(x, y, z) =
x2+yz
2xz
3xy
. Find the Jacobian matrix of fat the point (1,2,3).
19
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
0
3y
,∂f
∂y =
z
0
3x
,∂f
∂z =
y
2x
0
.
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x =
2(1)
0
3(2)
=
2
0
6
,
∂f
∂y =
3
0
3(1)
=
3
0
3
,
∂f
∂z =
2
2(1)
0
=
2
2
0
.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
206
303
220
.
Question 25
Question
Let f(x, y, z)=(x2+y2+z2, xyz, x +y+z). Calculate the Jacobian matrix of
f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f
∂x
For the first component: x2+y2+z2, we have:
∂(x2+y2+z2)
∂x = 2x
For the second component: xyz, we have:
∂(xyz)
∂x =yz
20
For the third component: x+y+z, we have:
∂(x+y+z)
∂x = 1
Therefore, the first row of the Jacobian matrix is 2x yz 1.
Step 2: Compute ∂f
∂y
Following the same process, we find:
∂f
∂y =2y xz 1
Step 3: Compute ∂f
∂z
Similarly, we find:
∂f
∂z =2z xy 1
Step 4: Assemble the Jacobian matrix
Thus, the Jacobian matrix of fis:
Jf=
2x yz 1
2y xz 1
2z xy 1
Question 26
Question
Let f(x, y)=(xy2, x2y, x +y) be a vector-valued function. Find the Jacobian
matrix of fat the point (2,1).
Solution
To find the Jacobian matrix of a vector-valued function, we find the partial
derivatives of each component function with respect to each variable. The Ja-
cobian matrix of fat the point (2,1) will be a 3 ×2 matrix.
Step 1: Find the partial derivatives of each component function.
Let’s denote u=xand v=y. Then,
∂f1
∂u =y2,∂f1
∂v = 2xy
∂f2
∂u = 2xy, ∂f2
∂v =x2
∂f3
∂u = 1,∂f3
∂v = 1
Step 2: Evaluate the partial derivatives at the point (2,1).At the
point (2,1), we have:
∂f1
∂u = 1,∂f1
∂v = 4
21
∂f2
∂u = 4,∂f2
∂v = 4
∂f3
∂u = 1,∂f3
∂v = 1
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat
(2,1) is:
Jf(2,1) =
1 4
4 4
1 1
.
Question 27
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+y2, yz).
Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z
Let f1(x, y, z) = x2+y2. Then,
∂f1
∂x = 2x, ∂f1
∂y = 2y, and ∂f1
∂z = 0.
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z
Let f2(x, y, z) = yz. Then,
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Put it all together to form the Jacobian matrix of f
The Jacobian matrix of fis given by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y.
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+y, xyz). Find the
Jacobian matrix of f.
22
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Assemble the Jacobian matrix:
Jf(x, y, z) = 2x1 0
yz xz xy
Question 29
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, z2−x). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the component functions of f.
Step 2: To find the Jacobian matrix of f, we first compute the partial
derivatives of fwith respect to x,y, and z:
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2z
Step 3: Now evaluate these partial derivatives at the point (1,2,3) to find
the Jacobian matrix at that point:
At (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2(3) = 6
Step 4: Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 2 3 2
−106
23
Question 30
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Compute the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y, yz), we have:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with f(x, y, z)=(x2+y, yz), we have:
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
The Jacobian matrix Jfis formed by arranging the partial derivatives in a
matrix. Therefore,
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0z y
So, the Jacobian matrix of fis 2x1 0
0z y.
24
Question 2
Question
Let f(x, y) = 3x2y
x+y2. Find the Jacobian matrix of f.
Solution
We want to find the Jacobian matrix of the function f(x, y) = 3x2y
x+y2.
Step 1: Calculate the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (3x2y)
∂
∂x (x+y2)
=6xy
1
∂f
∂y ="∂
∂y (3x2y)
∂
∂y (x+y2)#
=3x2
2y
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
where f1and f2are the components of f.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
Jf=6xy 1
3x22y
Therefore, the Jacobian matrix of fis
Jf=6xy 1
3x22y
Question 3
Question
Let f(x, y, z) = (x2+y, 2y+z, 3z+x) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,2,3).
2
Solution
To find the Jacobian matrix of fat the point (1,2,3), we must first define the
function fand then compute its partial derivatives with respect to each variable
at that point.
Step 1: Define the function f
f(x, y, z)=(x2+y, 2y+z, 3z+x)
Step 2: Compute partial derivatives of fwith respect to each
variable The Jacobian matrix of fat (1,2,3) is given by:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Where ∂f1
∂x is the partial derivative of the first component of fwith respect
to x, and so on.
