MATH 117 - ELEMENTS OF
MATHEMATICS - Jacobian matrix
Question Bank - Set 6
Liberty University
Question 1
Question
Let f(x, y) = 2x2y
x3+ 3y. Find the Jacobian matrix of fat the point (1,2).
Solution
To find the Jacobian matrix of fat the point (1,2), we first need to find the
partial derivatives and then evaluate them at the given point.
Step 1: Find the partial derivatives Let f(x, y) = u(x, y)
v(x, y). The
partial derivatives of uand vare: ∂u
∂x = 4xy,∂u
∂y = 2x2,∂v
∂x = 3x2, and ∂v
∂y = 3.
Step 2: Evaluate the partial derivatives at the point (1,2) At (1,2),
we have: ∂u
∂x = 4 ×1×2 = 8, ∂u
∂y = 2 ×12= 2, ∂v
∂x = 3 ×12= 3, and ∂v
∂y = 3.
Step 3: Construct the Jacobian matrix The Jacobian matrix of fat
(1,2) is:
Jf(1,2) = 8 2
3 3
Question 2
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Find the
Jacobian matrix of f.
Solution
Step 1: Calculate 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 Jacobian matrix. The Jacobian matrix of fis given
by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2xy x20
0z y
Therefore, the Jacobian matrix of fis
Jf=2xy x20
0z y
.
Question 3
Question
Let f(x, y) = x2+ 2y
3x−y. Find the Jacobian matrix of fat the point (1,−1).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (x2+ 2y)
∂
∂x (3x−y)=2x
3
∂f
∂y = ∂
∂y (x2+ 2y)
∂
∂y (3x−y)!=2
−1
2
Step 2: Evaluate the partial derivatives at the point (1,−1).
∂f
∂x (1,−1) = 2(1)
3=2
3
∂f
∂y (1,−1) = 2
−1
Step 3: Assemble the Jacobian matrix at the given point.
Jf(1,−1) = ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=2 2
3−1
Therefore, the Jacobian matrix of fat the point (1,−1) is 2 2
3−1.
Question 4
Question
Let f:R2→R2be a function defined by f(x, y)=(x2y, xy2). Find the
Jacobian matrix of fat the point (2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f1
∂x = 2xy and ∂f2
∂x =y2
∂f1
∂y =x2and ∂f2
∂y = 2xy
Step 2: Evaluate the partial derivatives at the point (2,3).
∂f1
∂x (2,3) = 2(2)(3) = 12,∂f2
∂x (2,3) = (3)2= 9
∂f1
∂y (2,3) = (2)2= 4,∂f2
∂y (2,3) = 2(2)(3) = 12
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat point
(2,3) is
Jf(2,3) = "∂f1
∂x (2,3) ∂f1
∂y (2,3)
∂f2
∂x (2,3) ∂f2
∂y (2,3)#=12 4
9 12
Question 5
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y2, yz). Find the
Jacobian matrix of f.
3
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of f
with respect to each variable and arrange them in a matrix.
Step 1: Calculate the partial derivatives of f.
∂f1
∂x = 2x
∂f1
∂y = 2y
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 2: Assemble the partial derivatives into a matrix. The Jacobian matrix
of fis:
Jf=2x2y0
0z y
Question 6
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2y, yz, xy2). 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 each variable.
Step 1: Compute the partial derivatives of f:
∂
∂x f1(x, y, z) = ∂
∂x (x2y) = 2xy
∂
∂y f1(x, y, z) = ∂
∂y (x2y) = x2
∂
∂z f1(x, y, z) = 0
4
Step 2: Repeat the process for the other component functions of f:
∂
∂x f2(x, y, z) = ∂
∂x (yz) = 0
∂
∂y f2(x, y, z) = ∂
∂y (yz) = z
∂
∂z f2(x, y, z) = y
∂
∂x f3(x, y, z) = ∂
∂x (xy2) = y2
∂
∂y f3(x, y, z) = ∂
∂y (xy2) = 2xy
∂
∂z f3(x, y, z) = 0
Step 3: Assemble the partial derivatives into the Jacobian matrix:
Jf(x, y, z) =
2xy x20
0z y
y22xy 0
Therefore, the Jacobian matrix of fis given by Jf(x, y, z) =
2xy x20
0z y
y22xy 0
.
Question 7
Question
Let f(x, y) = (2x3y, x2y2). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f1
∂x and ∂f1
∂y for the first component function f1(x, y) =
2x3y.
∂f1
∂x = 6x2yand ∂f1
∂y = 2x3
Step 2: Compute ∂f2
∂x and ∂f2
∂y for the second component function f2(x, y) =
x2y2.
∂f2
∂x = 2xy2and ∂f2
∂y = 2x2y
5
Step 3: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis:
6x2y2x3
2xy22x2y
Question 8
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2). Find the Jacobian matrix of f.
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x0 0
∂f
∂y =0 2y0
∂f
∂z =002z
Step 2: Arrange the partial derivatives as columns of the Jacobian matrix.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Jf=
2x0 0
0 2y0
0 0 2z
Question 9
Question
Let f(x, y, z)=(x2+y2, xy +z, x +y+z). Find the Jacobian matrix of fat
the point (1,−1,2).
Solution
To find the Jacobian matrix of fat the point (1,−1,2), we need to compute
the partial derivatives of each component function with respect to each variable
at this point.
