MATH 117 - ELEMENTS OF
MATHEMATICS - Jacobian matrix
Question Bank - Set 7
Liberty University
Question 1
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 (1,2,3), we need to compute the partial
derivatives of fwith respect to x,y, and z, and evaluate them at (1,2,3).
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0,∂f2
∂y = 2y, ∂f2
∂z = 2z
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x = 2x, ∂f3
∂y = 0,∂f3
∂z = 2z
Step 4: Evaluate the partial derivatives at (1,2,3) to find the Jacobian
matrix.
Jf(1,2,3) =
2(1) 2(2) 0
0 2(2) 2(3)
2(1) 0 2(3)
=
240
046
206
Therefore, the Jacobian matrix of fat the point (1,2,3) is
240
046
206
.
Question 2
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+yz, y2+xz). 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 need to compute the
partial derivatives of each component function of fwith respect to each input
variable at the given point.
Let’s start by computing the partial derivatives:
∂
∂x (x2+yz)=2x, ∂
∂y (x2+yz) = z, ∂
∂z (x2+yz) = y,
∂
∂x (y2+xz) = z, ∂
∂y (y2+xz)=2y, ∂
∂z (y2+xz) = x.
Evaluate these partial derivatives at (1,2,3):
∂
∂x (x2+yz)(1,2,3)
= 2(1) = 2,∂
∂y (x2+yz)(1,2,3)
= 3,∂
∂z (x2+yz)(1,2,3)
= 2,
∂
∂x (y2+xz)(1,2,3)
= 3,∂
∂y (y2+xz)(1,2,3)
= 4,∂
∂z (y2+xz)(1,2,3)
= 1.
Therefore, the Jacobian matrix of fat the point (1,2,3) is given by:
232
341.
Question 3
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2z+y, 2xyz). 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.
2
Step 1: Compute the partial derivatives of f. Let f1(x, y, z) = x2z+yand
f2(x, y, z)=2xyz. We calculate the partial derivatives of f1and f2:
∂f1
∂x = 2xz, ∂f1
∂y = 1,∂f1
∂z =x2,
∂f2
∂x = 2yz, ∂f2
∂y = 2xz, ∂f2
∂z = 2xy.
Step 2: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xz 1x2
2yz 2xz 2xy.
Therefore, the Jacobian matrix of fis 2xz 1x2
2yz 2xz 2xy.
Question 4
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2y, zex, xyz). Find the
Jacobian matrix of f.
Solution
Step 1: Write out the function component-wise.
Given f(x, y, z) = (x2y, zex, xyz), we have
f(x, y, z) =
x2y
zex
xyz
Step 2: Find the partial derivatives of each component of f.
Compute the partial derivatives as follows:
∂f
∂x =
∂
∂x (x2y)
∂
∂x (zex)
∂
∂x (xyz)
=
2xy
zex
yz
∂f
∂y =
∂
∂y (x2y)
∂
∂y (zex)
∂
∂y (xyz)
=
x2
0
xz
∂f
∂z =
∂
∂z (x2y)
∂
∂z (zex)
∂
∂z (xyz)
=
0
ex
xy
3
Step 3: Assemble the partial derivatives into a matrix.
The Jacobian matrix of fis:
Jf=
2xy x20
zex0ex
yz xz xy
Question 5
Question
Let f:R2→R2be a differentiable function defined by f(x, y) = (3x2y, x +y).
Find the Jacobian matrix of fat the point (1,2).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmat a point ais an m×n
matrix whose entries are the partial derivatives of the components of f. In this
case, f(x, y) = (3x2y, x +y), so the Jacobian matrix Jf(a) is given by
Jf(a) = ∂f1
∂x (a)∂f1
∂y (a)
∂f2
∂x (a)∂f2
∂y (a)!
Step 2: Compute the partial derivatives of f:
∂f1
∂x (x, y)=6xy and ∂f1
∂y (x, y)=3x2
∂f2
∂x (x, y) = 1 and ∂f2
∂y (x, y) = 1
Step 3: Evaluate the Jacobian matrix at the point (1,2):
Jf(1,2) = ∂f1
∂x (1,2) ∂f1
∂y (1,2)
∂f2
∂x (1,2) ∂f2
∂y (1,2)!
=12 3
1 1
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (4x2+yz, 3y−z).
Compute the Jacobian matrix of f.
4
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each input variable. The Jacobian matrix
of fwill then be a 2 ×3 matrix.
Step 1: Compute the partial derivatives of f:
∂f1
∂x =∂
∂x (4x2+yz) = 8x
∂f1
∂y =∂
∂y (4x2+yz) = z
∂f1
∂z =∂
∂z (4x2+yz) = y
∂f2
∂x =∂
∂x (3y−z)=0
∂f2
∂y =∂
∂y (3y−z)=3
∂f2
∂z =∂
∂z (3y−z) = −1
Step 2: Assemble the Jacobian matrix: The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=8x z y
0 3 −1
Question 7
Question
Let f:R2→R2be defined by f(x, y)=(x2y, sin(xy)). Calculate the Jacobian
matrix of fat the point (2, π).
Solution
To calculate the Jacobian matrix of fat the point (2, π), we need to find the
partial derivatives of fwith respect to xand y.
Step 1: Find ∂f1∂x and ∂f1
∂y
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
5
Step 2: Find ∂f2∂x and ∂f2
∂y
∂f2
∂x =∂
∂x (sin(xy)) = ycos(xy)
∂f2
∂y =∂
∂y (sin(xy)) = xcos(xy)
Step 3: Evaluate the Jacobian matrix at (2, π) Substitute x= 2 and
y=πinto the partial derivatives:
∂f1
∂x (2, π) = 2(2)(π)=4π
∂f1
∂y (2, π)=22= 4
∂f2
∂x (2, π) = πcos(2π) = −π
∂f2
∂y (2, π) = 2 cos(2π)=2
Thus, the Jacobian matrix of fat (2, π) is:
Jf(2, π) = 4π4
−π2
Question 8
Question
Let f:R3→R2be a differentiable function given by f(x, y, z)=(x2y, y2z).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix Jfof f, we need to calculate the partial derivatives
of fwith respect to each variable.
Step 1: Calculate ∂f
∂x :
∂f
∂x =∂(x2y)
∂x ,∂(y2z)
∂x = (2xy, 0)
Step 2: Calculate ∂f
∂y :
∂f
∂y =∂(x2y)
∂y ,∂(y2z)
∂y = (x2,2yz)
Step 3: Calculate ∂f
∂z :
∂f
∂z =∂(x2y)
∂z ,∂(y2z)
∂z = (0, y2)
6
Therefore, the Jacobian matrix Jfof fis:
Jf=
2xy 0
x22yz
0y2
Question 9
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2+ 2y, y2+ 3z, z2+ 4x).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to x,y, and z.
Step 1: Partial Derivatives with respect to xWe calculate the partial
derivatives of fwith respect to x:
∂f1
∂x =∂
∂x (x2+ 2y)=2x
∂f2
∂x =∂
∂x (y2+ 3z) = 0
∂f3
∂x =∂
∂x (z2+ 4x) = 4
Step 2: Partial Derivatives with respect to yWe calculate the partial
derivatives of fwith respect to y:
∂f1
∂y =∂
∂y (x2+ 2y)=2
∂f2
∂y =∂
∂y (y2+ 3z) = 2y
∂f3
∂y =∂
∂y (z2+ 4x) = 0
Step 3: Partial Derivatives with respect to zWe calculate the partial
derivatives of fwith respect to z:
∂f1
∂z =∂
∂z (x2+ 2y) = 0
∂f2
∂z =∂
∂z (y2+ 3z)=3
∂f3
∂z =∂
∂z (z2+ 4x) = 2z
7
Step 4: Assemble the Jacobian Matrix The Jacobian matrix of fis
given by the matrix of its partial derivatives:
Jac(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2 0
0 2y3
4 0 2z
Question 10
Question
Let f:R3→R3be defined by f(x, y, z) = (x2+yz, y2+zx, z2+xy). Calculate
the Jacobian matrix of fat the point (1,2,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂f1
∂x
∂f2
∂x
∂f3
∂x
,∂f
∂y =
∂f1
∂y
∂f2
∂y
∂f3
∂y
,∂f
∂z =
∂f1
∂z
∂f2
∂z
∂f3
∂z
Step 2: Compute the partial derivatives.
∂f
∂x =
2x z y
z0x
0y2z
,∂f
∂y =
z2y x
0 2y z
x z 0
,∂f
∂z =
y0 2z
0x y
2x y 0
Step 3: Evaluate the partial derivatives at the point (1,2,2).
∂f
∂x (1,2,2) =
222
201
024
,∂f
∂y (1,2,2) =
241
042
120
,∂f
∂z (1,2,2) =
204
012
220
Therefore, the Jacobian matrix of fat the point (1,2,2) is:
Jf(1,2,2) =
222
201
024
Question 11
Question
Consider the transformation given by the equations:
x=u2−v2
y= 2uv
Find the Jacobian matrix of the transformation at the point (u, v) = (1,1).
8
Solution
We are given the transformation equations:
x=u2−v2
y= 2uv
The Jacobian matrix of a transformation is given by:
J=∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
Step 1: Find ∂x ∂u and ∂x
∂v .
∂x
∂u =∂
∂u (u2−v2)=2u
∂x
∂v =∂
∂v (u2−v2) = −2v
Step 2: Find ∂y ∂u and ∂y
∂v .