Step 3: Compute the partial derivatives
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2,∂f2
∂z = 1
∂f3
∂x = 1,∂f3
∂y = 0,∂f3
∂z = 3
Step 4: Evaluate the Jacobian matrix at (1,2,3) Substitute x= 1,
y= 2, and z= 3 into the partial derivatives to find the Jacobian matrix at
(1,2,3):
Jf(1,2,3) =
210
021
103
Question 4
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2) be a vector-valued function. Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to the variables x,y, and z. The Jacobian
matrix will then be the matrix of these partial derivatives.
3
Step 1: Compute the partial derivatives of each component function.
For the first component function, f1(x, y, z) = x2+y2, we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
For the second component function, f2(x, y, z) = y2+z2, we have:
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
For the third component function, f3(x, y, z) = z2+x2, we have:
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix.
The Jacobian matrix of fis given by:
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y0
0 2y2z
2x0 2z
Question 5
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, ex+y). Find the Jacobian
matrix of fat the point (1,0,1).
Solution
To find the Jacobian matrix of fat the point (1,0,1), we need to compute the
partial derivatives of fwith respect to each variable at that point.
Step 1: Compute the partial derivatives of fLet F(x, y, z) = (u, v) =
f(x, y, z). Then, the Jacobian matrix of fis given by:
Jf= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
To find the partial derivatives, we have:
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
∂v
∂x =ex+y,∂v
∂y =ex+y,∂v
∂z = 0
4
Step 2: Evaluate the partial derivatives at the point (1,0,1) Substi-
tute x= 1, y= 0, and z= 1 into the partial derivatives:
∂u
∂x = 2(1) = 2,∂u
∂y = 1,∂u
∂z = 0
∂v
∂x =e1+0 =e, ∂v
∂y =e1+0 =e, ∂v
∂z = 0
Step 3: Construct the Jacobian matrix at the point (1,0,1) The
Jacobian matrix of fat the point (1,0,1) is:
Jf(1,0,1) = 210
e e 0
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, yz). Calculate the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet u=x2yand v=yz.
Then, the components of fare uand v. We will compute the partial derivatives
of uand vwith respect to x,y, and z.
∂u
∂x =∂(x2y)
∂x = 2xy
∂u
∂y =∂(x2y)
∂y =x2
∂u
∂z = 0
∂v
∂x = 0
∂v
∂y =∂(yz)
∂y =z
∂v
∂z =∂(yz)
∂z =y
5
Step 2: Assemble the Jacobian matrix The Jacobian matrix of fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the computed partial derivatives into the matrix:
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of fis:
Jf=2xy x20
0z y
Question 7
Question
Let f:R3→R2be a differentiable function given by f(x, y, z) = (xy +z, x2+
yz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f:R3→R2, we need to compute the partial
derivatives of each component function.
Step 1: Compute the partial derivatives of f: Let u=xy+zand v=x2+yz.
We have: ∂u
∂x =y, ∂u
∂y =x, ∂u
∂z = 1
and ∂v
∂x = 2x, ∂v
∂y =z, ∂v
∂z =y
Step 2: Assemble the Jacobian matrix: The Jacobian matrix Jfof fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the partial derivatives into the Jacobian matrix:
Jf=y x 1
2x z y
Hence, the Jacobian matrix of fis:
Jf=y x 1
2x z y
6
Question 8
Question
Consider the following system of equations:
(x2+ 2y2= 6
xy = 3
Find the Jacobian matrix of the system at the point (x, y) = (1,3).
Solution
To find the Jacobian matrix of the system at the point (x, y) = (1,3), we first
need to find the partial derivatives of the two equations.
Step 1: Find the partial derivatives
For the first equation x2+ 2y2= 6, we have:
∂
∂x (x2+ 2y2)=2x
∂
∂y (x2+ 2y2)=4y
For the second equation xy = 3, we have:
∂
∂x (xy) = y
∂
∂y (xy) = x
Step 2: Evaluate the partial derivatives at (x, y) = (1,3)
Evaluating at (x, y) = (1,3), we get:
∂
∂x (x2+ 2y2)(1,3)
= 2 ×1=2
∂
∂y (x2+ 2y2)(1,3)
= 4 ×3 = 12
∂
∂x (xy)(1,3)
= 3
∂
∂y (xy)(1,3)
= 1
7
Step 3: Assemble the Jacobian matrix The Jacobian matrix is formed
by arranging the partial derivatives in the following order:
J="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute the evaluated partial derivatives into the Jacobian matrix:
J=2 12
3 1
Therefore, the Jacobian matrix of the system at the point (x, y) = (1,3) is
2 12
3 1 .
Question 9
Question
Let f(x, y, z) =
x2y
ysin(z)
xyz
. Compute the Jacobian matrix of fwith respect to
x,y, and z.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂(x2y)
∂x = 2xy,
∂(x2y)
∂y =x2,
∂(ysin(z))
∂y = sin(z),
∂(ysin(z))
∂z =ycos(z),
∂(xyz)
∂x =yz,
∂(xyz)
∂y =xz,
∂(xyz)
∂z =xy.
Step 2: Assemble the partial derivatives into the Jacobian matrix of f.
