MATH 117 - ELEMENTS OF
MATHEMATICS - Jacobian matrix
Question Bank - Set 2
Liberty University
Question 1
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+y, xyz).
Compute the Jacobian matrix of f.
Solution
The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix whose
entries are the partial derivatives of the components of f. For the function
f(x, y, z)=(x2+y, xyz), the Jacobian matrix is given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 1: Calculate partial derivatives
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
∂f1
∂z = 0
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 2: Assemble the Jacobian matrix Substitute the partial deriva-
tives into the Jacobian matrix:
Jf=2x1 0
yz xz xy
Question 2
Question
Let f(x) = 3x2
1+ 2x2
x1+ 4x3
2be a vector-valued function. Find the Jacobian matrix
of f.
Solution
Step 1: The Jacobian matrix of fis defined as the matrix of all first-order partial
derivatives of the components of f:
J(f) = "∂f1
∂x1
∂f1
∂x2
∂f2
∂x1
∂f2
∂x2#
Step 2: To find the partial derivatives, we differentiate each component of f
with respect to each variable:
∂f1
∂x1
= 6x1,∂f1
∂x2
= 2
∂f2
∂x1
= 1,∂f2
∂x2
= 12x2
2
Step 3: Now, we can construct the Jacobian matrix using the computed
partial derivatives:
J(f) = 6x12
1 12x2
2
Therefore, the Jacobian matrix of fis 6x12
1 12x2
2.
Question 3
Question
Let f:R3→R2be defined by f(x, y, z) = (2x2+yz, xz). Find the Jacobian
matrix Jfof f.
2
Solution
Step 1: Write out the components of the function f:
f(x, y, z) = 2x2+yz
xz
Step 2: Determine the partial derivatives of each component of fwith respect
to x, y, z:
∂f1
∂x = 4x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 3: Assemble the partial derivatives into the Jacobian matrix Jf:
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=4x z y
z0x
Therefore, the Jacobian matrix of fis:
Jf=4x z y
z0x
Question 4
Question
Let f(x, y, z)=(xy2z, x2yz, xyz2) be a vector-valued function. Find the Jaco-
bian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we need to compute the
partial derivatives of each component function at that point and organize them
into a matrix.
Step 1: Calculate the partial derivatives
Let’s find the partial derivative of f1(x, y, z) = xy2zwith respect to x,y,
and zat the point (1,2,3).
∂f1
∂x =y2z= 22·3 = 12
∂f1
∂y = 2xyz = 2 ·1·3=6
∂f1
∂z =xy2= 1 ·22= 4
3
Now, find the partial derivative of f2(x, y, z) = x2yz with respect to x,y,
and zat the point (1,2,3).
∂f2
∂x = 2xyz = 2 ·1·3=6
∂f2
∂y =x2z= 12·3=3
∂f2
∂z =x2y= 12·2=2
Lastly, find the partial derivative of f3(x, y, z) = xyz2with respect to x,y,
and zat the point (1,2,3).
∂f3
∂x =yz2= 2 ·32= 18
∂f3
∂y =xz2= 1 ·32= 9
∂f3
∂z = 2xyz = 2 ·1·3=6
Step 2: Organize the partial derivatives into the Jacobian matrix
The Jacobian matrix of fat the point (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
=
12 6 4
6 3 2
18 9 6
Question 5
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y2−
xz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
fwith respect to each variable and arrange them into a matrix.
4
Step 1: Compute the partial derivatives of f(x, y, z).
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
∂f2
∂x =−z
∂f2
∂y = 2y
∂f2
∂z =−x
Step 2: Arrange the partial derivatives into a matrix. The Jacobian matrix
of fis given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
−z2y−x
Therefore, the Jacobian matrix of fis 2x z y
−z2y−x.
Question 6
Question
Let f(x, y, z) =
x2+y2
yz
z2
. Compute the Jacobian matrix of fat the point
(1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we need to compute the
partial derivatives of each component function with respect to each variable,
then evaluate them at the given point.
Step 1: Compute the partial derivatives of f.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 2z
5
Step 2: Evaluate the partial derivatives at (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) = 0,∂f2
∂y (1,2,3) = 3,∂f2
∂z (1,2,3) = 2
∂f3
∂x (1,2,3) = 0,∂f3
∂y (1,2,3) = 0,∂f3
∂z (1,2,3) = 2(3) = 6
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fat (1,2,3)
is given by
Jf(1,2,3) =
240
032
006
.
Question 7
Question
Let f(x, y, z) = x3yz +exy sin(z) where x= 2t,y=t2, and z=t. Compute the
Jacobian matrix of fwith respect to tat t= 1.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x3yz +exy sin(z))
= 3x2yz +yexy sin(z)
= 12t2t2t+t2e2tsin(t)
= 12t5+t2e2tsin(t)
∂f
∂y =∂
∂y (x3yz +exy sin(z))
=x3z+xexy sin(z)
= 8t3t+ 2te2tsin(t)
= 8t4+ 2te2tsin(t)
∂f
∂z =∂
∂z (x3yz +exy sin(z))
=x3y+exy cos(z)
= 8t4+e2tcos(t)
6
Step 2: Compute the Jacobian matrix Jf(t) by arranging the partial deriva-
tives in a matrix form.
Jf(t) = 12t5+t2e2tsin(t) 8t4+ 2te2tsin(t) 8t4+e2tcos(t)
Step 3: Evaluate the Jacobian matrix at t= 1.
Jf(1) = 12(1)5+ (1)2e2(1) sin(1) 8(1)4+ 2(1)e2(1) sin(1) 8(1)4+e2(1) cos(1)
Jf(1) = 12 + e2sin(1) 8 + 2e2sin(1) 8 + e2cos(1)
Question 8
Question
Find the Jacobian matrix of the transformation T:R3→R3given by:
T(x, y, z) = (x2+y, 2y−z, 3z)
Solution
To find the Jacobian matrix of the transformation T, we need to compute the
partial derivatives of each component function.
Step 1: Compute ∂T
∂x . For the first component function, T1(x, y, z) = x2+y,
we have: ∂T1
∂x = 2x
Step 2: Compute ∂T
∂y . For the second component function, T2(x, y, z) =
2y−z, we have:
∂T2
∂y = 2
Step 3: Compute ∂T
∂z . For the third component function, T3(x, y, z)=3z,
we have: ∂T3
∂z = 3
Therefore, the Jacobian matrix of the transformation T:R3→R3is:
2 0 0
0 2 −1
0 0 3
Question 9
Question
Let f:R3→R3be a mapping defined by f(x, y, z)=(x2+y2, x +z, xyz). Find
the Jacobian matrix of f.
7
Solution
To find the Jacobian matrix of f, we need to calculate the partial derivatives of
each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2+y2.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function f2(x, y, z) =
x+z.∂f2
∂x = 1,∂f2
∂y = 0,∂f2
∂z = 1
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for the third component function f3(x, y, z) =
xyz.
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Assemble the partial derivatives into a 3 ×3 matrix to obtain the
Jacobian matrix of f:
Jf =
2x2y0
1 0 1
yz xz xy
Question 10
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2y, yz, z3). Find the
Jacobian matrix of fat the point (1,2,−1).
Solution
Given the function f:R3→R3defined by f(x, y, z) = (x2y, yz, z3), let’s find
the Jacobian matrix of fat the point (1,2,−1).
Step 1: Calculate the partial derivatives The Jacobian matrix of fis
given by
J=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
where f1=x2y,f2=yz, and f3=z3.
Calculating the partial derivatives:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
8
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 3z2
Step 2: Evaluate the Jacobian matrix at the point (1,2,−1) Substi-
tute x= 1, y = 2, z =−1 into the partial derivatives calculated above to find
the values of the Jacobian matrix at the point (1,2,−1).
J(1,2,−1) =
2(1)(2) 1 0
0−1 2
0 0 −3
=
4 1 0
0−1 2
0 0 −3
Therefore, the Jacobian matrix of fat the point (1,2,−1) is
J(1,2,−1) =
4 1 0
0−1 2
0 0 −3
Question 11
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(exsin(yz), x2+
y2). Find the Jacobian matrix of fat the point (0,0,0).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x =∂
∂x (exsin(yz)) = exsin(yz)
∂f1
∂y =∂
∂y (exsin(yz)) = zexcos(yz)
∂f1
∂z =∂
∂z (exsin(yz)) = yexcos(yz)
Step 2: Calculate the partial derivatives of fwith respect to xand y.