∂y
∂u =∂
∂u (2uv)=2v
∂y
∂v =∂
∂v (2uv) = 2u
Step 3: Evaluate the Jacobian matrix at (u, v) = (1,1).Substitute
u= 1 and v= 1 into the partial derivatives:
J=2−2
2 2
Therefore, the Jacobian matrix at the point (u, v) = (1,1) is:
J=2−2
2 2
Question 12
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,−1).
9
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable:
∂f1
∂x =∂
∂x (x2+yz)=2x,
∂f1
∂y =∂
∂y (x2+yz) = z,
∂f1
∂z =∂
∂z (x2+yz) = y,
∂f2
∂x =∂
∂x (y2−xz) = −z,
∂f2
∂y =∂
∂y (y2−xz)=2y,
∂f2
∂z =∂
∂z (y2−xz) = −x.
Step 2: Evaluate the partial derivatives at the point (1,2,−1):
∂f1
∂x (1,2,−1)
= 2(1) = 2,
∂f1
∂y (1,2,−1)
= (−1) = −1,
∂f1
∂z (1,2,−1)
= (2) = 2,
∂f2
∂x (1,2,−1)
= (−(−1)) = 1,
∂f2
∂y (1,2,−1)
= 2(2) = 4,
∂f2
∂z (1,2,−1)
=−1.
Step 3: Assemble the Jacobian matrix of fat (1,2,−1):
Jf(1,2,−1) = 2−1 2
1 4 −1.
Question 13
Question
Let f:R3→R3be defined by f(x, y, z) = (x2y, yz, z2). Compute the Jacobian
matrix of f.
10
Solution
To compute the Jacobian matrix of f, we first need to find the partial derivatives
of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
∂f1
∂z =∂
∂z (x2y)=0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (yz)=0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (z2)=0
∂f3
∂y =∂
∂y (z2)=0
∂f3
∂z =∂
∂z (z2) = 2z
Step 4: Assemble the Jacobian matrix. The Jacobian matrix Jfof fis
given by
Jf=
2xy x20
0z y
0 0 2z
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z) = (3x+ 2y, xz). Find the
Jacobian matrix of f.
11
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each variable. The Jacobian matrix of f
is given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #.
Step 1: Calculate the partial derivatives of f
For f(x, y, z) = (3x+ 2y, xz), we have:
∂f1
∂x = 3,
∂f1
∂y = 2,
∂f1
∂z = 0,
∂f2
∂x =z,
∂f2
∂y = 0,
∂f2
∂z =x.
Step 2: Construct the Jacobian matrix Jf
Plugging in the computed partial derivatives, the Jacobian matrix Jfis:
Jf=3 2 0
z0x.
Therefore, the Jacobian matrix of the function f(x, y, z) = (3x+ 2y, xz) is:
Jf=3 2 0
z0x.
Question 15
Question
Let f(x, y, z) =
x2y+y2z
xyz
xz2
. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Step 1: Find the partial derivatives of f.
∂f
∂x =2xy y2z2
12
∂f
∂y =x22xz 0
∂f
∂z =0xy 2xz
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
=449
∂f
∂y (1,2,3)
=160
∂f
∂z (1,2,3)
=026
Step 3: Assemble the Jacobian matrix.
Jf(1,2,3) =
449
160
026
Question 16
Question
Consider the transformation T:R3→R3given by:
T(x, y, z)=(x2+y, y2+z, z2+x)
Determine the Jacobian matrix of T.
13
Solution
Step 1: Compute the partial derivatives of Twith respect to each variable x, y, z:
∂T1
∂x = 2x
∂T1
∂y = 1
∂T1
∂z = 0
∂T2
∂x = 0
∂T2
∂y = 2y
∂T2
∂z = 1
∂T3
∂x = 1
∂T3
∂y = 0
∂T3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix J(T):
J(T) =
∂T1
∂x
∂T1
∂y
∂T1
∂z
∂T2
∂x
∂T2
∂y
∂T2
∂z
∂T3
∂x
∂T3
∂y
∂T3
∂z
=
2x1 0
0 2y1
1 0 2z
Question 17
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, xyz). Find the
Jacobian matrix of f.
Solution
Step 1: To find the Jacobian matrix of f, we first need to compute the partial
derivatives of fwith respect to each variable.
Step 2: Let’s find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z in the Jacobian matrix.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0.
14
Step 3: Next, let’s find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z in the Jacobian matrix.
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy.
Step 4: Finally, we construct the Jacobian matrix of fusing these partial
derivatives.
Jac f=2x1 0
yz xz xy.
Question 18
Question
Let f(x, y, z) = (xy, yz, xz) be a vector field. Find the Jacobian matrix Jfof f
at the point (2,−1,3).
Solution
To find the Jacobian matrix Jfof f, we need to compute the partial derivatives
of each component function with respect to each variable at the given point
(2,−1,3).
Step 1: Compute ∂f1∂x, ∂f1
∂y ,∂f1
∂z
∂f1
∂x =y=−1
∂f1
∂y =x= 2
∂f1
∂z = 0
Step 2: Compute ∂f2∂x, ∂f2
∂y ,∂f2
∂z
∂f2
∂x = 0
∂f2
∂y =z= 3
∂f2
∂z =y=−1
Step 3: Compute ∂f3∂x, ∂f3
∂y ,∂f3
∂z
∂f3
∂x =z= 3
∂f3
∂y = 0
∂f3
∂z =x= 2
15
Step 4: Assemble the Jacobian matrix JfThe Jacobian matrix is given
by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Plugging in the computed partial derivatives at (2,−1,3), we get
Jf(2,−1,3) =
−1 2 0
0 3 −1
302
Question 19
Question
Let f:R3→R2be a function given by f(x, y, z)=(x2y, yz). Find the Jacobian
matrix of f.
Solution
Step 1: Compute the partial derivatives of f. The Jacobian matrix is formed
by arranging the partial derivatives as the rows of a matrix.
Step 2: The partial derivatives of fare:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
and ∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 3: Arrange the partial derivatives in a matrix to form the Jacobian
matrix of f:
Jf(x, y, z) = 2xy x20
0z y
Question 20
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2yz, xy2z, xyz2).
Compute the Jacobian matrix of f.
16
Solution
To compute the Jacobian matrix of f, we will find the partial derivatives of each
component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2yz)=2xyz
∂f1
∂y =∂
∂y (x2yz) = x2z
∂f1
∂z =∂
∂z (x2yz) = x2y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy2z) = y2z
∂f2
∂y =∂
∂y (xy2z)=2xyz
∂f2
∂z =∂
∂z (xy2z) = xy2
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz2) = yz2
∂f3
∂y =∂
∂y (xyz2) = xz2
∂f3
∂z =∂
∂z (xyz2) = 2xyz
Step 4: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis given by
Jf=
2xyz x2z x2y
y2z2xyz xy2
yz2xz22xyz
Question 21
Question
Let f:R3→R3be a mapping given by f(x, y, z) = (2xy, x2+ 3z, y2). Find the
Jacobian matrix of fat the point (1,2,0).
17
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x =∂
∂x (2xy)=2y
∂f2
∂x =∂
∂x (x2+ 3z) = 2x
∂f3
∂x =∂
∂x (y2) = 0
∂f1
∂y =∂
∂y (2xy)=2x
∂f2
∂y =∂
∂y (x2+ 3z) = 0
∂f3
∂y =∂
∂y (y2)=2y
∂f1
∂z =∂
∂z (2xy)=0
∂f2
∂z =∂
∂z (x2+ 3z)=3
∂f3
∂z =∂
∂z (y2)=0
18
Step 2: Evaluate the partial derivatives at the point (1,2,0).
∂f1
∂x (1,2,0)
= 2(2) = 4
∂f2
∂x (1,2,0)
= 2(1) = 2
∂f3
∂x (1,2,0)
= 0
∂f1
∂y (1,2,0)
= 2(1) = 2
∂f2
∂y (1,2,0)
= 0
∂f3
∂y (1,2,0)
= 2(2) = 4
∂f1
∂z (1,2,0)
= 0
∂f2
∂z (1,2,0)
= 3
∂f3
∂z (1,2,0)
= 0
Step 3: Assemble the Jacobian matrix of fat (1,2,0).
Jf(1,2,0) =
420
203
040
Question 22
Question
Consider the function f:R3→R2given by f(x, y, z)=(x2+yz, xyz). Deter-
mine 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) =
19
x2+yz.
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = xyz.
∂f2
∂x =yz
∂f2
∂y =xz
∂f2
∂z =xy
Step 3: Construct the Jacobian matrix of fby arranging the partial deriva-
tives in a matrix.
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis Jf=2x z y
yz xz xy.
Question 23
Question
Let f:R3→R3be a differentiable function defined by f(x, y, z)=(x2+
yz, xy −z2, xz +y2). Find the Jacobian matrix of f.
Solution
Step 1: Write the components of f. The components of fare f1(x, y, z) =
x2+yz,f2(x, y, z) = xy −z2, and f3(x, y, z) = xz +y2.
Step 2: Compute the partial derivatives. To find the Jacobian matrix of f,
we need to compute the partial derivatives of each component of fwith respect
20
to x,y, and z.
∂f1
∂x =∂
∂x (x2+yz)=2x,
∂f1
∂y =∂
∂y (x2+yz) = z,
∂f1
∂z =∂
∂z (x2+yz) = y,
∂f2
∂x =∂
∂x (xy −z2) = y,
∂f2
∂y =∂
∂y (xy −z2) = x,
∂f2
∂z =∂
∂z (xy −z2) = −2z,
∂f3
∂x =∂
∂x (xz +y2) = z,
∂f3
∂y =∂
∂y (xz +y2)=2y,
∂f3
∂z =∂
∂z (xz +y2) = x.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf(x, y, z) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x z y
y x −2z
z2y x
.