Jacobian matrix of f=
2xy x20
0 sin(z)ycos(z)
yz xz xy
.
8
Question 10
Question
Let f(x, y, z) =
exyz
sin(xy) + z2
xyz
. Find the Jacobian matrix of f.
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
yzexyz
ycos(xy)
yz
,
∂f
∂y =
xzexyz
xcos(xy)
xz
,
∂f
∂z =
yxexyz
0
xy
.
Step 2: Construct the Jacobian matrix by arranging the partial derivatives.
Jacobian matrix J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Therefore, the Jacobian matrix of fis given by:
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Question 11
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2y, eyz , xyz).
Find the Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable (x, y, z).
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for f1(x, y, z) = x2y:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for f2(x, y, z) = eyz:
∂f2
∂x = 0,∂f2
∂y =zeyz,∂f2
∂z =yeyz
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for f3(x, y, z) = xyz:
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Construct the Jacobian matrix of fusing the partial derivatives
computed above:
Jacobian matrix =
2xy x20
0zeyz yeyz
yz xz xy
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, xz). Calculate the
Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x2y),∂
∂x (xz)= (2xy, z)
∂f
∂y =∂
∂y (x2y),∂
∂y (xz)= (x2,0)
∂f
∂z =∂
∂z (x2y),∂
∂z (xz)= (0, x)
10
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1)(2),3) = (2,3)
∂f
∂y (1,2,3)
= (12,0) = (1,0)
∂f
∂z (1,2,3)
= (0,1) = (0,1)
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
The Jacobian matrix of fat the point (1,2,3) is:
2 3
1 0
0 1
Question 13
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+z, yz).
Find the Jacobian matrix of fat the point (1,−2,3).
Solution
Step 1: Compute the partial derivatives of the function f.The Jacobian matrix
of a function is the matrix of all its first-order partial derivatives. For a function
f:Rn→Rm, the Jacobian matrix is an m×nmatrix.
Given f(x, y, z)=(x2+z, yz), the partial derivatives of fare:
∂f
∂x = (2x, 0) and ∂f
∂y = (0, z) and ∂f
∂z = (1, y)
Step 2: Evaluate the partial derivatives at the point (1,−2,3).Now, evaluate
these partial derivatives at the given point (1,−2,3):
∂f
∂x (1,−2,3) = (2(1),0) = (2,0)
∂f
∂y (1,−2,3) = (0,3) = (0,3)
∂f
∂z (1,−2,3) = (1,−2) = (1,−2)
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat the
point (1,−2,3) is:
Jf(1,−2,3) =
2 0
0 3
1−2
11
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2y, yz). Compute the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2y.
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jfusing the computed partial deriva-
tives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Question 15
Question
Let f:R3→R2be given by f(x, y, z) = (x2, yz). Determine the Jacobian
matrix of f.
Solution
To determine the Jacobian matrix of f, we need to compute the partial deriva-
tives of each component function.
12
Step 1: Compute ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2)=2x,
∂f1
∂y = 0,
∂f1
∂z = 0.
Step 2: Compute ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,
∂f2
∂y =z,
∂f2
∂z =y.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y.
Therefore, the Jacobian matrix of fis 2x0 0
0z y.
Question 16
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xyz). Compute the Jacobian
matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x, y, z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
13
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix Jfof fusing the computed partial
derivatives:
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis 2x z y
yz xz xy.
Question 17
Question
Let f(x, y, z) = (x+y+z, xy, xyz). Compute the Jacobian matrix Jfof fat
the point (1,2,3).
Solution
To compute the Jacobian matrix Jfof the function f(x, y, z)=(x+y+
z, xy, xyz) at the point (1,2,3), we need to find the partial derivatives of each
component of fwith respect to x,y, and zat the given point.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x+y+z) = 1,
∂f1
∂y =∂
∂y (x+y+z) = 1,
∂f1
∂z =∂
∂z (x+y+z)=1.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy) = y= 2,
∂f2
∂y =∂
∂y (xy) = x= 1,
∂f2
∂z = 0.
14
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz) = yz = 6,
∂f3
∂y =∂
∂y (xyz) = xz = 3,
∂f3
∂z =∂
∂z (xyz) = xy = 2.
Step 4: Assemble the Jacobian matrix Jf.
Jf=
111
210
632
Question 18
Question
Let f(x, y, z) = (x2+y2, xy +z, xz −y). Find the Jacobian matrix of fat the
point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 1
∂f3
∂x =z, ∂f3
∂y =−1,∂f3
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 2(2) = 4,∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 1,∂f2
∂z (1,2,3) = 1
∂f3
∂x (1,2,3) = 3,∂f3
∂y (1,2,3) = −1,∂f3
∂z (1,2,3) = 1
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
Jf(1,2,3) =
240
211
3−1 1
15
Question 19
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y −
z). Determine the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x z
∂f
∂y =z1
∂f
∂z =y−1
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2 3
∂f
∂y (1,2,3) = 3 1
∂f
∂z (1,2,3) = 2−1
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2 3 2
3 1 −1
Question 20
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xz). Calculate the Jacobian
matrix of fat the point (1,2,3).