∂f2
∂x =∂
∂x (x2+y2)=2x
∂f2
∂y =∂
∂y (x2+y2)=2y
∂f2
∂z = 0
9
Step 3: Evaluate the Jacobian matrix of fat (0,0,0). The Jacobian matrix
is given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Substitute the partial derivatives we calculated earlier:
Jf=100
000
Therefore, the Jacobian matrix of fat (0,0,0) is:
Jf=100
000
Question 12
Question
Let f(x, y, z) =
x2+y2
y2+z2
z2+x2
. Find the Jacobian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we first need to compute
the partial derivatives of each component of fwith respect to x,y, and zat the
given point.
Step 1: Compute the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Evaluate the partial derivatives at (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 2(2) = 4,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2(2) = 4,∂f2
∂z = 2(3) = 6
∂f3
∂x = 2(1) = 2,∂f3
∂y = 0,∂f3
∂z = 2(3) = 6
10
Step 3: Construct the Jacobian matrix:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
240
046
206
Question 13
Question
Let f(x, y) = (2x2+y, 3xy) and g(u, v)=(u+v, 2u−v). Determine the Jacobian
matrix of the composition g◦f.
Solution
Step 1: Find the composition g◦f.
(g◦f)(x, y) = g(f(x, y))
=g(2x2+y, 3xy)
= (2x2+y+ 3xy, 2(2x2+y)−3xy)
= (2x2+ 3xy +y, 4x2+ 2y−3xy)
= (2x2+ 3xy +y, 4x2+y−xy)
Step 2: Find the Jacobian matrix of g◦f. The Jacobian matrix of a function
h:Rn→Rmis an m×nmatrix where the i-th row and j-th column entry is the
partial derivative of the i-th component of hwith respect to the j-th variable.
For the function h(x, y) = (2x2+ 3xy +y, 4x2+y−xy), the Jacobian matrix
is given by
Jg◦f="∂h1
∂x
∂h1
∂y
∂h2
∂x
∂h2
∂y #=4x+ 3y2x+ 3
8x−y1
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Determine the
Jacobian matrix of fat the point (2,1,−3).
11
Solution
Step 1: Compute the partial derivatives of f:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0,
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y.
Step 2: Evaluate the partial derivatives at the point (2,1,−3):
∂f1
∂x (2,1,−3) = 2(2)(1) = 4,∂f1
∂y (2,1,−3) = 22= 4,∂f1
∂z (2,1,−3) = 0,
∂f2
∂x (2,1,−3) = 0,∂f2
∂y (2,1,−3) = −3,∂f2
∂z (2,1,−3) = 1.
Step 3: Assemble the Jacobian matrix of fat (2,1,−3):
Jf(2,1,−3) = 440
0−3 1.
Question 15
Question
Find the Jacobian matrix of the following system of equations:
x= 2u2+v2
y=uv + 3u
z=u−v
Solution
Step 1: To find the Jacobian matrix, we need to compute the partial derivatives
of each equation with respect to each variable.
Step 2: Let’s begin by finding ∂(x,y,z)
∂(u,v).
Step 3: Compute the partial derivatives of xwith respect to uand v:
∂x
∂u = 4u, ∂x
∂v = 2v
Step 4: Compute the partial derivatives of ywith respect to uand v:
∂y
∂u =v+ 3,∂y
∂v =u
Step 5: Compute the partial derivatives of zwith respect to uand v:
∂z
∂u = 1,∂z
∂v =−1
12
Step 6: Assemble the partial derivatives into the Jacobian matrix:
J=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
=
4u2v
v+ 3 u
1−1
Question 16
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, y2+ 2z). Find the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we compute the partial derivatives of each
component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2.
Step 3: Assemble the Jacobian matrix of f. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0 2y2.
Therefore, the Jacobian matrix of fis
Jf=2x1 0
0 2y2.
Question 17
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Find the Jacobian
matrix of f.
13
Solution
Step 1: We first need to determine the partial derivatives of f.
∂f
∂x =∂f1
∂x
∂f2
∂x
∂f
∂y ="∂f1
∂y
∂f2
∂y #
∂f
∂z =∂f1
∂z
∂f2
∂z
Step 2: Calculate the partial derivatives of f.
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2x1 0
0z y
Question 18
Question
Find the Jacobian matrix of the following transformation:
T(x, y)=(x2+y2, xy)
14
Solution
To find the Jacobian matrix of the transformation T(x, y) = (x2+y2, xy), we
need to calculate the partial derivatives of the components of the transformation.
Step 1: Find ∂f
∂x and ∂f
∂y for the first component function f(x, y) = x2+y2.
∂f
∂x = 2xand ∂f
∂y = 2y
Step 2: Find ∂g
∂x and ∂g
∂y for the second component function g(x, y) = xy.
∂g
∂x =yand ∂g
∂y =x
Step 3: Assemble the Jacobian matrix using the partial derivatives calcu-
lated in Step 1 and Step 2. The Jacobian matrix is given by:
J="∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y #=2x2y
y x
Therefore, the Jacobian matrix of the transformation T(x, y)=(x2+y2, xy)
is J=2x2y
y x .
Question 19
Question
Let f(x, y, z) =
x2y
yz
x+z
. Find the Jacobian matrix Jf(x, y, z) of f.
Solution
To find the Jacobian matrix Jf(x, y, z) of f, we must compute the partial deriva-
tives of each component of fwith respect to x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y ,∂f1
∂z :
f1(x, y, z) = x2y
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
15
Step 2: Compute ∂f2
∂x ,∂f2
∂y ,∂f2
∂z :
f2(x, y, z) = yz
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Compute ∂f3
∂x ,∂f3
∂y ,∂f3
∂z :
f3(x, y, z) = x+z
∂f3
∂x = 1
∂f3
∂y = 0
∂f3
∂z = 1
Step 4: Assemble the partial derivatives into the Jacobian matrix:
Jf(x, y, z) =
2xy x20
0z y
1 0 1
Question 20
Question
Let f:R3→R3be defined by f(x, y, z)=(x2+y2, y2+z2, z2+x2). 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 of fwith respect to x,y, and z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
16
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 4: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by:
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Substitute the partial derivatives into the matrix to get:
Jf=
2x2y0
0 2y2z
2x0 2z
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x3, y2z). Calculate the
Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis a matrix of partial
derivatives. Given f(x, y, z)=(x3, y2z), we need to find the partial derivatives
of each component function:
For the first component function, f1(x, y, z) = x3, the partial derivatives are:
∂f1
∂x = 3x2,∂f1
∂y = 0,and ∂f1
∂z = 0.
For the second component function, f2(x, y, z) = y2z, the partial derivatives
are: ∂f2
∂x = 0,∂f2
∂y = 2yz, and ∂f2
∂z =y2.
Step 2: Evaluate the partial derivatives at the point (1,2,−1):
∂f1
∂x (1,2,−1) = 3(1)2= 3,∂f1
∂y (1,2,−1) = 0,∂f1
∂z (1,2,−1) = 0,
∂f2
∂x (1,2,−1) = 0,∂f2
∂y (1,2,−1) = 2(2)(−1) = −4,∂f2
∂z (1,2,−1) = (2)2= 4.
Step 3: Assemble the Jacobian matrix Jfusing the partial derivatives:
Jf(1,2,−1) = 300
0−4 4.
17
Therefore, the Jacobian matrix of fat the point (1,2,−1) is 300
0−4 4.
Question 22
Question
Find the Jacobian matrix of the transformation given by x=u+v, y =u−v, z =
uv.
Solution
To find the Jacobian matrix of a transformation, we need to compute the partial
derivatives of each resulting variable with respect to each input variable.
Step 1: Find ∂x
∂u ,∂x
∂v ,∂y
∂u ,∂y
∂v ,∂z
∂u , and ∂z
∂v .
Step 2: Compute the partial derivatives:
∂x
∂u =∂
∂u (u+v)=1,∂x
∂v =∂
∂v (u+v)=1,
∂y
∂u =∂
∂u (u−v)=1,∂y
∂v =∂
∂v (u−v) = −1,
∂z
∂u =∂
∂u (uv) = v, ∂z
∂v =∂
∂v (uv) = u.
Step 3: Assemble the Jacobian matrix:
J=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
=
1 1
1−1
v u
.
Question 23
Question
Let f(x, y, z) =
ex2+y2
sin(xy)
xyz
be a vector-valued function. Determine the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f(x, y, z), we need to compute the partial deriva-
tives of each component function with respect to x,y, and z.
Step 1: Compute the partial derivatives of fwith respect to x.