Question 24
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+y, xy +z, x +z2).
Find the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z:
∂f1
∂x =∂
∂x (x2+y) = 2x,
∂f2
∂x =∂
∂x (xy +z) = y,
∂f3
∂x =∂
∂x (x+z2)=1.
21
Step 2: Compute the partial derivatives of fwith respect to y,x, and z:
∂f1
∂y =∂
∂y (x2+y)=1,
∂f2
∂y =∂
∂y (xy +z) = x,
∂f3
∂y =∂
∂y (x+z2) = 0.
Step 3: Compute the partial derivatives of fwith respect to z,x, and y:
∂f1
∂z =∂
∂z (x2+y)=0,
∂f2
∂z =∂
∂z (xy +z)=1,
∂f3
∂z =∂
∂z (x+z2)=2z.
Step 4: Evaluate the Jacobian matrix of fat the point (1,2,3):
Jf(1,2,3) =
210
120
106
.
Question 25
Question
Let f:R2→R2be defined by f(x, y) = (x2+y, xy2). Find the Jacobian matrix
of fat the point (1,2).
Solution
Step 1: Find the partial derivatives of f(x, y) with respect to xand y:
∂f
∂x =2x1
y22xy
∂f
∂y =0 1
2xy x2
Step 2: Evaluate the Jacobian matrix of fat the point (1,2) by substituting
x= 1 and y= 2 into the partial derivative matrices:
Jf(1,2) = 2(1) 1
222(1)(2)=2 1
4 4
Therefore, the Jacobian matrix of fat the point (1,2) is 2 1
4 4.
22
Question 26
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+yz, 2x+z). Calculate
the Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the components of f. In this case,
f:R3→R2, so the Jacobian matrix will be a 2 ×3 matrix.
Step 2: To find the Jacobian matrix of f, we need to calculate the partial
derivatives of the components of f. Let’s start by finding the partial derivatives
of f:
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
∂f2
∂x =∂
∂x (2x+z)=2
∂f2
∂y = 0
∂f2
∂z =∂
∂z (2x+z)=1
Step 3: Now we can construct the Jacobian matrix of f:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
2 0 1
Therefore, the Jacobian matrix of fis
Jf=2x z y
2 0 1
Question 27
Question
Let f(x, y, z) =
3x+y+z
x−2y+z
2x+y−3z
. Determine the Jacobian matrix of f.
23
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 input
variable.
Step 1: Find ∂f1
∂x .
∂f1
∂x =∂
∂x (3x+y+z)=3.
Step 2: Find ∂f1
∂y .
∂f1
∂y =∂
∂y (3x+y+z)=1.
Step 3: Find ∂f1
∂z .
∂f1
∂z =∂
∂z (3x+y+z)=1.
Step 4: Find ∂f2
∂x .
∂f2
∂x =∂
∂x (x−2y+z)=1.
Step 5: Find ∂f2
∂y .
∂f2
∂y =∂
∂y (x−2y+z) = −2.
Step 6: Find ∂f2
∂z .
∂f2
∂z =∂
∂z (x−2y+z)=1.
Step 7: Find ∂f3
∂x .
∂f3
∂x =∂
∂x (2x+y−3z)=2.
Step 8: Find ∂f3
∂y .
∂f3
∂y =∂
∂y (2x+y−3z)=1.
Step 9: Find ∂f3
∂z .
∂f3
∂z =∂
∂z (2x+y−3z) = −3.
Therefore, the Jacobian matrix of fis:
Jf(x, y, z) =
3 1 1
1−2 1
2 1 −3
.
24
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+yz, xz). Compute
the Jacobian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of a function, we need to compute the partial
derivatives of each component of the function.
Step 1: Compute partial derivatives Let u=x2+yz and v=xz. The
Jacobian matrix of fis given by
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
We will now compute the partial derivatives of uand vwith respect to x,y,
and z.
Step 2: Compute ∂u ∂x,∂u
∂y ,∂u
∂z
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
Step 3: Compute ∂v ∂x,∂v
∂y ,∂v
∂z
∂v
∂x =z, ∂v
∂y = 0,∂v
∂z =x
Step 4: Evaluate the Jacobian matrix at (1,2,3) Now, we substitute
x= 1, y= 2, and z= 3 into the partial derivatives we computed:
Jf(1,2,3) = 232
301
Therefore, the Jacobian matrix of fat the point (1,2,3) is
Jf(1,2,3) = 232
301
Question 29
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y, xz). Find the
Jacobian matrix of fat the point (1,−1,2).
25
Solution
Step 1: Write the function fin terms of its component functions.
The function fcan be written as f(x, y, z) = (2x+y, xz)=(f1(x, y, z), f2(x, y, z))
where f1(x, y, z)=2x+yand f2(x, y, z) = xz.
Step 2: Compute the partial derivatives of f1and f2.
The partial derivatives of f1are:
∂f1
∂x = 2,∂f1
∂y = 1,and ∂f1
∂z = 0
The partial derivatives of f2are:
∂f2
∂x =z, ∂f2
∂y = 0,and ∂f2
∂z =x
Step 3: Evaluate the Jacobian matrix of fat the point (1,−1,2).
The Jacobian matrix of fat the point (1,−1,2) is given by
Jf(1,−1,2) = "∂f1
∂x (1,−1,2) ∂f1
∂y (1,−1,2) ∂f1
∂z (1,−1,2)
∂f2
∂x (1,−1,2) ∂f2
∂y (1,−1,2) ∂f2
∂z (1,−1,2)#
Now substitute the values of the partial derivatives:
Jf(1,−1,2) = 210
201
Question 30
Question
Let f(x, y, z)=(x2, y2, z2). Compute the Jacobian matrix of fat the point
(1,2,3).
Solution
To compute the Jacobian matrix of a function f:Rn→Rm, we need to find
the matrix of all partial derivatives of fat the given point.
Step 1: Find the partial derivatives of f. Calculate the partial derivatives
of fas follows:
∂f
∂x =
∂x2
∂x
∂y2
∂x
∂z2
∂x
=
2x
0
0
∂f
∂y =
∂x2
∂y
∂y2
∂y
∂z2
∂y
=
0
2y
0
26
Question 2
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+yz, y2+xz). 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 need to compute the
partial derivatives of each component function of fwith respect to each input
variable at the given point.
Let’s start by computing the partial derivatives:
∂
∂x (x2+yz)=2x, ∂
∂y (x2+yz) = z, ∂
∂z (x2+yz) = y,
∂
∂x (y2+xz) = z, ∂
∂y (y2+xz)=2y, ∂
∂z (y2+xz) = x.
Evaluate these partial derivatives at (1,2,3):
∂
∂x (x2+yz)(1,2,3)
= 2(1) = 2,∂
∂y (x2+yz)(1,2,3)
= 3,∂
∂z (x2+yz)(1,2,3)
= 2,
∂
∂x (y2+xz)(1,2,3)
= 3,∂
∂y (y2+xz)(1,2,3)
= 4,∂
∂z (y2+xz)(1,2,3)
= 1.
Therefore, the Jacobian matrix of fat the point (1,2,3) is given by:
232
341.
Question 3
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2z+y, 2xyz). 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.
2
Step 1: Compute the partial derivatives of f. Let f1(x, y, z) = x2z+yand
f2(x, y, z)=2xyz. We calculate the partial derivatives of f1and f2:
∂f1
∂x = 2xz, ∂f1
∂y = 1,∂f1
∂z =x2,
∂f2
∂x = 2yz, ∂f2
∂y = 2xz, ∂f2
∂z = 2xy.
Step 2: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xz 1x2
2yz 2xz 2xy.
Therefore, the Jacobian matrix of fis 2xz 1x2
2yz 2xz 2xy.
Question 4
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2y, zex, xyz). Find the
Jacobian matrix of f.
Solution
Step 1: Write out the function component-wise.
Given f(x, y, z) = (x2y, zex, xyz), we have
f(x, y, z) =
x2y
zex
xyz
Step 2: Find the partial derivatives of each component of f.
Compute the partial derivatives as follows:
∂f
∂x =
∂
∂x (x2y)
∂
∂x (zex)
∂
∂x (xyz)
=
2xy
zex
yz
∂f
∂y =
∂
∂y (x2y)
∂
∂y (zex)
∂
∂y (xyz)
=
x2
0
xz
∂f
∂z =
∂
∂z (x2y)
∂
∂z (zex)
∂
∂z (xyz)
=
0
ex
xy
3
Step 3: Assemble the partial derivatives into a matrix.
The Jacobian matrix of fis:
Jf=
2xy x20
zex0ex
yz xz xy
Question 5
Question
Let f:R2→R2be a differentiable function defined by f(x, y) = (3x2y, x +y).
Find the Jacobian matrix of fat the point (1,2).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmat a point ais an m×n
matrix whose entries are the partial derivatives of the components of f. In this
case, f(x, y) = (3x2y, x +y), so the Jacobian matrix Jf(a) is given by
Jf(a) = ∂f1
∂x (a)∂f1
∂y (a)
∂f2
∂x (a)∂f2
∂y (a)!