Solution
To calculate the Jacobian matrix of fat the point (1,2,3), we first need to find
the partial derivatives of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y.
16
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x.
Step 3: Evaluate the partial derivatives at (1,2,3).
Plugging in (x, y, z) = (1,2,3), we have:
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2,
∂f2
∂x = 3,∂f2
∂y = 0,∂f2
∂z = 1.
Step 4: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=232
301.
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2+y, y2−z). Compute
the Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: Calculate the partial derivatives of fwith respect to each variable.
f1(x, y, z) = ∂
∂x (x2+y) = 2x,
f2(x, y, z) = ∂
∂y (x2+y) = 1,
f3(x, y, z) = ∂
∂z (x2+y)=0,
f4(x, y, z) = ∂
∂x (y2−z)=0,
f5(x, y, z) = ∂
∂y (y2−z)=2y,
f6(x, y, z) = ∂
∂z (y2−z) = −1.
17
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
f1(1,2,−1) = 2(1) = 2,
f2(1,2,−1) = 1,
f3(1,2,−1) = 0,
f4(1,2,−1) = 0,
f5(1,2,−1) = 2(2) = 4,
f6(1,2,−1) = −1.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in a 2 ×3 matrix.
Jf(1,2,−1) = 2 1 0
0 4 −1.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, z3). Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f1
∂x = 2xand ∂f2
∂x = 0
Step 2: Compute the partial derivatives of fwith respect to y:
∂f1
∂y =zand ∂f2
∂y = 0
Step 3: Compute the partial derivatives of fwith respect to z:
∂f1
∂z =yand ∂f2
∂z = 3z2
Step 4: Assemble the partial derivatives into the Jacobian matrix J(f):
J(f) = 2x z
y3z2
Therefore, the Jacobian matrix of the function f(x, y, z) = (x2+yz, z3) is
given by
J(f) = 2x z
y3z2
.
18
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Given f(x, y, z) = x2y+y2z+z2x, the Jacobian matrix of fis given by:
Jf=h∂f
∂x
∂f
∂y
∂f
∂z i
Step 1: Find ∂f
∂x .
∂f
∂x = 2xy +z2
Step 2: Find ∂f
∂y .
∂f
∂y =x2+ 2yz
Step 3: Find ∂f
∂z .
∂f
∂z =y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3). Substitute
x= 1, y= 2, z= 3 into the partial derivatives calculated in steps 1-3:
∂f
∂x = 2(1)(2) + (3)2= 4 + 9 = 13
∂f
∂y = (1)2+ 2(2)(3) = 1 + 12 = 13
∂f
∂z = (2)2+ 2(3)(1) = 4 + 6 = 10
Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 13 13 10
Question 24
Question
Let f(x, y, z) =
x2+yz
2xz
3xy
. Find the Jacobian matrix of fat the point (1,2,3).
19
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
0
3y
,∂f
∂y =
z
0
3x
,∂f
∂z =
y
2x
0
.
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x =
2(1)
0
3(2)
=
2
0
6
,
∂f
∂y =
3
0
3(1)
=
3
0
3
,
∂f
∂z =
2
2(1)
0
=
2
2
0
.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
206
303
220
.
Question 25
Question
Let f(x, y, z)=(x2+y2+z2, xyz, x +y+z). Calculate the Jacobian matrix of
f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f
∂x
For the first component: x2+y2+z2, we have:
∂(x2+y2+z2)
∂x = 2x
For the second component: xyz, we have:
∂(xyz)
∂x =yz
20
For the third component: x+y+z, we have:
∂(x+y+z)
∂x = 1
Therefore, the first row of the Jacobian matrix is 2x yz 1.
Step 2: Compute ∂f
∂y
Following the same process, we find:
∂f
∂y =2y xz 1
Step 3: Compute ∂f
∂z
Similarly, we find:
∂f
∂z =2z xy 1
Step 4: Assemble the Jacobian matrix
Thus, the Jacobian matrix of fis:
Jf=
2x yz 1
2y xz 1
2z xy 1
Question 26
Question
Let f(x, y)=(xy2, x2y, x +y) be a vector-valued function. Find the Jacobian
matrix of fat the point (2,1).
Solution
To find the Jacobian matrix of a vector-valued function, we find the partial
derivatives of each component function with respect to each variable. The Ja-
cobian matrix of fat the point (2,1) will be a 3 ×2 matrix.
Step 1: Find the partial derivatives of each component function.
Let’s denote u=xand v=y. Then,
∂f1
∂u =y2,∂f1
∂v = 2xy
∂f2
∂u = 2xy, ∂f2
∂v =x2
∂f3
∂u = 1,∂f3
∂v = 1
Step 2: Evaluate the partial derivatives at the point (2,1).At the
point (2,1), we have:
∂f1
∂u = 1,∂f1
∂v = 4
21
∂f2
∂u = 4,∂f2
∂v = 4
∂f3
∂u = 1,∂f3
∂v = 1
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat
(2,1) is:
Jf(2,1) =
1 4
4 4
1 1
.