∂f
∂x =
∂
∂x (ex2+y2)
∂
∂x (sin(xy))
∂
∂x (xyz)
=
2xex2+y2
ycos(xy)
yz
18
Step 2: Compute the partial derivatives of fwith respect to y.
∂f
∂y =
∂
∂y (ex2+y2)
∂
∂y (sin(xy))
∂
∂y (xyz)
=
2yex2+y2
xcos(xy)
xz
Step 3: Compute the partial derivatives of fwith respect to z.
∂f
∂z =
∂
∂z (ex2+y2)
∂
∂z (sin(xy))
∂
∂z (xyz)
=
0
0
xy
Therefore, the Jacobian matrix of fis given by
Jf(x, y, z) =
2xex2+y22yex2+y20
ycos(xy)xcos(xy) 0
yz xz xy
Question 24
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+yz, exy ).
Find the Jacobian matrix of fat the point (1,2,0).
Solution
To find the Jacobian matrix of fat (1,2,0), we compute the partial derivatives
of each component function with respect to each variable and then arrange them
in a matrix.
Let f(x, y, z)=(u(x, y, z), v(x, y, z)), where u(x, y, z) = x2+yz and v(x, y, z) =
exy.
Step 1: Calculate the partial derivatives of u:
∂u
∂x = 2x,
∂u
∂y =z,
∂u
∂z =y.
Step 2: Calculate the partial derivatives of v:
∂v
∂x =yexy ,
∂v
∂y =xexy,
∂v
∂z = 0.
19
Step 3: Evaluate the partial derivatives at the point (1,2,0):
∂u
∂x (1,2,0) = 2(1) = 2,
∂u
∂y (1,2,0) = 0,
∂u
∂z (1,2,0) = 2,
∂v
∂x (1,2,0) = 2e2,
∂v
∂y (1,2,0) = e,
∂v
∂z (1,2,0) = 0.
Step 4: Assemble the Jacobian matrix:
Jf(1,2,0) = "∂u
∂x (1,2,0) ∂u
∂y (1,2,0) ∂u
∂z (1,2,0)
∂v
∂x (1,2,0) ∂v
∂y (1,2,0) ∂v
∂z (1,2,0)#=2 0 2
2e2e0.
So, the Jacobian matrix of fat (1,2,0) is 2 0 2
2e2e0.
Question 25
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). 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.
20
Step 1: Find the partial derivatives of f:
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 2: Assemble the partial derivatives into the Jacobian matrix of f:
Jac(f) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Therefore, the Jacobian matrix of fis 2xy x20
0z y.
Question 26
Question
Let f:R3→R3be a function defined by f(x, y, z)=(xyz, x2y, z3). Compute
the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =yz, ∂f1
∂y =xz, ∂f1
∂z =xy
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 2xy, ∂f2
∂y =x2,∂f2
∂z = 0
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 3z2
21
Step 4: Assemble the Jacobian matrix of f. The Jacobian matrix of fis:
Jf=
yz xz xy
2xy x20
0 0 3z2
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+yz, y2+xz). 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 of
partial derivatives. In this case, since f:R3→R2, the Jacobian matrix will be
a 2 ×3 matrix.
Step 2: First, we find the partial derivatives of fwith respect to x,y, and
z.∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =z, ∂f2
∂y = 2y, ∂f2
∂z =x
Step 3: Evaluate these partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 3,∂f1
∂z (1,2,3) = 2
∂f2
∂x (1,2,3) = 3,∂f2
∂y (1,2,3) = 4,∂f2
∂z (1,2,3) = 1
Step 4: Assemble these partial derivatives into the Jacobian matrix.
Jacobian matrix of fat (1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=232
341
Question 28
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+y, y2−
z). Find the Jacobian matrix of fat the point (1,2,3).
22
Solution
To find the Jacobian matrix of fat (1,2,3), we first need to determine the
partial derivatives of each component function.
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0,
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z =−1.
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) = 1,∂f1
∂z (1,2,3) = 0,
∂f2
∂x (1,2,3) = 0,∂f2
∂y (1,2,3) = 2 ·2=4,∂f2
∂z (1,2,3) = −1.
Step 3: Construct the Jacobian matrix of fat (1,2,3). The Jacobian
matrix is given by
Jf(1,2,3) = 2 1 0
0 4 −1.
Question 29
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y, 1
2yz). 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 with respect to each variable. The Jacobian matrix of
fwill be a 2 ×3 matrix.
Step 1: Compute the partial derivatives of f
Let’s first write fin component form:
f(x, y, z) = x2+y
1
2yz
Now, we will find the partial derivatives of each component function with
respect to each variable.
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
23
∂f1
∂z = 0
(since f1does not depend on z)
∂f2
∂x = 0
(since f2does not depend on x)
∂f2
∂y = 0
(since f2does not depend on y)
∂f2
∂z =∂
∂z (1
2yz) = 1
2y
Step 2: Assemble the Jacobian matrix
The Jacobian matrix of fis formed by arranging these partial derivatives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0 0 1
2y
Therefore, the Jacobian matrix of fis:
Jf=2x1 0
0 0 1
2y
Question 30
Question
Consider the following system of equations:
x2+y2= 25
xy + 2z= 10
z3+xyz = 40
Compute the Jacobian matrix at the point (3,4,2).
Solution
Step 1: Define the system of equations as a function: Let’s define the system of
equations as a vector-valued function F(x):
F(x) =
f1(x, y, z)
f2(x, y, z)
f3(x, y, z)
=
x2+y2−25
xy + 2z−10
z3+xyz −40
24
Step 2: Assemble the Jacobian matrix Substitute the partial deriva-
tives into the Jacobian matrix:
Jf=2x1 0
yz xz xy
Question 2
Question
Let f(x) = 3x2
1+ 2x2
x1+ 4x3
2be a vector-valued function. Find the Jacobian matrix
of f.
Solution
Step 1: The Jacobian matrix of fis defined as the matrix of all first-order partial
derivatives of the components of f:
J(f) = "∂f1
∂x1
∂f1
∂x2
∂f2
∂x1
∂f2
∂x2#
Step 2: To find the partial derivatives, we differentiate each component of f
with respect to each variable:
∂f1
∂x1
= 6x1,∂f1
∂x2
= 2
∂f2
∂x1
= 1,∂f2
∂x2
= 12x2
2
Step 3: Now, we can construct the Jacobian matrix using the computed
partial derivatives:
J(f) = 6x12
1 12x2
2
Therefore, the Jacobian matrix of fis 6x12
1 12x2
2.
Question 3
Question
Let f:R3→R2be defined by f(x, y, z) = (2x2+yz, xz). Find the Jacobian
matrix Jfof f.
2
Solution
Step 1: Write out the components of the function f:
f(x, y, z) = 2x2+yz
xz
Step 2: Determine the partial derivatives of each component of fwith respect
to x, y, z:
∂f1
∂x = 4x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 3: Assemble the partial derivatives into the Jacobian matrix Jf:
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=4x z y
z0x
Therefore, the Jacobian matrix of fis:
Jf=4x z y
z0x
Question 4
Question
Let f(x, y, z)=(xy2z, x2yz, xyz2) be a vector-valued function. Find the Jaco-
bian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we need to compute the
partial derivatives of each component function at that point and organize them
into a matrix.
Step 1: Calculate the partial derivatives
Let’s find the partial derivative of f1(x, y, z) = xy2zwith respect to x,y,
and zat the point (1,2,3).
∂f1
∂x =y2z= 22·3 = 12
∂f1
∂y = 2xyz = 2 ·1·3=6
∂f1
∂z =xy2= 1 ·22= 4
3
Now, find the partial derivative of f2(x, y, z) = x2yz with respect to x,y,
and zat the point (1,2,3).
∂f2
∂x = 2xyz = 2 ·1·3=6
∂f2
∂y =x2z= 12·3=3
∂f2
∂z =x2y= 12·2=2
Lastly, find the partial derivative of f3(x, y, z) = xyz2with respect to x,y,
and zat the point (1,2,3).
∂f3
∂x =yz2= 2 ·32= 18
∂f3
∂y =xz2= 1 ·32= 9
∂f3
∂z = 2xyz = 2 ·1·3=6
Step 2: Organize the partial derivatives into the Jacobian matrix
The Jacobian matrix of fat the point (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
=
12 6 4
6 3 2
18 9 6
Question 5
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y2−
xz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
fwith respect to each variable and arrange them into a matrix.
4
Step 1: Compute the partial derivatives of f(x, y, z).