Step 2: Compute the partial derivatives of f:
∂f1
∂x (x, y)=6xy and ∂f1
∂y (x, y)=3x2
∂f2
∂x (x, y) = 1 and ∂f2
∂y (x, y) = 1
Step 3: Evaluate the Jacobian matrix at the point (1,2):
Jf(1,2) = ∂f1
∂x (1,2) ∂f1
∂y (1,2)
∂f2
∂x (1,2) ∂f2
∂y (1,2)!
=12 3
1 1
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (4x2+yz, 3y−z).
Compute the Jacobian matrix of f.
4
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each input variable. The Jacobian matrix
of fwill then be a 2 ×3 matrix.
Step 1: Compute the partial derivatives of f:
∂f1
∂x =∂
∂x (4x2+yz) = 8x
∂f1
∂y =∂
∂y (4x2+yz) = z
∂f1
∂z =∂
∂z (4x2+yz) = y
∂f2
∂x =∂
∂x (3y−z)=0
∂f2
∂y =∂
∂y (3y−z)=3
∂f2
∂z =∂
∂z (3y−z) = −1
Step 2: Assemble the Jacobian matrix: The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=8x z y
0 3 −1
Question 7
Question
Let f:R2→R2be defined by f(x, y)=(x2y, sin(xy)). Calculate the Jacobian
matrix of fat the point (2, π).
Solution
To calculate the Jacobian matrix of fat the point (2, π), we need to find the
partial derivatives of fwith respect to xand y.
Step 1: Find ∂f1∂x and ∂f1
∂y
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
5
Step 2: Find ∂f2∂x and ∂f2
∂y
∂f2
∂x =∂
∂x (sin(xy)) = ycos(xy)
∂f2
∂y =∂
∂y (sin(xy)) = xcos(xy)
Step 3: Evaluate the Jacobian matrix at (2, π) Substitute x= 2 and
y=πinto the partial derivatives:
∂f1
∂x (2, π) = 2(2)(π)=4π
∂f1
∂y (2, π)=22= 4
∂f2
∂x (2, π) = πcos(2π) = −π
∂f2
∂y (2, π) = 2 cos(2π)=2
Thus, the Jacobian matrix of fat (2, π) is:
Jf(2, π) = 4π4
−π2
Question 8
Question
Let f:R3→R2be a differentiable function given by f(x, y, z)=(x2y, y2z).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix Jfof f, we need to calculate the partial derivatives
of fwith respect to each variable.
Step 1: Calculate ∂f
∂x :
∂f
∂x =∂(x2y)
∂x ,∂(y2z)
∂x = (2xy, 0)
Step 2: Calculate ∂f
∂y :
∂f
∂y =∂(x2y)
∂y ,∂(y2z)
∂y = (x2,2yz)
Step 3: Calculate ∂f
∂z :
∂f
∂z =∂(x2y)
∂z ,∂(y2z)
∂z = (0, y2)
6
Therefore, the Jacobian matrix Jfof fis:
Jf=
2xy 0
x22yz
0y2
Question 9
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2+ 2y, y2+ 3z, z2+ 4x).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to x,y, and z.
Step 1: Partial Derivatives with respect to xWe calculate the partial
derivatives of fwith respect to x:
∂f1
∂x =∂
∂x (x2+ 2y)=2x
∂f2
∂x =∂
∂x (y2+ 3z) = 0
∂f3
∂x =∂
∂x (z2+ 4x) = 4
Step 2: Partial Derivatives with respect to yWe calculate the partial
derivatives of fwith respect to y:
∂f1
∂y =∂
∂y (x2+ 2y)=2
∂f2
∂y =∂
∂y (y2+ 3z) = 2y
∂f3
∂y =∂
∂y (z2+ 4x) = 0
Step 3: Partial Derivatives with respect to zWe calculate the partial
derivatives of fwith respect to z:
∂f1
∂z =∂
∂z (x2+ 2y) = 0
∂f2
∂z =∂
∂z (y2+ 3z)=3
∂f3
∂z =∂
∂z (z2+ 4x) = 2z
7
Step 4: Assemble the Jacobian Matrix The Jacobian matrix of fis
given by the matrix of its partial derivatives:
Jac(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2 0
0 2y3
4 0 2z
Question 10
Question
Let f:R3→R3be defined by f(x, y, z) = (x2+yz, y2+zx, z2+xy). Calculate
the Jacobian matrix of fat the point (1,2,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂f1
∂x
∂f2
∂x
∂f3
∂x
,∂f
∂y =
∂f1
∂y
∂f2
∂y
∂f3
∂y
,∂f
∂z =
∂f1
∂z
∂f2
∂z
∂f3
∂z
Step 2: Compute the partial derivatives.
∂f
∂x =
2x z y
z0x
0y2z
,∂f
∂y =
z2y x
0 2y z
x z 0
,∂f
∂z =
y0 2z
0x y
2x y 0
Step 3: Evaluate the partial derivatives at the point (1,2,2).
∂f
∂x (1,2,2) =
222
201
024
,∂f
∂y (1,2,2) =
241
042
120
,∂f
∂z (1,2,2) =
204
012
220
Therefore, the Jacobian matrix of fat the point (1,2,2) is:
Jf(1,2,2) =
222
201
024
Question 11
Question
Consider the transformation given by the equations:
x=u2−v2
y= 2uv
Find the Jacobian matrix of the transformation at the point (u, v) = (1,1).
8
Solution
We are given the transformation equations:
x=u2−v2
y= 2uv
The Jacobian matrix of a transformation is given by:
J=∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
Step 1: Find ∂x ∂u and ∂x
∂v .
∂x
∂u =∂
∂u (u2−v2)=2u
∂x
∂v =∂
∂v (u2−v2) = −2v
Step 2: Find ∂y ∂u and ∂y
∂v .
∂y
∂u =∂
∂u (2uv)=2v
∂y
∂v =∂
∂v (2uv) = 2u
Step 3: Evaluate the Jacobian matrix at (u, v) = (1,1).Substitute
u= 1 and v= 1 into the partial derivatives:
J=2−2
2 2
Therefore, the Jacobian matrix at the point (u, v) = (1,1) is:
J=2−2
2 2
Question 12
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,−1).
9
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable:
∂f1
∂x =∂
∂x (x2+yz)=2x,
∂f1
∂y =∂
∂y (x2+yz) = z,
∂f1
∂z =∂
∂z (x2+yz) = y,
∂f2
∂x =∂
∂x (y2−xz) = −z,
∂f2
∂y =∂
∂y (y2−xz)=2y,
∂f2
∂z =∂
∂z (y2−xz) = −x.
Step 2: Evaluate the partial derivatives at the point (1,2,−1):
∂f1
∂x (1,2,−1)
= 2(1) = 2,
∂f1
∂y (1,2,−1)
= (−1) = −1,
∂f1
∂z (1,2,−1)
= (2) = 2,
∂f2
∂x (1,2,−1)
= (−(−1)) = 1,
∂f2
∂y (1,2,−1)
= 2(2) = 4,
∂f2
∂z (1,2,−1)
=−1.
Step 3: Assemble the Jacobian matrix of fat (1,2,−1):
Jf(1,2,−1) = 2−1 2
1 4 −1.
Question 13
Question
Let f:R3→R3be defined by f(x, y, z) = (x2y, yz, z2). Compute the Jacobian
matrix of f.
10
Solution
To compute the Jacobian matrix of f, we first need to find the partial derivatives
of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
∂f1
∂z =∂
∂z (x2y)=0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (yz)=0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (z2)=0
∂f3
∂y =∂
∂y (z2)=0
∂f3
∂z =∂
∂z (z2) = 2z
Step 4: Assemble the Jacobian matrix. The Jacobian matrix Jfof fis
given by
Jf=
2xy x20
0z y
0 0 2z
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z) = (3x+ 2y, xz). Find the
Jacobian matrix of f.
11
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each variable. The Jacobian matrix of f
is given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #.
Step 1: Calculate the partial derivatives of f
For f(x, y, z) = (3x+ 2y, xz), we have:
∂f1
∂x = 3,
∂f1
∂y = 2,
∂f1
∂z = 0,
∂f2
∂x =z,
∂f2
∂y = 0,
∂f2
∂z =x.
Step 2: Construct the Jacobian matrix Jf
Plugging in the computed partial derivatives, the Jacobian matrix Jfis:
Jf=3 2 0
z0x.
Therefore, the Jacobian matrix of the function f(x, y, z) = (3x+ 2y, xz) is:
Jf=3 2 0
z0x.
Question 15
Question
Let f(x, y, z) =
x2y+y2z
xyz
xz2
. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Step 1: Find the partial derivatives of f.
∂f
∂x =2xy y2z2
12
∂f
∂y =x22xz 0
∂f
∂z =0xy 2xz
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
=449
∂f
∂y (1,2,3)
=160
∂f
∂z (1,2,3)
=026
Step 3: Assemble the Jacobian matrix.
Jf(1,2,3) =
449
160
026
Question 16
Question
Consider the transformation T:R3→R3given by:
T(x, y, z)=(x2+y, y2+z, z2+x)
Determine the Jacobian matrix of T.
13
Solution
Step 1: Compute the partial derivatives of Twith respect to each variable x, y, z:
∂T1
∂x = 2x
∂T1
∂y = 1
∂T1
∂z = 0
∂T2
∂x = 0
∂T2
∂y = 2y
∂T2
∂z = 1
∂T3
∂x = 1
∂T3
∂y = 0
∂T3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix J(T):
J(T) =
∂T1
∂x
∂T1
∂y
∂T1
∂z
∂T2
∂x
∂T2
∂y
∂T2
∂z
∂T3
∂x
∂T3
∂y
∂T3
∂z
=
2x1 0
0 2y1
1 0 2z
Question 17
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, xyz). Find the
Jacobian matrix of f.