Question 27
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+y2, yz).
Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z
Let f1(x, y, z) = x2+y2. Then,
∂f1
∂x = 2x, ∂f1
∂y = 2y, and ∂f1
∂z = 0.
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z
Let f2(x, y, z) = yz. Then,
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Put it all together to form the Jacobian matrix of f
The Jacobian matrix of fis given by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y.
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+y, xyz). Find the
Jacobian matrix of f.
22
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Assemble the Jacobian matrix:
Jf(x, y, z) = 2x1 0
yz xz xy
Question 29
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, z2−x). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the component functions of f.
Step 2: To find the Jacobian matrix of f, we first compute the partial
derivatives of fwith respect to x,y, and z:
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2z
Step 3: Now evaluate these partial derivatives at the point (1,2,3) to find
the Jacobian matrix at that point:
At (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2(3) = 6
Step 4: Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 2 3 2
−106
23
Question 30
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Compute the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y, yz), we have:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with f(x, y, z)=(x2+y, yz), we have:
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
The Jacobian matrix Jfis formed by arranging the partial derivatives in a
matrix. Therefore,
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0z y
So, the Jacobian matrix of fis 2x1 0
0z y.
24
Question 2
Question
Let f(x, y) = 3x2y
x+y2. Find the Jacobian matrix of f.
Solution
We want to find the Jacobian matrix of the function f(x, y) = 3x2y
x+y2.
Step 1: Calculate the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (3x2y)
∂
∂x (x+y2)
=6xy
1
∂f
∂y ="∂
∂y (3x2y)
∂
∂y (x+y2)#
=3x2
2y
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
where f1and f2are the components of f.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
Jf=6xy 1
3x22y
Therefore, the Jacobian matrix of fis
Jf=6xy 1
3x22y
Question 3
Question
Let f(x, y, z) = (x2+y, 2y+z, 3z+x) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,2,3).
2
Solution
To find the Jacobian matrix of fat the point (1,2,3), we must first define the
function fand then compute its partial derivatives with respect to each variable
at that point.
Step 1: Define the function f
f(x, y, z)=(x2+y, 2y+z, 3z+x)
Step 2: Compute partial derivatives of fwith respect to each
variable The Jacobian matrix of fat (1,2,3) is given by:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Where ∂f1
∂x is the partial derivative of the first component of fwith respect
to x, and so on.
Step 3: Compute the partial derivatives
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2,∂f2
∂z = 1
∂f3
∂x = 1,∂f3
∂y = 0,∂f3
∂z = 3
Step 4: Evaluate the Jacobian matrix at (1,2,3) Substitute x= 1,
y= 2, and z= 3 into the partial derivatives to find the Jacobian matrix at
(1,2,3):
Jf(1,2,3) =
210
021
103
Question 4
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2) be a vector-valued function. Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to the variables x,y, and z. The Jacobian
matrix will then be the matrix of these partial derivatives.
3
Step 1: Compute the partial derivatives of each component function.
For the first component function, f1(x, y, z) = x2+y2, we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
For the second component function, f2(x, y, z) = y2+z2, we have:
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
For the third component function, f3(x, y, z) = z2+x2, we have:
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix.
The Jacobian matrix of fis given by:
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y0
0 2y2z
2x0 2z
Question 5
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, ex+y). Find the Jacobian
matrix of fat the point (1,0,1).
Solution
To find the Jacobian matrix of fat the point (1,0,1), we need to compute the
partial derivatives of fwith respect to each variable at that point.
Step 1: Compute the partial derivatives of fLet F(x, y, z) = (u, v) =
f(x, y, z). Then, the Jacobian matrix of fis given by:
Jf= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
To find the partial derivatives, we have:
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
∂v
∂x =ex+y,∂v
∂y =ex+y,∂v
∂z = 0
4
Step 2: Evaluate the partial derivatives at the point (1,0,1) Substi-
tute x= 1, y= 0, and z= 1 into the partial derivatives:
∂u
∂x = 2(1) = 2,∂u
∂y = 1,∂u
∂z = 0
∂v
∂x =e1+0 =e, ∂v
∂y =e1+0 =e, ∂v
∂z = 0
Step 3: Construct the Jacobian matrix at the point (1,0,1) The
Jacobian matrix of fat the point (1,0,1) is:
Jf(1,0,1) = 210
e e 0
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, yz). Calculate the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet u=x2yand v=yz.
Then, the components of fare uand v. We will compute the partial derivatives
of uand vwith respect to x,y, and z.