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
∂f2
∂x =−z
∂f2
∂y = 2y
∂f2
∂z =−x
Step 2: Arrange the partial derivatives into a matrix. The Jacobian matrix
of fis given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
−z2y−x
Therefore, the Jacobian matrix of fis 2x z y
−z2y−x.
Question 6
Question
Let f(x, y, z) =
x2+y2
yz
z2
. Compute the Jacobian matrix of fat the point
(1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we need to compute the
partial derivatives of each component function with respect to each variable,
then evaluate them at the given point.
Step 1: Compute the partial derivatives of f.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 2z
5
Step 2: Evaluate the partial derivatives at (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) = 0,∂f2
∂y (1,2,3) = 3,∂f2
∂z (1,2,3) = 2
∂f3
∂x (1,2,3) = 0,∂f3
∂y (1,2,3) = 0,∂f3
∂z (1,2,3) = 2(3) = 6
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fat (1,2,3)
is given by
Jf(1,2,3) =
240
032
006
.
Question 7
Question
Let f(x, y, z) = x3yz +exy sin(z) where x= 2t,y=t2, and z=t. Compute the
Jacobian matrix of fwith respect to tat t= 1.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x3yz +exy sin(z))
= 3x2yz +yexy sin(z)
= 12t2t2t+t2e2tsin(t)
= 12t5+t2e2tsin(t)
∂f
∂y =∂
∂y (x3yz +exy sin(z))
=x3z+xexy sin(z)
= 8t3t+ 2te2tsin(t)
= 8t4+ 2te2tsin(t)
∂f
∂z =∂
∂z (x3yz +exy sin(z))
=x3y+exy cos(z)
= 8t4+e2tcos(t)
6
Step 2: Compute the Jacobian matrix Jf(t) by arranging the partial deriva-
tives in a matrix form.
Jf(t) = 12t5+t2e2tsin(t) 8t4+ 2te2tsin(t) 8t4+e2tcos(t)
Step 3: Evaluate the Jacobian matrix at t= 1.
Jf(1) = 12(1)5+ (1)2e2(1) sin(1) 8(1)4+ 2(1)e2(1) sin(1) 8(1)4+e2(1) cos(1)
Jf(1) = 12 + e2sin(1) 8 + 2e2sin(1) 8 + e2cos(1)
Question 8
Question
Find the Jacobian matrix of the transformation T:R3→R3given by:
T(x, y, z) = (x2+y, 2y−z, 3z)
Solution
To find the Jacobian matrix of the transformation T, we need to compute the
partial derivatives of each component function.
Step 1: Compute ∂T
∂x . For the first component function, T1(x, y, z) = x2+y,
we have: ∂T1
∂x = 2x
Step 2: Compute ∂T
∂y . For the second component function, T2(x, y, z) =
2y−z, we have:
∂T2
∂y = 2
Step 3: Compute ∂T
∂z . For the third component function, T3(x, y, z)=3z,
we have: ∂T3
∂z = 3
Therefore, the Jacobian matrix of the transformation T:R3→R3is:
2 0 0
0 2 −1
0 0 3
Question 9
Question
Let f:R3→R3be a mapping defined by f(x, y, z)=(x2+y2, x +z, xyz). Find
the Jacobian matrix of f.
7
Solution
To find the Jacobian matrix of f, we need to calculate the partial derivatives of
each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2+y2.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function f2(x, y, z) =
x+z.∂f2
∂x = 1,∂f2
∂y = 0,∂f2
∂z = 1
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for the third component function f3(x, y, z) =
xyz.
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Assemble the partial derivatives into a 3 ×3 matrix to obtain the
Jacobian matrix of f:
Jf =
2x2y0
1 0 1
yz xz xy
Question 10
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2y, yz, z3). Find the
Jacobian matrix of fat the point (1,2,−1).
Solution
Given the function f:R3→R3defined by f(x, y, z) = (x2y, yz, z3), let’s find
the Jacobian matrix of fat the point (1,2,−1).
Step 1: Calculate the partial derivatives The Jacobian matrix of fis
given by
J=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
where f1=x2y,f2=yz, and f3=z3.
Calculating the partial derivatives:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
8
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 3z2
Step 2: Evaluate the Jacobian matrix at the point (1,2,−1) Substi-
tute x= 1, y = 2, z =−1 into the partial derivatives calculated above to find
the values of the Jacobian matrix at the point (1,2,−1).
J(1,2,−1) =
2(1)(2) 1 0
0−1 2
0 0 −3
=
4 1 0
0−1 2
0 0 −3
Therefore, the Jacobian matrix of fat the point (1,2,−1) is
J(1,2,−1) =
4 1 0
0−1 2
0 0 −3
Question 11
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(exsin(yz), x2+
y2). Find the Jacobian matrix of fat the point (0,0,0).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x =∂
∂x (exsin(yz)) = exsin(yz)
∂f1
∂y =∂
∂y (exsin(yz)) = zexcos(yz)
∂f1
∂z =∂
∂z (exsin(yz)) = yexcos(yz)
Step 2: Calculate the partial derivatives of fwith respect to xand y.
∂f2
∂x =∂
∂x (x2+y2)=2x
∂f2
∂y =∂
∂y (x2+y2)=2y
∂f2
∂z = 0
9
Step 3: Evaluate the Jacobian matrix of fat (0,0,0). The Jacobian matrix
is given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Substitute the partial derivatives we calculated earlier:
Jf=100
000
Therefore, the Jacobian matrix of fat (0,0,0) is:
Jf=100
000
Question 12
Question
Let f(x, y, z) =
x2+y2
y2+z2
z2+x2
. Find the Jacobian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we first need to compute
the partial derivatives of each component of fwith respect to x,y, and zat the
given point.
Step 1: Compute the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Evaluate the partial derivatives at (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 2(2) = 4,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2(2) = 4,∂f2
∂z = 2(3) = 6
∂f3
∂x = 2(1) = 2,∂f3
∂y = 0,∂f3
∂z = 2(3) = 6
10
Step 3: Construct the Jacobian matrix:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
240
046
206
Question 13
Question
Let f(x, y) = (2x2+y, 3xy) and g(u, v)=(u+v, 2u−v). Determine the Jacobian
matrix of the composition g◦f.
Solution
Step 1: Find the composition g◦f.
(g◦f)(x, y) = g(f(x, y))
=g(2x2+y, 3xy)
= (2x2+y+ 3xy, 2(2x2+y)−3xy)
= (2x2+ 3xy +y, 4x2+ 2y−3xy)
= (2x2+ 3xy +y, 4x2+y−xy)
Step 2: Find the Jacobian matrix of g◦f. The Jacobian matrix of a function
h:Rn→Rmis an m×nmatrix where the i-th row and j-th column entry is the
partial derivative of the i-th component of hwith respect to the j-th variable.
For the function h(x, y) = (2x2+ 3xy +y, 4x2+y−xy), the Jacobian matrix
is given by
Jg◦f="∂h1
∂x
∂h1
∂y
∂h2
∂x
∂h2
∂y #=4x+ 3y2x+ 3
8x−y1
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Determine the
Jacobian matrix of fat the point (2,1,−3).
11
Solution
Step 1: Compute the partial derivatives of f:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0,
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y.
Step 2: Evaluate the partial derivatives at the point (2,1,−3):
∂f1
∂x (2,1,−3) = 2(2)(1) = 4,∂f1
∂y (2,1,−3) = 22= 4,∂f1
∂z (2,1,−3) = 0,
∂f2
∂x (2,1,−3) = 0,∂f2
∂y (2,1,−3) = −3,∂f2
∂z (2,1,−3) = 1.
Step 3: Assemble the Jacobian matrix of fat (2,1,−3):
Jf(2,1,−3) = 440
0−3 1.
Question 15
Question
Find the Jacobian matrix of the following system of equations:
x= 2u2+v2
y=uv + 3u
z=u−v
Solution
Step 1: To find the Jacobian matrix, we need to compute the partial derivatives
of each equation with respect to each variable.
Step 2: Let’s begin by finding ∂(x,y,z)
∂(u,v).
Step 3: Compute the partial derivatives of xwith respect to uand v:
∂x
∂u = 4u, ∂x
∂v = 2v
Step 4: Compute the partial derivatives of ywith respect to uand v:
∂y
∂u =v+ 3,∂y
∂v =u
Step 5: Compute the partial derivatives of zwith respect to uand v:
∂z
∂u = 1,∂z
∂v =−1
12
Step 6: Assemble the partial derivatives into the Jacobian matrix:
J=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
=
4u2v
v+ 3 u
1−1
Question 16
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, y2+ 2z). Find the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we compute the partial derivatives of each
component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2.