Solution
Step 1: To find the Jacobian matrix of f, we first need to compute the partial
derivatives of fwith respect to each variable.
Step 2: Let’s find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z in the Jacobian matrix.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0.
14
Step 3: Next, let’s find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z in the Jacobian matrix.
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy.
Step 4: Finally, we construct the Jacobian matrix of fusing these partial
derivatives.
Jac f=2x1 0
yz xz xy.
Question 18
Question
Let f(x, y, z) = (xy, yz, xz) be a vector field. Find the Jacobian matrix Jfof f
at the point (2,−1,3).
Solution
To find the Jacobian matrix Jfof f, we need to compute the partial derivatives
of each component function with respect to each variable at the given point
(2,−1,3).
Step 1: Compute ∂f1∂x, ∂f1
∂y ,∂f1
∂z
∂f1
∂x =y=−1
∂f1
∂y =x= 2
∂f1
∂z = 0
Step 2: Compute ∂f2∂x, ∂f2
∂y ,∂f2
∂z
∂f2
∂x = 0
∂f2
∂y =z= 3
∂f2
∂z =y=−1
Step 3: Compute ∂f3∂x, ∂f3
∂y ,∂f3
∂z
∂f3
∂x =z= 3
∂f3
∂y = 0
∂f3
∂z =x= 2
15
Step 4: Assemble the Jacobian matrix JfThe Jacobian matrix is given
by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Plugging in the computed partial derivatives at (2,−1,3), we get
Jf(2,−1,3) =
−1 2 0
0 3 −1
302
Question 19
Question
Let f:R3→R2be a function given by f(x, y, z)=(x2y, yz). Find the Jacobian
matrix of f.
Solution
Step 1: Compute the partial derivatives of f. The Jacobian matrix is formed
by arranging the partial derivatives as the rows of a matrix.
Step 2: The partial derivatives of fare:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
and ∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 3: Arrange the partial derivatives in a matrix to form the Jacobian
matrix of f:
Jf(x, y, z) = 2xy x20
0z y
Question 20
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2yz, xy2z, xyz2).
Compute the Jacobian matrix of f.
16
Solution
To compute the Jacobian matrix of f, we will find the partial derivatives of each
component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2yz)=2xyz
∂f1
∂y =∂
∂y (x2yz) = x2z
∂f1
∂z =∂
∂z (x2yz) = x2y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy2z) = y2z
∂f2
∂y =∂
∂y (xy2z)=2xyz
∂f2
∂z =∂
∂z (xy2z) = xy2
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz2) = yz2
∂f3
∂y =∂
∂y (xyz2) = xz2
∂f3
∂z =∂
∂z (xyz2) = 2xyz
Step 4: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis given by
Jf=
2xyz x2z x2y
y2z2xyz xy2
yz2xz22xyz
Question 21
Question
Let f:R3→R3be a mapping given by f(x, y, z) = (2xy, x2+ 3z, y2). Find the
Jacobian matrix of fat the point (1,2,0).
17
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x =∂
∂x (2xy)=2y
∂f2
∂x =∂
∂x (x2+ 3z) = 2x
∂f3
∂x =∂
∂x (y2) = 0
∂f1
∂y =∂
∂y (2xy)=2x
∂f2
∂y =∂
∂y (x2+ 3z) = 0
∂f3
∂y =∂
∂y (y2)=2y
∂f1
∂z =∂
∂z (2xy)=0
∂f2
∂z =∂
∂z (x2+ 3z)=3
∂f3
∂z =∂
∂z (y2)=0
18
Step 2: Evaluate the partial derivatives at the point (1,2,0).
∂f1
∂x (1,2,0)
= 2(2) = 4
∂f2
∂x (1,2,0)
= 2(1) = 2
∂f3
∂x (1,2,0)
= 0
∂f1
∂y (1,2,0)
= 2(1) = 2
∂f2
∂y (1,2,0)
= 0
∂f3
∂y (1,2,0)
= 2(2) = 4
∂f1
∂z (1,2,0)
= 0
∂f2
∂z (1,2,0)
= 3
∂f3
∂z (1,2,0)
= 0
Step 3: Assemble the Jacobian matrix of fat (1,2,0).
Jf(1,2,0) =
420
203
040
Question 22
Question
Consider the function f:R3→R2given by f(x, y, z)=(x2+yz, xyz). Deter-
mine 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) =
19
x2+yz.
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = xyz.
∂f2
∂x =yz
∂f2
∂y =xz
∂f2
∂z =xy
Step 3: Construct the Jacobian matrix of fby arranging the partial deriva-
tives in a matrix.
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis Jf=2x z y
yz xz xy.
Question 23
Question
Let f:R3→R3be a differentiable function defined by f(x, y, z)=(x2+
yz, xy −z2, xz +y2). Find the Jacobian matrix of f.
Solution
Step 1: Write the components of f. The components of fare f1(x, y, z) =
x2+yz,f2(x, y, z) = xy −z2, and f3(x, y, z) = xz +y2.
Step 2: Compute the partial derivatives. To find the Jacobian matrix of f,
we need to compute the partial derivatives of each component of fwith respect
20
to x,y, and z.
∂f1
∂x =∂
∂x (x2+yz)=2x,
∂f1
∂y =∂
∂y (x2+yz) = z,
∂f1
∂z =∂
∂z (x2+yz) = y,
∂f2
∂x =∂
∂x (xy −z2) = y,
∂f2
∂y =∂
∂y (xy −z2) = x,
∂f2
∂z =∂
∂z (xy −z2) = −2z,
∂f3
∂x =∂
∂x (xz +y2) = z,
∂f3
∂y =∂
∂y (xz +y2)=2y,
∂f3
∂z =∂
∂z (xz +y2) = x.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf(x, y, z) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x z y
y x −2z
z2y x
.
Question 24
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+y, xy +z, x +z2).
Find the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z:
∂f1
∂x =∂
∂x (x2+y) = 2x,
∂f2
∂x =∂
∂x (xy +z) = y,
∂f3
∂x =∂
∂x (x+z2)=1.
21
Step 2: Compute the partial derivatives of fwith respect to y,x, and z:
∂f1
∂y =∂
∂y (x2+y)=1,
∂f2
∂y =∂
∂y (xy +z) = x,
∂f3
∂y =∂
∂y (x+z2) = 0.
Step 3: Compute the partial derivatives of fwith respect to z,x, and y:
∂f1
∂z =∂
∂z (x2+y)=0,
∂f2
∂z =∂
∂z (xy +z)=1,
∂f3
∂z =∂
∂z (x+z2)=2z.
Step 4: Evaluate the Jacobian matrix of fat the point (1,2,3):
Jf(1,2,3) =
210
120
106
.
Question 25
Question
Let f:R2→R2be defined by f(x, y) = (x2+y, xy2). Find the Jacobian matrix
of fat the point (1,2).
Solution
Step 1: Find the partial derivatives of f(x, y) with respect to xand y:
∂f
∂x =2x1
y22xy
∂f
∂y =0 1
2xy x2
Step 2: Evaluate the Jacobian matrix of fat the point (1,2) by substituting
x= 1 and y= 2 into the partial derivative matrices:
Jf(1,2) = 2(1) 1
222(1)(2)=2 1
4 4
Therefore, the Jacobian matrix of fat the point (1,2) is 2 1
4 4.
22
Question 26
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+yz, 2x+z). Calculate
the Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the components of f. In this case,
f:R3→R2, so the Jacobian matrix will be a 2 ×3 matrix.
Step 2: To find the Jacobian matrix of f, we need to calculate the partial
derivatives of the components of f. Let’s start by finding the partial derivatives
of f:
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
∂f2
∂x =∂
∂x (2x+z)=2
∂f2
∂y = 0
∂f2
∂z =∂
∂z (2x+z)=1
Step 3: Now we can construct the Jacobian matrix of f:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
2 0 1
Therefore, the Jacobian matrix of fis
Jf=2x z y
2 0 1
Question 27
Question
Let f(x, y, z) =
3x+y+z
x−2y+z
2x+y−3z
. Determine the Jacobian matrix of f.
23
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 input
variable.
Step 1: Find ∂f1
∂x .
∂f1
∂x =∂
∂x (3x+y+z)=3.
Step 2: Find ∂f1
∂y .
∂f1
∂y =∂
∂y (3x+y+z)=1.
Step 3: Find ∂f1
∂z .
∂f1
∂z =∂
∂z (3x+y+z)=1.
Step 4: Find ∂f2
∂x .
∂f2
∂x =∂
∂x (x−2y+z)=1.
Step 5: Find ∂f2
∂y .
∂f2
∂y =∂
∂y (x−2y+z) = −2.
Step 6: Find ∂f2
∂z .
∂f2
∂z =∂
∂z (x−2y+z)=1.
Step 7: Find ∂f3
∂x .
∂f3
∂x =∂
∂x (2x+y−3z)=2.
Step 8: Find ∂f3
∂y .
∂f3
∂y =∂
∂y (2x+y−3z)=1.
Step 9: Find ∂f3
∂z .
∂f3
∂z =∂
∂z (2x+y−3z) = −3.
Therefore, the Jacobian matrix of fis:
Jf(x, y, z) =
3 1 1
1−2 1
2 1 −3
.
24
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+yz, xz). Compute
the Jacobian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of a function, we need to compute the partial
derivatives of each component of the function.