∂u
∂x =∂(x2y)
∂x = 2xy
∂u
∂y =∂(x2y)
∂y =x2
∂u
∂z = 0
∂v
∂x = 0
∂v
∂y =∂(yz)
∂y =z
∂v
∂z =∂(yz)
∂z =y
5
Step 2: Assemble the Jacobian matrix The Jacobian matrix of fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the computed partial derivatives into the matrix:
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of fis:
Jf=2xy x20
0z y
Question 7
Question
Let f:R3→R2be a differentiable function given by f(x, y, z) = (xy +z, x2+
yz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f:R3→R2, we need to compute the partial
derivatives of each component function.
Step 1: Compute the partial derivatives of f: Let u=xy+zand v=x2+yz.
We have: ∂u
∂x =y, ∂u
∂y =x, ∂u
∂z = 1
and ∂v
∂x = 2x, ∂v
∂y =z, ∂v
∂z =y
Step 2: Assemble the Jacobian matrix: The Jacobian matrix Jfof fis
given by:
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
Substitute the partial derivatives into the Jacobian matrix:
Jf=y x 1
2x z y
Hence, the Jacobian matrix of fis:
Jf=y x 1
2x z y
6
Question 8
Question
Consider the following system of equations:
(x2+ 2y2= 6
xy = 3
Find the Jacobian matrix of the system at the point (x, y) = (1,3).
Solution
To find the Jacobian matrix of the system at the point (x, y) = (1,3), we first
need to find the partial derivatives of the two equations.
Step 1: Find the partial derivatives
For the first equation x2+ 2y2= 6, we have:
∂
∂x (x2+ 2y2)=2x
∂
∂y (x2+ 2y2)=4y
For the second equation xy = 3, we have:
∂
∂x (xy) = y
∂
∂y (xy) = x
Step 2: Evaluate the partial derivatives at (x, y) = (1,3)
Evaluating at (x, y) = (1,3), we get:
∂
∂x (x2+ 2y2)(1,3)
= 2 ×1=2
∂
∂y (x2+ 2y2)(1,3)
= 4 ×3 = 12
∂
∂x (xy)(1,3)
= 3
∂
∂y (xy)(1,3)
= 1
7
Step 3: Assemble the Jacobian matrix The Jacobian matrix is formed
by arranging the partial derivatives in the following order:
J="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute the evaluated partial derivatives into the Jacobian matrix:
J=2 12
3 1
Therefore, the Jacobian matrix of the system at the point (x, y) = (1,3) is
2 12
3 1 .
Question 9
Question
Let f(x, y, z) =
x2y
ysin(z)
xyz
. Compute the Jacobian matrix of fwith respect to
x,y, and z.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂(x2y)
∂x = 2xy,
∂(x2y)
∂y =x2,
∂(ysin(z))
∂y = sin(z),
∂(ysin(z))
∂z =ycos(z),
∂(xyz)
∂x =yz,
∂(xyz)
∂y =xz,
∂(xyz)
∂z =xy.
Step 2: Assemble the partial derivatives into the Jacobian matrix of f.
Jacobian matrix of f=
2xy x20
0 sin(z)ycos(z)
yz xz xy
.
8
Question 10
Question
Let f(x, y, z) =
exyz
sin(xy) + z2
xyz
. Find the Jacobian matrix of f.
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
yzexyz
ycos(xy)
yz
,
∂f
∂y =
xzexyz
xcos(xy)
xz
,
∂f
∂z =
yxexyz
0
xy
.
Step 2: Construct the Jacobian matrix by arranging the partial derivatives.
Jacobian matrix J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
.
Step 3: Substitute the partial derivatives into the Jacobian matrix.
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Therefore, the Jacobian matrix of fis given by:
J(f) =
yzexyz xzexyz yxexyz
ycos(xy)xcos(xy) 0
yz xz xy
.
Question 11
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2y, eyz , xyz).
Find the Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable (x, y, z).
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for f1(x, y, z) = x2y:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for f2(x, y, z) = eyz:
∂f2
∂x = 0,∂f2
∂y =zeyz,∂f2
∂z =yeyz
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for f3(x, y, z) = xyz:
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Construct the Jacobian matrix of fusing the partial derivatives
computed above:
Jacobian matrix =
2xy x20
0zeyz yeyz
yz xz xy
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y, xz). Calculate the
Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x2y),∂
∂x (xz)= (2xy, z)
∂f
∂y =∂
∂y (x2y),∂
∂y (xz)= (x2,0)
∂f
∂z =∂
∂z (x2y),∂
∂z (xz)= (0, x)
10
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1)(2),3) = (2,3)
∂f
∂y (1,2,3)
= (12,0) = (1,0)
∂f
∂z (1,2,3)
= (0,1) = (0,1)
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
The Jacobian matrix of fat the point (1,2,3) is:
2 3
1 0
0 1
Question 13
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+z, yz).
Find the Jacobian matrix of fat the point (1,−2,3).
Solution
Step 1: Compute the partial derivatives of the function f.The Jacobian matrix
of a function is the matrix of all its first-order partial derivatives. For a function
f:Rn→Rm, the Jacobian matrix is an m×nmatrix.