Step 3: Assemble the Jacobian matrix of f. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0 2y2.
Therefore, the Jacobian matrix of fis
Jf=2x1 0
0 2y2.
Question 17
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Find the Jacobian
matrix of f.
13
Solution
Step 1: We first need to determine the partial derivatives of f.
∂f
∂x =∂f1
∂x
∂f2
∂x
∂f
∂y ="∂f1
∂y
∂f2
∂y #
∂f
∂z =∂f1
∂z
∂f2
∂z
Step 2: Calculate the partial derivatives of f.
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2x1 0
0z y
Question 18
Question
Find the Jacobian matrix of the following transformation:
T(x, y)=(x2+y2, xy)
14
Solution
To find the Jacobian matrix of the transformation T(x, y) = (x2+y2, xy), we
need to calculate the partial derivatives of the components of the transformation.
Step 1: Find ∂f
∂x and ∂f
∂y for the first component function f(x, y) = x2+y2.
∂f
∂x = 2xand ∂f
∂y = 2y
Step 2: Find ∂g
∂x and ∂g
∂y for the second component function g(x, y) = xy.
∂g
∂x =yand ∂g
∂y =x
Step 3: Assemble the Jacobian matrix using the partial derivatives calcu-
lated in Step 1 and Step 2. The Jacobian matrix is given by:
J="∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y #=2x2y
y x
Therefore, the Jacobian matrix of the transformation T(x, y)=(x2+y2, xy)
is J=2x2y
y x .
Question 19
Question
Let f(x, y, z) =
x2y
yz
x+z
. Find the Jacobian matrix Jf(x, y, z) of f.
Solution
To find the Jacobian matrix Jf(x, y, z) of f, we must compute the partial deriva-
tives of each component of fwith respect to x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y ,∂f1
∂z :
f1(x, y, z) = x2y
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
15
Step 2: Compute ∂f2
∂x ,∂f2
∂y ,∂f2
∂z :
f2(x, y, z) = yz
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Compute ∂f3
∂x ,∂f3
∂y ,∂f3
∂z :
f3(x, y, z) = x+z
∂f3
∂x = 1
∂f3
∂y = 0
∂f3
∂z = 1
Step 4: Assemble the partial derivatives into the Jacobian matrix:
Jf(x, y, z) =
2xy x20
0z y
1 0 1
Question 20
Question
Let f:R3→R3be defined by f(x, y, z)=(x2+y2, y2+z2, z2+x2). 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 of fwith respect to x,y, and z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
16
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 4: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by:
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Substitute the partial derivatives into the matrix to get:
Jf=
2x2y0
0 2y2z
2x0 2z
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x3, y2z). Calculate the
Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis a matrix of partial
derivatives. Given f(x, y, z)=(x3, y2z), we need to find the partial derivatives
of each component function:
For the first component function, f1(x, y, z) = x3, the partial derivatives are:
∂f1
∂x = 3x2,∂f1
∂y = 0,and ∂f1
∂z = 0.
For the second component function, f2(x, y, z) = y2z, the partial derivatives
are: ∂f2
∂x = 0,∂f2
∂y = 2yz, and ∂f2
∂z =y2.
Step 2: Evaluate the partial derivatives at the point (1,2,−1):
∂f1
∂x (1,2,−1) = 3(1)2= 3,∂f1
∂y (1,2,−1) = 0,∂f1
∂z (1,2,−1) = 0,
∂f2
∂x (1,2,−1) = 0,∂f2
∂y (1,2,−1) = 2(2)(−1) = −4,∂f2
∂z (1,2,−1) = (2)2= 4.
Step 3: Assemble the Jacobian matrix Jfusing the partial derivatives:
Jf(1,2,−1) = 300
0−4 4.
17
Therefore, the Jacobian matrix of fat the point (1,2,−1) is 300
0−4 4.
Question 22
Question
Find the Jacobian matrix of the transformation given by x=u+v, y =u−v, z =
uv.
Solution
To find the Jacobian matrix of a transformation, we need to compute the partial
derivatives of each resulting variable with respect to each input variable.
Step 1: Find ∂x
∂u ,∂x
∂v ,∂y
∂u ,∂y
∂v ,∂z
∂u , and ∂z
∂v .
Step 2: Compute the partial derivatives:
∂x
∂u =∂
∂u (u+v)=1,∂x
∂v =∂
∂v (u+v)=1,
∂y
∂u =∂
∂u (u−v)=1,∂y
∂v =∂
∂v (u−v) = −1,
∂z
∂u =∂
∂u (uv) = v, ∂z
∂v =∂
∂v (uv) = u.
Step 3: Assemble the Jacobian matrix:
J=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
=
1 1
1−1
v u
.
Question 23
Question
Let f(x, y, z) =
ex2+y2
sin(xy)
xyz
be a vector-valued function. Determine the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f(x, y, z), we need to compute the partial deriva-
tives of each component function with respect to x,y, and z.
Step 1: Compute the partial derivatives of fwith respect to x.
∂f
∂x =
∂
∂x (ex2+y2)
∂
∂x (sin(xy))
∂
∂x (xyz)
=
2xex2+y2
ycos(xy)
yz
18
Step 2: Compute the partial derivatives of fwith respect to y.
∂f
∂y =
∂
∂y (ex2+y2)
∂
∂y (sin(xy))
∂
∂y (xyz)
=
2yex2+y2
xcos(xy)
xz
Step 3: Compute the partial derivatives of fwith respect to z.
∂f
∂z =
∂
∂z (ex2+y2)
∂
∂z (sin(xy))
∂
∂z (xyz)
=
0
0
xy
Therefore, the Jacobian matrix of fis given by
Jf(x, y, z) =
2xex2+y22yex2+y20
ycos(xy)xcos(xy) 0
yz xz xy
Question 24
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+yz, exy ).
Find the Jacobian matrix of fat the point (1,2,0).
Solution
To find the Jacobian matrix of fat (1,2,0), we compute the partial derivatives
of each component function with respect to each variable and then arrange them
in a matrix.
Let f(x, y, z)=(u(x, y, z), v(x, y, z)), where u(x, y, z) = x2+yz and v(x, y, z) =
exy.
Step 1: Calculate the partial derivatives of u:
∂u
∂x = 2x,
∂u
∂y =z,
∂u
∂z =y.
Step 2: Calculate the partial derivatives of v:
∂v
∂x =yexy ,
∂v
∂y =xexy,
∂v
∂z = 0.
19
Step 3: Evaluate the partial derivatives at the point (1,2,0):
∂u
∂x (1,2,0) = 2(1) = 2,
∂u
∂y (1,2,0) = 0,
∂u
∂z (1,2,0) = 2,
∂v
∂x (1,2,0) = 2e2,
∂v
∂y (1,2,0) = e,
∂v
∂z (1,2,0) = 0.
Step 4: Assemble the Jacobian matrix:
Jf(1,2,0) = "∂u
∂x (1,2,0) ∂u
∂y (1,2,0) ∂u
∂z (1,2,0)
∂v
∂x (1,2,0) ∂v
∂y (1,2,0) ∂v
∂z (1,2,0)#=2 0 2
2e2e0.
So, the Jacobian matrix of fat (1,2,0) is 2 0 2
2e2e0.
Question 25
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). 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.
20
Step 1: Find the partial derivatives of f:
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 2: Assemble the partial derivatives into the Jacobian matrix of f:
Jac(f) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Therefore, the Jacobian matrix of fis 2xy x20
0z y.
Question 26
Question
Let f:R3→R3be a function defined by f(x, y, z)=(xyz, x2y, z3). Compute
the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =yz, ∂f1
∂y =xz, ∂f1
∂z =xy
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 2xy, ∂f2
∂y =x2,∂f2
∂z = 0
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 3z2
21
Step 4: Assemble the Jacobian matrix of f. The Jacobian matrix of fis:
Jf=
yz xz xy
2xy x20
0 0 3z2
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+yz, y2+xz). 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 of
partial derivatives. In this case, since f:R3→R2, the Jacobian matrix will be
a 2 ×3 matrix.
Step 2: First, we find the partial derivatives of fwith respect to x,y, and
z.∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =z, ∂f2
∂y = 2y, ∂f2
∂z =x
Step 3: Evaluate these partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 3,∂f1
∂z (1,2,3) = 2
∂f2
∂x (1,2,3) = 3,∂f2
∂y (1,2,3) = 4,∂f2
∂z (1,2,3) = 1
Step 4: Assemble these partial derivatives into the Jacobian matrix.