Step 1: Compute partial derivatives Let u=x2+yz and v=xz. The
Jacobian matrix of fis given by
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
We will now compute the partial derivatives of uand vwith respect to x,y,
and z.
Step 2: Compute ∂u ∂x,∂u
∂y ,∂u
∂z
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
Step 3: Compute ∂v ∂x,∂v
∂y ,∂v
∂z
∂v
∂x =z, ∂v
∂y = 0,∂v
∂z =x
Step 4: Evaluate the Jacobian matrix at (1,2,3) Now, we substitute
x= 1, y= 2, and z= 3 into the partial derivatives we computed:
Jf(1,2,3) = 232
301
Therefore, the Jacobian matrix of fat the point (1,2,3) is
Jf(1,2,3) = 232
301
Question 29
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y, xz). Find the
Jacobian matrix of fat the point (1,−1,2).
25
Solution
Step 1: Write the function fin terms of its component functions.
The function fcan be written as f(x, y, z) = (2x+y, xz)=(f1(x, y, z), f2(x, y, z))
where f1(x, y, z)=2x+yand f2(x, y, z) = xz.
Step 2: Compute the partial derivatives of f1and f2.
The partial derivatives of f1are:
∂f1
∂x = 2,∂f1
∂y = 1,and ∂f1
∂z = 0
The partial derivatives of f2are:
∂f2
∂x =z, ∂f2
∂y = 0,and ∂f2
∂z =x
Step 3: Evaluate the Jacobian matrix of fat the point (1,−1,2).
The Jacobian matrix of fat the point (1,−1,2) is given by
Jf(1,−1,2) = "∂f1
∂x (1,−1,2) ∂f1
∂y (1,−1,2) ∂f1
∂z (1,−1,2)
∂f2
∂x (1,−1,2) ∂f2
∂y (1,−1,2) ∂f2
∂z (1,−1,2)#
Now substitute the values of the partial derivatives:
Jf(1,−1,2) = 210
201
Question 30
Question
Let f(x, y, z)=(x2, y2, z2). Compute the Jacobian matrix of fat the point
(1,2,3).
Solution
To compute the Jacobian matrix of a function f:Rn→Rm, we need to find
the matrix of all partial derivatives of fat the given point.
Step 1: Find the partial derivatives of f. Calculate the partial derivatives
of fas follows:
∂f
∂x =
∂x2
∂x
∂y2
∂x
∂z2
∂x
=
2x
0
0
∂f
∂y =
∂x2
∂y
∂y2
∂y
∂z2
∂y
=
0
2y
0
26
Question 2
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+yz, y2+xz). 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 need to compute the
partial derivatives of each component function of fwith respect to each input
variable at the given point.
Let’s start by computing the partial derivatives:
∂
∂x (x2+yz)=2x, ∂
∂y (x2+yz) = z, ∂
∂z (x2+yz) = y,
∂
∂x (y2+xz) = z, ∂
∂y (y2+xz)=2y, ∂
∂z (y2+xz) = x.
Evaluate these partial derivatives at (1,2,3):
∂
∂x (x2+yz)(1,2,3)
= 2(1) = 2,∂
∂y (x2+yz)(1,2,3)
= 3,∂
∂z (x2+yz)(1,2,3)
= 2,
∂
∂x (y2+xz)(1,2,3)
= 3,∂
∂y (y2+xz)(1,2,3)
= 4,∂
∂z (y2+xz)(1,2,3)
= 1.
Therefore, the Jacobian matrix of fat the point (1,2,3) is given by:
232
341.
Question 3
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2z+y, 2xyz). 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.
2
Step 1: Compute the partial derivatives of f. Let f1(x, y, z) = x2z+yand
f2(x, y, z)=2xyz. We calculate the partial derivatives of f1and f2:
∂f1
∂x = 2xz, ∂f1
∂y = 1,∂f1
∂z =x2,
∂f2
∂x = 2yz, ∂f2
∂y = 2xz, ∂f2
∂z = 2xy.
Step 2: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2xz 1x2
2yz 2xz 2xy.
Therefore, the Jacobian matrix of fis 2xz 1x2
2yz 2xz 2xy.
Question 4
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2y, zex, xyz). Find the
Jacobian matrix of f.
Solution
Step 1: Write out the function component-wise.
Given f(x, y, z) = (x2y, zex, xyz), we have
f(x, y, z) =
x2y
zex
xyz
Step 2: Find the partial derivatives of each component of f.
Compute the partial derivatives as follows:
∂f
∂x =
∂
∂x (x2y)
∂
∂x (zex)
∂
∂x (xyz)
=
2xy
zex
yz
∂f
∂y =
∂
∂y (x2y)
∂
∂y (zex)
∂
∂y (xyz)
=
x2
0
xz
∂f
∂z =
∂
∂z (x2y)
∂
∂z (zex)
∂
∂z (xyz)
=
0
ex
xy
3
Step 3: Assemble the partial derivatives into a matrix.
The Jacobian matrix of fis:
Jf=
2xy x20
zex0ex
yz xz xy
Question 5
Question
Let f:R2→R2be a differentiable function defined by f(x, y) = (3x2y, x +y).
Find the Jacobian matrix of fat the point (1,2).
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmat a point ais an m×n
matrix whose entries are the partial derivatives of the components of f. In this
case, f(x, y) = (3x2y, x +y), so the Jacobian matrix Jf(a) is given by
Jf(a) = ∂f1
∂x (a)∂f1
∂y (a)
∂f2
∂x (a)∂f2
∂y (a)!
Step 2: Compute the partial derivatives of f:
∂f1
∂x (x, y)=6xy and ∂f1
∂y (x, y)=3x2
∂f2
∂x (x, y) = 1 and ∂f2
∂y (x, y) = 1
Step 3: Evaluate the Jacobian matrix at the point (1,2):
Jf(1,2) = ∂f1
∂x (1,2) ∂f1
∂y (1,2)
∂f2
∂x (1,2) ∂f2
∂y (1,2)!
=12 3
1 1
Question 6
Question
Let f:R3→R2be a function defined by f(x, y, z) = (4x2+yz, 3y−z).
Compute the Jacobian matrix of f.
4
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each input variable. The Jacobian matrix
of fwill then be a 2 ×3 matrix.
Step 1: Compute the partial derivatives of f:
∂f1
∂x =∂
∂x (4x2+yz) = 8x
∂f1
∂y =∂
∂y (4x2+yz) = z
∂f1
∂z =∂
∂z (4x2+yz) = y
∂f2
∂x =∂
∂x (3y−z)=0
∂f2
∂y =∂
∂y (3y−z)=3
∂f2
∂z =∂
∂z (3y−z) = −1
Step 2: Assemble the Jacobian matrix: The Jacobian matrix of fis given
by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=8x z y
0 3 −1
Question 7
Question
Let f:R2→R2be defined by f(x, y)=(x2y, sin(xy)). Calculate the Jacobian
matrix of fat the point (2, π).
Solution
To calculate the Jacobian matrix of fat the point (2, π), we need to find the
partial derivatives of fwith respect to xand y.
Step 1: Find ∂f1∂x and ∂f1
∂y
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
5
Step 2: Find ∂f2∂x and ∂f2
∂y
∂f2
∂x =∂
∂x (sin(xy)) = ycos(xy)
∂f2
∂y =∂
∂y (sin(xy)) = xcos(xy)
Step 3: Evaluate the Jacobian matrix at (2, π) Substitute x= 2 and
y=πinto the partial derivatives:
∂f1
∂x (2, π) = 2(2)(π)=4π
∂f1
∂y (2, π)=22= 4
∂f2
∂x (2, π) = πcos(2π) = −π
∂f2
∂y (2, π) = 2 cos(2π)=2
Thus, the Jacobian matrix of fat (2, π) is:
Jf(2, π) = 4π4
−π2
Question 8
Question
Let f:R3→R2be a differentiable function given by f(x, y, z)=(x2y, y2z).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix Jfof f, we need to calculate the partial derivatives
of fwith respect to each variable.
Step 1: Calculate ∂f
∂x :
∂f
∂x =∂(x2y)
∂x ,∂(y2z)
∂x = (2xy, 0)
Step 2: Calculate ∂f
∂y :
∂f
∂y =∂(x2y)
∂y ,∂(y2z)
∂y = (x2,2yz)
Step 3: Calculate ∂f
∂z :
∂f
∂z =∂(x2y)
∂z ,∂(y2z)
∂z = (0, y2)
6
Therefore, the Jacobian matrix Jfof fis:
Jf=
2xy 0
x22yz
0y2
Question 9
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2+ 2y, y2+ 3z, z2+ 4x).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to x,y, and z.
Step 1: Partial Derivatives with respect to xWe calculate the partial
derivatives of fwith respect to x:
∂f1
∂x =∂
∂x (x2+ 2y)=2x
∂f2
∂x =∂
∂x (y2+ 3z) = 0
∂f3
∂x =∂
∂x (z2+ 4x) = 4
Step 2: Partial Derivatives with respect to yWe calculate the partial
derivatives of fwith respect to y:
∂f1
∂y =∂
∂y (x2+ 2y)=2
∂f2
∂y =∂
∂y (y2+ 3z) = 2y
∂f3
∂y =∂
∂y (z2+ 4x) = 0
Step 3: Partial Derivatives with respect to zWe calculate the partial
derivatives of fwith respect to z:
∂f1
∂z =∂
∂z (x2+ 2y) = 0
∂f2
∂z =∂
∂z (y2+ 3z)=3
∂f3
∂z =∂
∂z (z2+ 4x) = 2z
7
Step 4: Assemble the Jacobian Matrix The Jacobian matrix of fis
given by the matrix of its partial derivatives:
Jac(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2 0
0 2y3
4 0 2z
Question 10
Question
Let f:R3→R3be defined by f(x, y, z) = (x2+yz, y2+zx, z2+xy). Calculate
the Jacobian matrix of fat the point (1,2,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂f1
∂x
∂f2
∂x
∂f3
∂x
,∂f
∂y =
∂f1
∂y
∂f2
∂y
∂f3
∂y
,∂f
∂z =
∂f1
∂z
∂f2
∂z
∂f3
∂z
Step 2: Compute the partial derivatives.