Given f(x, y, z)=(x2+z, yz), the partial derivatives of fare:
∂f
∂x = (2x, 0) and ∂f
∂y = (0, z) and ∂f
∂z = (1, y)
Step 2: Evaluate the partial derivatives at the point (1,−2,3).Now, evaluate
these partial derivatives at the given point (1,−2,3):
∂f
∂x (1,−2,3) = (2(1),0) = (2,0)
∂f
∂y (1,−2,3) = (0,3) = (0,3)
∂f
∂z (1,−2,3) = (1,−2) = (1,−2)
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat the
point (1,−2,3) is:
Jf(1,−2,3) =
2 0
0 3
1−2
11
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2y, yz). Compute the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2y.
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jfusing the computed partial deriva-
tives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Question 15
Question
Let f:R3→R2be given by f(x, y, z) = (x2, yz). Determine the Jacobian
matrix of f.
Solution
To determine the Jacobian matrix of f, we need to compute the partial deriva-
tives of each component function.
12
Step 1: Compute ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2)=2x,
∂f1
∂y = 0,
∂f1
∂z = 0.
Step 2: Compute ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,
∂f2
∂y =z,
∂f2
∂z =y.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y.
Therefore, the Jacobian matrix of fis 2x0 0
0z y.
Question 16
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xyz). Compute the Jacobian
matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x, y, z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
13
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix Jfof fusing the computed partial
derivatives:
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis 2x z y
yz xz xy.
Question 17
Question
Let f(x, y, z) = (x+y+z, xy, xyz). Compute the Jacobian matrix Jfof fat
the point (1,2,3).
Solution
To compute the Jacobian matrix Jfof the function f(x, y, z)=(x+y+
z, xy, xyz) at the point (1,2,3), we need to find the partial derivatives of each
component of fwith respect to x,y, and zat the given point.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x+y+z) = 1,
∂f1
∂y =∂
∂y (x+y+z) = 1,
∂f1
∂z =∂
∂z (x+y+z)=1.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy) = y= 2,
∂f2
∂y =∂
∂y (xy) = x= 1,
∂f2
∂z = 0.
14
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz) = yz = 6,
∂f3
∂y =∂
∂y (xyz) = xz = 3,
∂f3
∂z =∂
∂z (xyz) = xy = 2.
Step 4: Assemble the Jacobian matrix Jf.
Jf=
111
210
632
Question 18
Question
Let f(x, y, z) = (x2+y2, xy +z, xz −y). Find the Jacobian matrix of fat the
point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 1
∂f3
∂x =z, ∂f3
∂y =−1,∂f3
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 2(2) = 4,∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 1,∂f2
∂z (1,2,3) = 1
∂f3
∂x (1,2,3) = 3,∂f3
∂y (1,2,3) = −1,∂f3
∂z (1,2,3) = 1
Step 3: Assemble the Jacobian matrix using the evaluated partial derivatives.
Jf(1,2,3) =
240
211
3−1 1
15
Question 19
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y −
z). Determine the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x z
∂f
∂y =z1
∂f
∂z =y−1
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2 3
∂f
∂y (1,2,3) = 3 1
∂f
∂z (1,2,3) = 2−1
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2 3 2
3 1 −1
Question 20
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, xz). Calculate the Jacobian
matrix of fat the point (1,2,3).
Solution
To calculate the Jacobian matrix of fat the point (1,2,3), we first need to find
the partial derivatives of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y.
16
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Given f(x, y, z) = (x2+yz, xz),
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x.
Step 3: Evaluate the partial derivatives at (1,2,3).
Plugging in (x, y, z) = (1,2,3), we have:
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2,
∂f2
∂x = 3,∂f2
∂y = 0,∂f2
∂z = 1.
Step 4: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=232
301.
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x2+y, y2−z). Compute
the Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: Calculate the partial derivatives of fwith respect to each variable.
f1(x, y, z) = ∂
∂x (x2+y) = 2x,
f2(x, y, z) = ∂
∂y (x2+y) = 1,
f3(x, y, z) = ∂
∂z (x2+y)=0,
f4(x, y, z) = ∂
∂x (y2−z)=0,
f5(x, y, z) = ∂
∂y (y2−z)=2y,
f6(x, y, z) = ∂
∂z (y2−z) = −1.
17
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
f1(1,2,−1) = 2(1) = 2,
f2(1,2,−1) = 1,
f3(1,2,−1) = 0,
f4(1,2,−1) = 0,
f5(1,2,−1) = 2(2) = 4,
f6(1,2,−1) = −1.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in a 2 ×3 matrix.
Jf(1,2,−1) = 2 1 0
0 4 −1.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, z3). Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f1
∂x = 2xand ∂f2
∂x = 0
Step 2: Compute the partial derivatives of fwith respect to y:
∂f1
∂y =zand ∂f2
∂y = 0
Step 3: Compute the partial derivatives of fwith respect to z:
∂f1
∂z =yand ∂f2
∂z = 3z2
Step 4: Assemble the partial derivatives into the Jacobian matrix J(f):
J(f) = 2x z
y3z2
Therefore, the Jacobian matrix of the function f(x, y, z) = (x2+yz, z3) is
given by
J(f) = 2x z
y3z2
.