Jacobian matrix of fat (1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=232
341
Question 28
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+y, y2−
z). Find the Jacobian matrix of fat the point (1,2,3).
22
Solution
To find the Jacobian matrix of fat (1,2,3), we first need to determine the
partial derivatives of each component function.
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0,
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z =−1.
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) = 1,∂f1
∂z (1,2,3) = 0,
∂f2
∂x (1,2,3) = 0,∂f2
∂y (1,2,3) = 2 ·2=4,∂f2
∂z (1,2,3) = −1.
Step 3: Construct the Jacobian matrix of fat (1,2,3). The Jacobian
matrix is given by
Jf(1,2,3) = 2 1 0
0 4 −1.
Question 29
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y, 1
2yz). 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 with respect to each variable. The Jacobian matrix of
fwill be a 2 ×3 matrix.
Step 1: Compute the partial derivatives of f
Let’s first write fin component form:
f(x, y, z) = x2+y
1
2yz
Now, we will find the partial derivatives of each component function with
respect to each variable.
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
23
∂f1
∂z = 0
(since f1does not depend on z)
∂f2
∂x = 0
(since f2does not depend on x)
∂f2
∂y = 0
(since f2does not depend on y)
∂f2
∂z =∂
∂z (1
2yz) = 1
2y
Step 2: Assemble the Jacobian matrix
The Jacobian matrix of fis formed by arranging these partial derivatives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0 0 1
2y
Therefore, the Jacobian matrix of fis:
Jf=2x1 0
0 0 1
2y
Question 30
Question
Consider the following system of equations:
x2+y2= 25
xy + 2z= 10
z3+xyz = 40
Compute the Jacobian matrix at the point (3,4,2).
Solution
Step 1: Define the system of equations as a function: Let’s define the system of
equations as a vector-valued function F(x):
F(x) =
f1(x, y, z)
f2(x, y, z)
f3(x, y, z)
=
x2+y2−25
xy + 2z−10
z3+xyz −40
24
Step 2: Assemble the Jacobian matrix Substitute the partial deriva-
tives into the Jacobian matrix:
Jf=2x1 0
yz xz xy
Question 2
Question
Let f(x) = 3x2
1+ 2x2
x1+ 4x3
2be a vector-valued function. Find the Jacobian matrix
of f.
Solution
Step 1: The Jacobian matrix of fis defined as the matrix of all first-order partial
derivatives of the components of f:
J(f) = "∂f1
∂x1
∂f1
∂x2
∂f2
∂x1
∂f2
∂x2#
Step 2: To find the partial derivatives, we differentiate each component of f
with respect to each variable:
∂f1
∂x1
= 6x1,∂f1
∂x2
= 2
∂f2
∂x1
= 1,∂f2
∂x2
= 12x2
2
Step 3: Now, we can construct the Jacobian matrix using the computed
partial derivatives:
J(f) = 6x12
1 12x2
2
Therefore, the Jacobian matrix of fis 6x12
1 12x2
2.
Question 3
Question
Let f:R3→R2be defined by f(x, y, z) = (2x2+yz, xz). Find the Jacobian
matrix Jfof f.
2
Solution
Step 1: Write out the components of the function f:
f(x, y, z) = 2x2+yz
xz
Step 2: Determine the partial derivatives of each component of fwith respect
to x, y, z:
∂f1
∂x = 4x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 3: Assemble the partial derivatives into the Jacobian matrix Jf:
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=4x z y
z0x
Therefore, the Jacobian matrix of fis:
Jf=4x z y
z0x
Question 4
Question
Let f(x, y, z)=(xy2z, x2yz, xyz2) be a vector-valued function. Find the Jaco-
bian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we need to compute the
partial derivatives of each component function at that point and organize them
into a matrix.
Step 1: Calculate the partial derivatives
Let’s find the partial derivative of f1(x, y, z) = xy2zwith respect to x,y,
and zat the point (1,2,3).
∂f1
∂x =y2z= 22·3 = 12
∂f1
∂y = 2xyz = 2 ·1·3=6
∂f1
∂z =xy2= 1 ·22= 4
3
Now, find the partial derivative of f2(x, y, z) = x2yz with respect to x,y,
and zat the point (1,2,3).
∂f2
∂x = 2xyz = 2 ·1·3=6
∂f2
∂y =x2z= 12·3=3
∂f2
∂z =x2y= 12·2=2
Lastly, find the partial derivative of f3(x, y, z) = xyz2with respect to x,y,
and zat the point (1,2,3).
∂f3
∂x =yz2= 2 ·32= 18
∂f3
∂y =xz2= 1 ·32= 9
∂f3
∂z = 2xyz = 2 ·1·3=6
Step 2: Organize the partial derivatives into the Jacobian matrix
The Jacobian matrix of fat the point (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
=
12 6 4
6 3 2
18 9 6
Question 5
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+yz, y2−
xz). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
fwith respect to each variable and arrange them into a matrix.
4
Step 1: Compute the partial derivatives of f(x, y, z).
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
∂f2
∂x =−z
∂f2
∂y = 2y
∂f2
∂z =−x
Step 2: Arrange the partial derivatives into a matrix. The Jacobian matrix
of fis given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
−z2y−x
Therefore, the Jacobian matrix of fis 2x z y
−z2y−x.
Question 6
Question
Let f(x, y, z) =
x2+y2
yz
z2
. Compute the Jacobian matrix of fat the point
(1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we need to compute the
partial derivatives of each component function with respect to each variable,
then evaluate them at the given point.
Step 1: Compute the partial derivatives of f.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 2z
5
Step 2: Evaluate the partial derivatives at (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) = 0,∂f2
∂y (1,2,3) = 3,∂f2
∂z (1,2,3) = 2
∂f3
∂x (1,2,3) = 0,∂f3
∂y (1,2,3) = 0,∂f3
∂z (1,2,3) = 2(3) = 6
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fat (1,2,3)
is given by
Jf(1,2,3) =
240
032
006
.
Question 7
Question
Let f(x, y, z) = x3yz +exy sin(z) where x= 2t,y=t2, and z=t. Compute the
Jacobian matrix of fwith respect to tat t= 1.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (x3yz +exy sin(z))
= 3x2yz +yexy sin(z)
= 12t2t2t+t2e2tsin(t)
= 12t5+t2e2tsin(t)
∂f
∂y =∂
∂y (x3yz +exy sin(z))
=x3z+xexy sin(z)
= 8t3t+ 2te2tsin(t)
= 8t4+ 2te2tsin(t)
∂f
∂z =∂
∂z (x3yz +exy sin(z))
=x3y+exy cos(z)
= 8t4+e2tcos(t)
6
Step 2: Compute the Jacobian matrix Jf(t) by arranging the partial deriva-
tives in a matrix form.
Jf(t) = 12t5+t2e2tsin(t) 8t4+ 2te2tsin(t) 8t4+e2tcos(t)
Step 3: Evaluate the Jacobian matrix at t= 1.
Jf(1) = 12(1)5+ (1)2e2(1) sin(1) 8(1)4+ 2(1)e2(1) sin(1) 8(1)4+e2(1) cos(1)
Jf(1) = 12 + e2sin(1) 8 + 2e2sin(1) 8 + e2cos(1)
Question 8
Question
Find the Jacobian matrix of the transformation T:R3→R3given by:
T(x, y, z) = (x2+y, 2y−z, 3z)
Solution
To find the Jacobian matrix of the transformation T, we need to compute the
partial derivatives of each component function.
Step 1: Compute ∂T
∂x . For the first component function, T1(x, y, z) = x2+y,
we have: ∂T1
∂x = 2x
Step 2: Compute ∂T
∂y . For the second component function, T2(x, y, z) =
2y−z, we have:
∂T2
∂y = 2
Step 3: Compute ∂T
∂z . For the third component function, T3(x, y, z)=3z,
we have: ∂T3
∂z = 3
Therefore, the Jacobian matrix of the transformation T:R3→R3is:
2 0 0
0 2 −1
0 0 3
Question 9
Question
Let f:R3→R3be a mapping defined by f(x, y, z)=(x2+y2, x +z, xyz). Find
the Jacobian matrix of f.
7
Solution
To find the Jacobian matrix of f, we need to calculate the partial derivatives of
each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function f1(x, y, z) =
x2+y2.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function f2(x, y, z) =
x+z.∂f2
∂x = 1,∂f2
∂y = 0,∂f2
∂z = 1
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z for the third component function f3(x, y, z) =
xyz.