∂f
∂x =
2x z y
z0x
0y2z
,∂f
∂y =
z2y x
0 2y z
x z 0
,∂f
∂z =
y0 2z
0x y
2x y 0
Step 3: Evaluate the partial derivatives at the point (1,2,2).
∂f
∂x (1,2,2) =
222
201
024
,∂f
∂y (1,2,2) =
241
042
120
,∂f
∂z (1,2,2) =
204
012
220
Therefore, the Jacobian matrix of fat the point (1,2,2) is:
Jf(1,2,2) =
222
201
024
Question 11
Question
Consider the transformation given by the equations:
x=u2−v2
y= 2uv
Find the Jacobian matrix of the transformation at the point (u, v) = (1,1).
8
Solution
We are given the transformation equations:
x=u2−v2
y= 2uv
The Jacobian matrix of a transformation is given by:
J=∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
Step 1: Find ∂x ∂u and ∂x
∂v .
∂x
∂u =∂
∂u (u2−v2)=2u
∂x
∂v =∂
∂v (u2−v2) = −2v
Step 2: Find ∂y ∂u and ∂y
∂v .
∂y
∂u =∂
∂u (2uv)=2v
∂y
∂v =∂
∂v (2uv) = 2u
Step 3: Evaluate the Jacobian matrix at (u, v) = (1,1).Substitute
u= 1 and v= 1 into the partial derivatives:
J=2−2
2 2
Therefore, the Jacobian matrix at the point (u, v) = (1,1) is:
J=2−2
2 2
Question 12
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,−1).
9
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable:
∂f1
∂x =∂
∂x (x2+yz)=2x,
∂f1
∂y =∂
∂y (x2+yz) = z,
∂f1
∂z =∂
∂z (x2+yz) = y,
∂f2
∂x =∂
∂x (y2−xz) = −z,
∂f2
∂y =∂
∂y (y2−xz)=2y,
∂f2
∂z =∂
∂z (y2−xz) = −x.
Step 2: Evaluate the partial derivatives at the point (1,2,−1):
∂f1
∂x (1,2,−1)
= 2(1) = 2,
∂f1
∂y (1,2,−1)
= (−1) = −1,
∂f1
∂z (1,2,−1)
= (2) = 2,
∂f2
∂x (1,2,−1)
= (−(−1)) = 1,
∂f2
∂y (1,2,−1)
= 2(2) = 4,
∂f2
∂z (1,2,−1)
=−1.
Step 3: Assemble the Jacobian matrix of fat (1,2,−1):
Jf(1,2,−1) = 2−1 2
1 4 −1.
Question 13
Question
Let f:R3→R3be defined by f(x, y, z) = (x2y, yz, z2). Compute the Jacobian
matrix of f.
10
Solution
To compute the Jacobian matrix of f, we first need to find the partial derivatives
of each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2y)=2xy
∂f1
∂y =∂
∂y (x2y) = x2
∂f1
∂z =∂
∂z (x2y)=0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (yz)=0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (z2)=0
∂f3
∂y =∂
∂y (z2)=0
∂f3
∂z =∂
∂z (z2) = 2z
Step 4: Assemble the Jacobian matrix. The Jacobian matrix Jfof fis
given by
Jf=
2xy x20
0z y
0 0 2z
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z) = (3x+ 2y, xz). Find the
Jacobian matrix of f.
11
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each variable. The Jacobian matrix of f
is given by
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #.
Step 1: Calculate the partial derivatives of f
For f(x, y, z) = (3x+ 2y, xz), we have:
∂f1
∂x = 3,
∂f1
∂y = 2,
∂f1
∂z = 0,
∂f2
∂x =z,
∂f2
∂y = 0,
∂f2
∂z =x.
Step 2: Construct the Jacobian matrix Jf
Plugging in the computed partial derivatives, the Jacobian matrix Jfis:
Jf=3 2 0
z0x.
Therefore, the Jacobian matrix of the function f(x, y, z) = (3x+ 2y, xz) is:
Jf=3 2 0
z0x.
Question 15
Question
Let f(x, y, z) =
x2y+y2z
xyz
xz2
. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
Step 1: Find the partial derivatives of f.
∂f
∂x =2xy y2z2
12
∂f
∂y =x22xz 0
∂f
∂z =0xy 2xz
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
=449
∂f
∂y (1,2,3)
=160
∂f
∂z (1,2,3)
=026
Step 3: Assemble the Jacobian matrix.
Jf(1,2,3) =
449
160
026
Question 16
Question
Consider the transformation T:R3→R3given by:
T(x, y, z)=(x2+y, y2+z, z2+x)
Determine the Jacobian matrix of T.
13
Solution
Step 1: Compute the partial derivatives of Twith respect to each variable x, y, z:
∂T1
∂x = 2x
∂T1
∂y = 1
∂T1
∂z = 0
∂T2
∂x = 0
∂T2
∂y = 2y
∂T2
∂z = 1
∂T3
∂x = 1
∂T3
∂y = 0
∂T3
∂z = 2z
Step 2: Assemble the partial derivatives into the Jacobian matrix J(T):
J(T) =
∂T1
∂x
∂T1
∂y
∂T1
∂z
∂T2
∂x
∂T2
∂y
∂T2
∂z
∂T3
∂x
∂T3
∂y
∂T3
∂z
=
2x1 0
0 2y1
1 0 2z
Question 17
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, xyz). Find the
Jacobian matrix of f.
Solution
Step 1: To find the Jacobian matrix of f, we first need to compute the partial
derivatives of fwith respect to each variable.
Step 2: Let’s find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z in the Jacobian matrix.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0.
14
Step 3: Next, let’s find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z in the Jacobian matrix.
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy.
Step 4: Finally, we construct the Jacobian matrix of fusing these partial
derivatives.
Jac f=2x1 0
yz xz xy.
Question 18
Question
Let f(x, y, z) = (xy, yz, xz) be a vector field. Find the Jacobian matrix Jfof f
at the point (2,−1,3).
Solution
To find the Jacobian matrix Jfof f, we need to compute the partial derivatives
of each component function with respect to each variable at the given point
(2,−1,3).
Step 1: Compute ∂f1∂x, ∂f1
∂y ,∂f1
∂z
∂f1
∂x =y=−1
∂f1
∂y =x= 2
∂f1
∂z = 0
Step 2: Compute ∂f2∂x, ∂f2
∂y ,∂f2
∂z
∂f2
∂x = 0
∂f2
∂y =z= 3
∂f2
∂z =y=−1
Step 3: Compute ∂f3∂x, ∂f3
∂y ,∂f3
∂z
∂f3
∂x =z= 3
∂f3
∂y = 0
∂f3
∂z =x= 2
15
Step 4: Assemble the Jacobian matrix JfThe Jacobian matrix is given
by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Plugging in the computed partial derivatives at (2,−1,3), we get
Jf(2,−1,3) =
−1 2 0
0 3 −1
302
Question 19
Question
Let f:R3→R2be a function given by f(x, y, z)=(x2y, yz). Find the Jacobian
matrix of f.
Solution
Step 1: Compute the partial derivatives of f. The Jacobian matrix is formed
by arranging the partial derivatives as the rows of a matrix.
Step 2: The partial derivatives of fare:
∂f1
∂x = 2xy, ∂f1
∂y =x2,∂f1
∂z = 0
and ∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 3: Arrange the partial derivatives in a matrix to form the Jacobian
matrix of f:
Jf(x, y, z) = 2xy x20
0z y
Question 20
Question
Let f:R3→R3be a vector-valued function defined by f(x, y, z)=(x2yz, xy2z, xyz2).
Compute the Jacobian matrix of f.
16
Solution
To compute the Jacobian matrix of f, we will find the partial derivatives of each
component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2yz)=2xyz
∂f1
∂y =∂
∂y (x2yz) = x2z
∂f1
∂z =∂
∂z (x2yz) = x2y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xy2z) = y2z
∂f2
∂y =∂
∂y (xy2z)=2xyz
∂f2
∂z =∂
∂z (xy2z) = xy2
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xyz2) = yz2
∂f3
∂y =∂
∂y (xyz2) = xz2
∂f3
∂z =∂
∂z (xyz2) = 2xyz
Step 4: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis given by
Jf=
2xyz x2z x2y
y2z2xyz xy2
yz2xz22xyz
Question 21
Question
Let f:R3→R3be a mapping given by f(x, y, z) = (2xy, x2+ 3z, y2). Find the
Jacobian matrix of fat the point (1,2,0).