18
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Given f(x, y, z) = x2y+y2z+z2x, the Jacobian matrix of fis given by:
Jf=h∂f
∂x
∂f
∂y
∂f
∂z i
Step 1: Find ∂f
∂x .
∂f
∂x = 2xy +z2
Step 2: Find ∂f
∂y .
∂f
∂y =x2+ 2yz
Step 3: Find ∂f
∂z .
∂f
∂z =y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3). Substitute
x= 1, y= 2, z= 3 into the partial derivatives calculated in steps 1-3:
∂f
∂x = 2(1)(2) + (3)2= 4 + 9 = 13
∂f
∂y = (1)2+ 2(2)(3) = 1 + 12 = 13
∂f
∂z = (2)2+ 2(3)(1) = 4 + 6 = 10
Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 13 13 10
Question 24
Question
Let f(x, y, z) =
x2+yz
2xz
3xy
. Find the Jacobian matrix of fat the point (1,2,3).
19
Solution
Step 1: Find the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
0
3y
,∂f
∂y =
z
0
3x
,∂f
∂z =
y
2x
0
.
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x =
2(1)
0
3(2)
=
2
0
6
,
∂f
∂y =
3
0
3(1)
=
3
0
3
,
∂f
∂z =
2
2(1)
0
=
2
2
0
.
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
206
303
220
.
Question 25
Question
Let f(x, y, z)=(x2+y2+z2, xyz, x +y+z). Calculate the Jacobian matrix of
f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f
∂x
For the first component: x2+y2+z2, we have:
∂(x2+y2+z2)
∂x = 2x
For the second component: xyz, we have:
∂(xyz)
∂x =yz
20
For the third component: x+y+z, we have:
∂(x+y+z)
∂x = 1
Therefore, the first row of the Jacobian matrix is 2x yz 1.
Step 2: Compute ∂f
∂y
Following the same process, we find:
∂f
∂y =2y xz 1
Step 3: Compute ∂f
∂z
Similarly, we find:
∂f
∂z =2z xy 1
Step 4: Assemble the Jacobian matrix
Thus, the Jacobian matrix of fis:
Jf=
2x yz 1
2y xz 1
2z xy 1
Question 26
Question
Let f(x, y)=(xy2, x2y, x +y) be a vector-valued function. Find the Jacobian
matrix of fat the point (2,1).
Solution
To find the Jacobian matrix of a vector-valued function, we find the partial
derivatives of each component function with respect to each variable. The Ja-
cobian matrix of fat the point (2,1) will be a 3 ×2 matrix.
Step 1: Find the partial derivatives of each component function.
Let’s denote u=xand v=y. Then,
∂f1
∂u =y2,∂f1
∂v = 2xy
∂f2
∂u = 2xy, ∂f2
∂v =x2
∂f3
∂u = 1,∂f3
∂v = 1
Step 2: Evaluate the partial derivatives at the point (2,1).At the
point (2,1), we have:
∂f1
∂u = 1,∂f1
∂v = 4
21
∂f2
∂u = 4,∂f2
∂v = 4
∂f3
∂u = 1,∂f3
∂v = 1
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat
(2,1) is:
Jf(2,1) =
1 4
4 4
1 1
.
Question 27
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+y2, yz).
Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component of fwith respect to each variable x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z
Let f1(x, y, z) = x2+y2. Then,
∂f1
∂x = 2x, ∂f1
∂y = 2y, and ∂f1
∂z = 0.
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z
Let f2(x, y, z) = yz. Then,
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Put it all together to form the Jacobian matrix of f
The Jacobian matrix of fis given by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y.
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+y, xyz). Find the
Jacobian matrix of f.
22
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Assemble the Jacobian matrix:
Jf(x, y, z) = 2x1 0
yz xz xy
Question 29
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+yz, z2−x). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the component functions of f.
Step 2: To find the Jacobian matrix of f, we first compute the partial
derivatives of fwith respect to x,y, and z:
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2z
Step 3: Now evaluate these partial derivatives at the point (1,2,3) to find
the Jacobian matrix at that point:
At (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 3,∂f1
∂z = 2
∂f2
∂x =−1,∂f2
∂y = 0,∂f2
∂z = 2(3) = 6
Step 4: Therefore, the Jacobian matrix of fat the point (1,2,3) is:
Jf(1,2,3) = 2 3 2
−106
23
Question 30
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Compute the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y, yz), we have:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with f(x, y, z)=(x2+y, yz), we have:
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
The Jacobian matrix Jfis formed by arranging the partial derivatives in a
matrix. Therefore,
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0z y
So, the Jacobian matrix of fis 2x1 0
0z y.
24