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z =xy
Step 4: Assemble the partial derivatives into a 3 ×3 matrix to obtain the
Jacobian matrix of f:
Jf =
2x2y0
1 0 1
yz xz xy
Question 10
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2y, yz, z3). Find the
Jacobian matrix of fat the point (1,2,−1).
Solution
Given the function f:R3→R3defined by f(x, y, z) = (x2y, yz, z3), let’s find
the Jacobian matrix of fat the point (1,2,−1).
Step 1: Calculate the partial derivatives The Jacobian matrix of fis
given by
J=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
where f1=x2y,f2=yz, and f3=z3.
Calculating the partial derivatives:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
8
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 3z2
Step 2: Evaluate the Jacobian matrix at the point (1,2,−1) Substi-
tute x= 1, y = 2, z =−1 into the partial derivatives calculated above to find
the values of the Jacobian matrix at the point (1,2,−1).
J(1,2,−1) =
2(1)(2) 1 0
0−1 2
0 0 −3
=
4 1 0
0−1 2
0 0 −3
Therefore, the Jacobian matrix of fat the point (1,2,−1) is
J(1,2,−1) =
4 1 0
0−1 2
0 0 −3
Question 11
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(exsin(yz), x2+
y2). Find the Jacobian matrix of fat the point (0,0,0).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x =∂
∂x (exsin(yz)) = exsin(yz)
∂f1
∂y =∂
∂y (exsin(yz)) = zexcos(yz)
∂f1
∂z =∂
∂z (exsin(yz)) = yexcos(yz)
Step 2: Calculate the partial derivatives of fwith respect to xand y.
∂f2
∂x =∂
∂x (x2+y2)=2x
∂f2
∂y =∂
∂y (x2+y2)=2y
∂f2
∂z = 0
9
Step 3: Evaluate the Jacobian matrix of fat (0,0,0). The Jacobian matrix
is given by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Substitute the partial derivatives we calculated earlier:
Jf=100
000
Therefore, the Jacobian matrix of fat (0,0,0) is:
Jf=100
000
Question 12
Question
Let f(x, y, z) =
x2+y2
y2+z2
z2+x2
. Find the Jacobian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of fat the point (1,2,3), we first need to compute
the partial derivatives of each component of fwith respect to x,y, and zat the
given point.
Step 1: Compute the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 2: Evaluate the partial derivatives at (1,2,3):
∂f1
∂x = 2(1) = 2,∂f1
∂y = 2(2) = 4,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y = 2(2) = 4,∂f2
∂z = 2(3) = 6
∂f3
∂x = 2(1) = 2,∂f3
∂y = 0,∂f3
∂z = 2(3) = 6
10
Step 3: Construct the Jacobian matrix:
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
240
046
206
Question 13
Question
Let f(x, y) = (2x2+y, 3xy) and g(u, v)=(u+v, 2u−v). Determine the Jacobian
matrix of the composition g◦f.
Solution
Step 1: Find the composition g◦f.
(g◦f)(x, y) = g(f(x, y))
=g(2x2+y, 3xy)
= (2x2+y+ 3xy, 2(2x2+y)−3xy)
= (2x2+ 3xy +y, 4x2+ 2y−3xy)
= (2x2+ 3xy +y, 4x2+y−xy)
Step 2: Find the Jacobian matrix of g◦f. The Jacobian matrix of a function
h:Rn→Rmis an m×nmatrix where the i-th row and j-th column entry is the
partial derivative of the i-th component of hwith respect to the j-th variable.
For the function h(x, y) = (2x2+ 3xy +y, 4x2+y−xy), the Jacobian matrix
is given by
Jg◦f="∂h1
∂x
∂h1
∂y
∂h2
∂x
∂h2
∂y #=4x+ 3y2x+ 3
8x−y1
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Determine the
Jacobian matrix of fat the point (2,1,−3).
11
Solution
Step 1: Compute the partial derivatives of f:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0,
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y.
Step 2: Evaluate the partial derivatives at the point (2,1,−3):
∂f1
∂x (2,1,−3) = 2(2)(1) = 4,∂f1
∂y (2,1,−3) = 22= 4,∂f1
∂z (2,1,−3) = 0,
∂f2
∂x (2,1,−3) = 0,∂f2
∂y (2,1,−3) = −3,∂f2
∂z (2,1,−3) = 1.
Step 3: Assemble the Jacobian matrix of fat (2,1,−3):
Jf(2,1,−3) = 440
0−3 1.
Question 15
Question
Find the Jacobian matrix of the following system of equations:
x= 2u2+v2
y=uv + 3u
z=u−v
Solution
Step 1: To find the Jacobian matrix, we need to compute the partial derivatives
of each equation with respect to each variable.
Step 2: Let’s begin by finding ∂(x,y,z)
∂(u,v).
Step 3: Compute the partial derivatives of xwith respect to uand v:
∂x
∂u = 4u, ∂x
∂v = 2v
Step 4: Compute the partial derivatives of ywith respect to uand v:
∂y
∂u =v+ 3,∂y
∂v =u
Step 5: Compute the partial derivatives of zwith respect to uand v:
∂z
∂u = 1,∂z
∂v =−1
12
Step 6: Assemble the partial derivatives into the Jacobian matrix:
J=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
=
4u2v
v+ 3 u
1−1
Question 16
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, y2+ 2z). Find the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f, we compute the partial derivatives of each
component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2.
Step 3: Assemble the Jacobian matrix of f. The Jacobian matrix of fis
given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0 2y2.
Therefore, the Jacobian matrix of fis
Jf=2x1 0
0 2y2.
Question 17
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Find the Jacobian
matrix of f.
13
Solution
Step 1: We first need to determine the partial derivatives of f.
∂f
∂x =∂f1
∂x
∂f2
∂x
∂f
∂y ="∂f1
∂y
∂f2
∂y #
∂f
∂z =∂f1
∂z
∂f2
∂z
Step 2: Calculate the partial derivatives of f.
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Assemble the Jacobian matrix Jf.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2x1 0
0z y
Question 18
Question
Find the Jacobian matrix of the following transformation:
T(x, y)=(x2+y2, xy)
14
Solution
To find the Jacobian matrix of the transformation T(x, y) = (x2+y2, xy), we
need to calculate the partial derivatives of the components of the transformation.
Step 1: Find ∂f
∂x and ∂f
∂y for the first component function f(x, y) = x2+y2.
∂f
∂x = 2xand ∂f
∂y = 2y
Step 2: Find ∂g
∂x and ∂g
∂y for the second component function g(x, y) = xy.
∂g
∂x =yand ∂g
∂y =x
Step 3: Assemble the Jacobian matrix using the partial derivatives calcu-
lated in Step 1 and Step 2. The Jacobian matrix is given by:
J="∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y #=2x2y
y x
Therefore, the Jacobian matrix of the transformation T(x, y)=(x2+y2, xy)
is J=2x2y
y x .
Question 19
Question
Let f(x, y, z) =
x2y
yz
x+z
. Find the Jacobian matrix Jf(x, y, z) of f.
Solution
To find the Jacobian matrix Jf(x, y, z) of f, we must compute the partial deriva-
tives of each component of fwith respect to x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y ,∂f1
∂z :
f1(x, y, z) = x2y
∂f1
∂x = 2xy
∂f1
∂y =x2
∂f1
∂z = 0
15
Step 2: Compute ∂f2
∂x ,∂f2
∂y ,∂f2
∂z :
f2(x, y, z) = yz
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 3: Compute ∂f3
∂x ,∂f3
∂y ,∂f3
∂z :
f3(x, y, z) = x+z
∂f3
∂x = 1
∂f3
∂y = 0
∂f3
∂z = 1
Step 4: Assemble the partial derivatives into the Jacobian matrix:
Jf(x, y, z) =
2xy x20
0z y
1 0 1
Question 20
Question
Let f:R3→R3be defined by f(x, y, z)=(x2+y2, y2+z2, z2+x2). 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 of fwith respect to x,y, and z.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
16
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 4: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by:
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Substitute the partial derivatives into the matrix to get:
Jf=
2x2y0
0 2y2z
2x0 2z
Question 21
Question
Let f:R3→R2be the function defined by f(x, y, z) = (x3, y2z). Calculate the
Jacobian matrix of fat the point (1,2,−1).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis a matrix of partial
derivatives. Given f(x, y, z)=(x3, y2z), we need to find the partial derivatives
of each component function:
For the first component function, f1(x, y, z) = x3, the partial derivatives are:
∂f1
∂x = 3x2,∂f1
∂y = 0,and ∂f1
∂z = 0.