17
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x =∂
∂x (2xy)=2y
∂f2
∂x =∂
∂x (x2+ 3z) = 2x
∂f3
∂x =∂
∂x (y2) = 0
∂f1
∂y =∂
∂y (2xy)=2x
∂f2
∂y =∂
∂y (x2+ 3z) = 0
∂f3
∂y =∂
∂y (y2)=2y
∂f1
∂z =∂
∂z (2xy)=0
∂f2
∂z =∂
∂z (x2+ 3z)=3
∂f3
∂z =∂
∂z (y2)=0
18
Step 2: Evaluate the partial derivatives at the point (1,2,0).
∂f1
∂x (1,2,0)
= 2(2) = 4
∂f2
∂x (1,2,0)
= 2(1) = 2
∂f3
∂x (1,2,0)
= 0
∂f1
∂y (1,2,0)
= 2(1) = 2
∂f2
∂y (1,2,0)
= 0
∂f3
∂y (1,2,0)
= 2(2) = 4
∂f1
∂z (1,2,0)
= 0
∂f2
∂z (1,2,0)
= 3
∂f3
∂z (1,2,0)
= 0
Step 3: Assemble the Jacobian matrix of fat (1,2,0).
Jf(1,2,0) =
420
203
040
Question 22
Question
Consider the function f:R3→R2given by f(x, y, z)=(x2+yz, xyz). Deter-
mine 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) =
19
x2+yz.
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = xyz.
∂f2
∂x =yz
∂f2
∂y =xz
∂f2
∂z =xy
Step 3: Construct the Jacobian matrix of fby arranging the partial deriva-
tives in a matrix.
Jf=2x z y
yz xz xy
Therefore, the Jacobian matrix of fis Jf=2x z y
yz xz xy.
Question 23
Question
Let f:R3→R3be a differentiable function defined by f(x, y, z)=(x2+
yz, xy −z2, xz +y2). Find the Jacobian matrix of f.
Solution
Step 1: Write the components of f. The components of fare f1(x, y, z) =
x2+yz,f2(x, y, z) = xy −z2, and f3(x, y, z) = xz +y2.
Step 2: Compute the partial derivatives. To find the Jacobian matrix of f,
we need to compute the partial derivatives of each component of fwith respect
20
to x,y, and z.
∂f1
∂x =∂
∂x (x2+yz)=2x,
∂f1
∂y =∂
∂y (x2+yz) = z,
∂f1
∂z =∂
∂z (x2+yz) = y,
∂f2
∂x =∂
∂x (xy −z2) = y,
∂f2
∂y =∂
∂y (xy −z2) = x,
∂f2
∂z =∂
∂z (xy −z2) = −2z,
∂f3
∂x =∂
∂x (xz +y2) = z,
∂f3
∂y =∂
∂y (xz +y2)=2y,
∂f3
∂z =∂
∂z (xz +y2) = x.
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis given
by
Jf(x, y, z) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x z y
y x −2z
z2y x
.
Question 24
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+y, xy +z, x +z2).
Find the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z:
∂f1
∂x =∂
∂x (x2+y) = 2x,
∂f2
∂x =∂
∂x (xy +z) = y,
∂f3
∂x =∂
∂x (x+z2)=1.
21
Step 2: Compute the partial derivatives of fwith respect to y,x, and z:
∂f1
∂y =∂
∂y (x2+y)=1,
∂f2
∂y =∂
∂y (xy +z) = x,
∂f3
∂y =∂
∂y (x+z2) = 0.
Step 3: Compute the partial derivatives of fwith respect to z,x, and y:
∂f1
∂z =∂
∂z (x2+y)=0,
∂f2
∂z =∂
∂z (xy +z)=1,
∂f3
∂z =∂
∂z (x+z2)=2z.
Step 4: Evaluate the Jacobian matrix of fat the point (1,2,3):
Jf(1,2,3) =
210
120
106
.
Question 25
Question
Let f:R2→R2be defined by f(x, y) = (x2+y, xy2). Find the Jacobian matrix
of fat the point (1,2).
Solution
Step 1: Find the partial derivatives of f(x, y) with respect to xand y:
∂f
∂x =2x1
y22xy
∂f
∂y =0 1
2xy x2
Step 2: Evaluate the Jacobian matrix of fat the point (1,2) by substituting
x= 1 and y= 2 into the partial derivative matrices:
Jf(1,2) = 2(1) 1
222(1)(2)=2 1
4 4
Therefore, the Jacobian matrix of fat the point (1,2) is 2 1
4 4.
22
Question 26
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+yz, 2x+z). Calculate
the Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix
whose entries are the partial derivatives of the components of f. In this case,
f:R3→R2, so the Jacobian matrix will be a 2 ×3 matrix.
Step 2: To find the Jacobian matrix of f, we need to calculate the partial
derivatives of the components of f. Let’s start by finding the partial derivatives
of f:
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
∂f2
∂x =∂
∂x (2x+z)=2
∂f2
∂y = 0
∂f2
∂z =∂
∂z (2x+z)=1
Step 3: Now we can construct the Jacobian matrix of f:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
2 0 1
Therefore, the Jacobian matrix of fis
Jf=2x z y
2 0 1
Question 27
Question
Let f(x, y, z) =
3x+y+z
x−2y+z
2x+y−3z
. Determine the Jacobian matrix of f.
23
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 input
variable.
Step 1: Find ∂f1
∂x .
∂f1
∂x =∂
∂x (3x+y+z)=3.
Step 2: Find ∂f1
∂y .
∂f1
∂y =∂
∂y (3x+y+z)=1.
Step 3: Find ∂f1
∂z .
∂f1
∂z =∂
∂z (3x+y+z)=1.
Step 4: Find ∂f2
∂x .
∂f2
∂x =∂
∂x (x−2y+z)=1.
Step 5: Find ∂f2
∂y .
∂f2
∂y =∂
∂y (x−2y+z) = −2.
Step 6: Find ∂f2
∂z .
∂f2
∂z =∂
∂z (x−2y+z)=1.
Step 7: Find ∂f3
∂x .
∂f3
∂x =∂
∂x (2x+y−3z)=2.
Step 8: Find ∂f3
∂y .
∂f3
∂y =∂
∂y (2x+y−3z)=1.
Step 9: Find ∂f3
∂z .
∂f3
∂z =∂
∂z (2x+y−3z) = −3.
Therefore, the Jacobian matrix of fis:
Jf(x, y, z) =
3 1 1
1−2 1
2 1 −3
.
24
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+yz, xz). Compute
the Jacobian matrix of fat the point (1,2,3).
Solution
To find the Jacobian matrix of a function, we need to compute the partial
derivatives of each component of the function.
Step 1: Compute partial derivatives Let u=x2+yz and v=xz. The
Jacobian matrix of fis given by
Jf="∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #
We will now compute the partial derivatives of uand vwith respect to x,y,
and z.
Step 2: Compute ∂u ∂x,∂u
∂y ,∂u
∂z
∂u
∂x = 2x, ∂u
∂y =z, ∂u
∂z =y
Step 3: Compute ∂v ∂x,∂v
∂y ,∂v
∂z
∂v
∂x =z, ∂v
∂y = 0,∂v
∂z =x
Step 4: Evaluate the Jacobian matrix at (1,2,3) Now, we substitute
x= 1, y= 2, and z= 3 into the partial derivatives we computed:
Jf(1,2,3) = 232
301
Therefore, the Jacobian matrix of fat the point (1,2,3) is
Jf(1,2,3) = 232
301
Question 29
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y, xz). Find the
Jacobian matrix of fat the point (1,−1,2).
25
Solution
Step 1: Write the function fin terms of its component functions.
The function fcan be written as f(x, y, z) = (2x+y, xz)=(f1(x, y, z), f2(x, y, z))
where f1(x, y, z)=2x+yand f2(x, y, z) = xz.
Step 2: Compute the partial derivatives of f1and f2.
The partial derivatives of f1are:
∂f1
∂x = 2,∂f1
∂y = 1,and ∂f1
∂z = 0
The partial derivatives of f2are:
∂f2
∂x =z, ∂f2
∂y = 0,and ∂f2
∂z =x
Step 3: Evaluate the Jacobian matrix of fat the point (1,−1,2).
The Jacobian matrix of fat the point (1,−1,2) is given by
Jf(1,−1,2) = "∂f1
∂x (1,−1,2) ∂f1
∂y (1,−1,2) ∂f1
∂z (1,−1,2)
∂f2
∂x (1,−1,2) ∂f2
∂y (1,−1,2) ∂f2
∂z (1,−1,2)#
Now substitute the values of the partial derivatives:
Jf(1,−1,2) = 210
201
Question 30
Question
Let f(x, y, z)=(x2, y2, z2). Compute the Jacobian matrix of fat the point
(1,2,3).
Solution
To compute the Jacobian matrix of a function f:Rn→Rm, we need to find
the matrix of all partial derivatives of fat the given point.
Step 1: Find the partial derivatives of f. Calculate the partial derivatives
of fas follows:
∂f
∂x =
∂x2
∂x
∂y2
∂x
∂z2
∂x
=
2x
0
0
∂f
∂y =
∂x2
∂y
∂y2
∂y
∂z2
∂y
=
0
2y
0
26
∂f
∂z =
∂x2
∂z
∂y2
∂z
∂z2
∂z
=
0
0
2z
Step 2: Evaluate the Jacobian matrix at the point (1,2,3). Substitute the
values x= 1, y= 2, and z= 3 into the partial derivatives calculated above to
find the Jacobian matrix at the point (1,2,3).
Jf(1,2,3) =
2(1) 0 0
0 2(2) 0
0 0 2(3)
=
200
040
006
Therefore, the Jacobian matrix of fat the point (1,2,3) is
200
040
006
.
27