For the second component function, f2(x, y, z) = y2z, the partial derivatives
are: ∂f2
∂x = 0,∂f2
∂y = 2yz, and ∂f2
∂z =y2.
Step 2: Evaluate the partial derivatives at the point (1,2,−1):
∂f1
∂x (1,2,−1) = 3(1)2= 3,∂f1
∂y (1,2,−1) = 0,∂f1
∂z (1,2,−1) = 0,
∂f2
∂x (1,2,−1) = 0,∂f2
∂y (1,2,−1) = 2(2)(−1) = −4,∂f2
∂z (1,2,−1) = (2)2= 4.
Step 3: Assemble the Jacobian matrix Jfusing the partial derivatives:
Jf(1,2,−1) = 300
0−4 4.
17
Therefore, the Jacobian matrix of fat the point (1,2,−1) is 300
0−4 4.
Question 22
Question
Find the Jacobian matrix of the transformation given by x=u+v, y =u−v, z =
uv.
Solution
To find the Jacobian matrix of a transformation, we need to compute the partial
derivatives of each resulting variable with respect to each input variable.
Step 1: Find ∂x
∂u ,∂x
∂v ,∂y
∂u ,∂y
∂v ,∂z
∂u , and ∂z
∂v .
Step 2: Compute the partial derivatives:
∂x
∂u =∂
∂u (u+v)=1,∂x
∂v =∂
∂v (u+v)=1,
∂y
∂u =∂
∂u (u−v)=1,∂y
∂v =∂
∂v (u−v) = −1,
∂z
∂u =∂
∂u (uv) = v, ∂z
∂v =∂
∂v (uv) = u.
Step 3: Assemble the Jacobian matrix:
J=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂z
∂u
∂z
∂v
=
1 1
1−1
v u
.
Question 23
Question
Let f(x, y, z) =
ex2+y2
sin(xy)
xyz
be a vector-valued function. Determine the Jacobian
matrix of f.
Solution
To find the Jacobian matrix of f(x, y, z), we need to compute the partial deriva-
tives of each component function with respect to x,y, and z.
Step 1: Compute the partial derivatives of fwith respect to x.
∂f
∂x =
∂
∂x (ex2+y2)
∂
∂x (sin(xy))
∂
∂x (xyz)
=
2xex2+y2
ycos(xy)
yz
18
Step 2: Compute the partial derivatives of fwith respect to y.
∂f
∂y =
∂
∂y (ex2+y2)
∂
∂y (sin(xy))
∂
∂y (xyz)
=
2yex2+y2
xcos(xy)
xz
Step 3: Compute the partial derivatives of fwith respect to z.
∂f
∂z =
∂
∂z (ex2+y2)
∂
∂z (sin(xy))
∂
∂z (xyz)
=
0
0
xy
Therefore, the Jacobian matrix of fis given by
Jf(x, y, z) =
2xex2+y22yex2+y20
ycos(xy)xcos(xy) 0
yz xz xy
Question 24
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+yz, exy ).
Find the Jacobian matrix of fat the point (1,2,0).
Solution
To find the Jacobian matrix of fat (1,2,0), we compute the partial derivatives
of each component function with respect to each variable and then arrange them
in a matrix.
Let f(x, y, z)=(u(x, y, z), v(x, y, z)), where u(x, y, z) = x2+yz and v(x, y, z) =
exy.
Step 1: Calculate the partial derivatives of u:
∂u
∂x = 2x,
∂u
∂y =z,
∂u
∂z =y.
Step 2: Calculate the partial derivatives of v:
∂v
∂x =yexy ,
∂v
∂y =xexy,
∂v
∂z = 0.
19
Step 3: Evaluate the partial derivatives at the point (1,2,0):
∂u
∂x (1,2,0) = 2(1) = 2,
∂u
∂y (1,2,0) = 0,
∂u
∂z (1,2,0) = 2,
∂v
∂x (1,2,0) = 2e2,
∂v
∂y (1,2,0) = e,
∂v
∂z (1,2,0) = 0.
Step 4: Assemble the Jacobian matrix:
Jf(1,2,0) = "∂u
∂x (1,2,0) ∂u
∂y (1,2,0) ∂u
∂z (1,2,0)
∂v
∂x (1,2,0) ∂v
∂y (1,2,0) ∂v
∂z (1,2,0)#=2 0 2
2e2e0.
So, the Jacobian matrix of fat (1,2,0) is 2 0 2
2e2e0.
Question 25
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). 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.
20
Step 1: Find the partial derivatives of f:
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 2: Assemble the partial derivatives into the Jacobian matrix of f:
Jac(f) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xy x20
0z y
Therefore, the Jacobian matrix of fis 2xy x20
0z y.
Question 26
Question
Let f:R3→R3be a function defined by f(x, y, z)=(xyz, x2y, z3). Compute
the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =yz, ∂f1
∂y =xz, ∂f1
∂z =xy
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 2xy, ∂f2
∂y =x2,∂f2
∂z = 0
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 0,∂f3
∂y = 0,∂f3
∂z = 3z2
21
Step 4: Assemble the Jacobian matrix of f. The Jacobian matrix of fis:
Jf=
yz xz xy
2xy x20
0 0 3z2
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+yz, y2+xz). 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 of
partial derivatives. In this case, since f:R3→R2, the Jacobian matrix will be
a 2 ×3 matrix.
Step 2: First, we find the partial derivatives of fwith respect to x,y, and
z.∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
∂f2
∂x =z, ∂f2
∂y = 2y, ∂f2
∂z =x
Step 3: Evaluate these partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 3,∂f1
∂z (1,2,3) = 2
∂f2
∂x (1,2,3) = 3,∂f2
∂y (1,2,3) = 4,∂f2
∂z (1,2,3) = 1
Step 4: Assemble these partial derivatives into the Jacobian matrix.
Jacobian matrix of fat (1,2,3) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=232
341
Question 28
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z) = (x2+y, y2−
z). Find the Jacobian matrix of fat the point (1,2,3).
22
Solution
To find the Jacobian matrix of fat (1,2,3), we first need to determine the
partial derivatives of each component function.
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0,
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z =−1.
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) = 1,∂f1
∂z (1,2,3) = 0,
∂f2
∂x (1,2,3) = 0,∂f2
∂y (1,2,3) = 2 ·2=4,∂f2
∂z (1,2,3) = −1.
Step 3: Construct the Jacobian matrix of fat (1,2,3). The Jacobian
matrix is given by
Jf(1,2,3) = 2 1 0
0 4 −1.
Question 29
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y, 1
2yz). 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 with respect to each variable. The Jacobian matrix of
fwill be a 2 ×3 matrix.
Step 1: Compute the partial derivatives of f
Let’s first write fin component form:
f(x, y, z) = x2+y
1
2yz
Now, we will find the partial derivatives of each component function with
respect to each variable.
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
23
∂f1
∂z = 0
(since f1does not depend on z)
∂f2
∂x = 0
(since f2does not depend on x)
∂f2
∂y = 0
(since f2does not depend on y)
∂f2
∂z =∂
∂z (1
2yz) = 1
2y
Step 2: Assemble the Jacobian matrix
The Jacobian matrix of fis formed by arranging these partial derivatives:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x1 0
0 0 1
2y
Therefore, the Jacobian matrix of fis:
Jf=2x1 0
0 0 1
2y
Question 30
Question
Consider the following system of equations:
x2+y2= 25
xy + 2z= 10
z3+xyz = 40
Compute the Jacobian matrix at the point (3,4,2).
Solution
Step 1: Define the system of equations as a function: Let’s define the system of
equations as a vector-valued function F(x):
F(x) =
f1(x, y, z)
f2(x, y, z)
f3(x, y, z)
=
x2+y2−25
xy + 2z−10
z3+xyz −40
24
Step 2: Compute the Jacobian matrix: The Jacobian matrix J(F) of Fis
given by:
J(F) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Step 3: Compute the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 2
∂f3
∂x =yz, ∂f3
∂y =xz, ∂f3
∂z = 3z2+xy
Step 4: Evaluate the Jacobian at the point (3,4,2): Substitute x= 3, y= 4,
and z= 2 into the partial derivatives computed in Step 3 to find the Jacobian
matrix at the point (3,4,2).
Step 5: Compute the Jacobian matrix at (3,4,2):
J(F)(3,4,2)
=
6 8 0
4 3 2
8 6 10
Therefore, the Jacobian matrix of Fat the point (3,4,2) is:
J(F)(3,4,2)
=
6 8 0
4 3 2
8 6 10
25