6
Step 1: Find ∂f1∂x,∂f1
∂y ,∂f1
∂z .
∂f1
∂x = 2x∂f1
∂y = 2y∂f1
∂z = 0
Step 2: Find ∂f2∂x,∂f2
∂y ,∂f2
∂z .
∂f2
∂x =y∂f2
∂y =x∂f2
∂z = 1
Step 3: Find ∂f3∂x,∂f3
∂y ,∂f3
∂z .
∂f3
∂x = 1 ∂f3
∂y = 1 ∂f3
∂z = 1
Step 4: Evaluate the Jacobian matrix at (1,−1,2).The Jacobian
matrix Jis given by
J=
2−2 0
−111
1 1 1
Thus, the Jacobian matrix of fat the point (1,−1,2) is:
2−2 0
−111
1 1 1
Question 10
Question
Let f:R3→R2be a mapping given by f(x, y, z)=(x2+yz, z2−xy). Calculate
the Jacobian matrix of f.
Solution
To calculate the Jacobian matrix of f, we need to find the partial derivatives of
the components of fwith respect to each variable.
Step 1: Partial derivatives of f
7
Let’s find the partial derivatives of each component function of f:
∂
∂x (x2+yz)=2x,
∂
∂y (x2+yz) = z,
∂
∂z (x2+yz) = y,
∂
∂x (z2−xy) = −y,
∂
∂y (z2−xy) = −x,
∂
∂z (z2−xy)=2z.
Step 2: Construct the Jacobian matrix
The Jacobian matrix of fis given by:
Jf(x, y, z) = "∂
∂x (x2+yz)∂
∂y (x2+yz)∂
∂z (x2+yz)
∂
∂x (z2−xy)∂
∂y (z2−xy)∂
∂z (z2−xy)#.
Plugging in the partial derivatives, we get:
Jf(x, y, z) = 2x z y
−y−x2z.
Therefore, the Jacobian matrix of the mapping fis:
Jf(x, y, z) = 2x z y
−y−x2z.
Question 11
Question
Find the Jacobian matrix of the following system of equations:
(x2+y= 3
2x−y2= 1
Solution
Step 1: To find the Jacobian matrix, we first need to compute the partial
derivatives of the given system of equations: Let f(x, y) = x2+y−3 and
g(x, y)=2x−y2−1, then we have fx= 2x,fy= 1, gx= 2, and gy=−2y.
Step 2: The Jacobian matrix of the system is given by:
J=fxfy
gxgy=2x1
2−2y
8
Step 3: Now, substitute x= 1 and y= 2 into the Jacobian matrix:
J(1,2) = 2(1) 1
2−2(2)=2 1
2−4
Therefore, the Jacobian matrix of the given system of equations at the point
(1,2) is 2 1
2−4.
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, ex+yz). Compute
the Jacobian matrix of fat the point (1,0,1).
Solution
To compute the Jacobian matrix of fat the point (1,0,1), we need to find the
partial derivatives of fwith respect to each variable at this point.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y),∂
∂x (ex+yz)
∂f
∂x = (2xy, ex)
Step 2: Find ∂f ∂y
∂f
∂y =∂
∂y (x2y),∂
∂y (ex+yz)
∂f
∂y = (x2, z)
Step 3: Find ∂f ∂z
∂f
∂z =∂
∂z (x2y),∂
∂z (ex+yz)
∂f
∂z = (0, y)
Step 4: Evaluate the Jacobian matrix at (1,0,1) Now, we evaluate the
Jacobian matrix at the point (1,0,1) by plugging in the values.
Jf(1,0,1) =
2(1)(0) e1
121
0 0
Jf(1,0,1) =
0e
1 1
0 0
9
Question 13
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+y2, xyz). Determine 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.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y2, xyz), we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Again, we have:
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Therefore, the Jacobian matrix of fis:
Jf(x, y, z) = 2x2y0
yz xz xy
Question 14
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2yz, xy2z, xyz2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: To find the Jacobian matrix of f, we first need to compute the partial
derivatives of fwith respect to x,y, and z.
Step 2: The partial derivatives are given by:
∂f
∂x =
2xyz
y2z
yz2
,∂f
∂y =
x2z
2xyz
xz2
,and ∂f
∂z =
x2y
xy2
2xyz
.
Step 3: Evaluate the partial derivatives at the point (1,2,3):
∂f
∂x (1,2,3) =
12
6
6
,∂f
∂y (1,2,3) =
6
12
3
,and ∂f
∂z (1,2,3) =
2
4
6
.
10
Step 4: The Jacobian matrix of fat (1,2,3) is the matrix formed by arrang-
ing the three partial derivatives evaluated at (1,2,3) as rows:
Jf(1,2,3) =
12 6 6
6 12 3
2 4 6
.
Question 15
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, y2+z). Compute the
Jacobian matrix of f.
Solution
Step 1: Write out the components of the function f.
f(x, y, z)=(x2+y, y2+z)
Step 2: Determine the partial derivatives of each component of fwith respect
to x,y, and z. For the first component:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
For the second component:
∂f2
∂x = 0
∂f2
∂y = 2y
∂f2
∂z = 1
Step 3: Assemble the Jacobian matrix using the partial derivatives calculated
above.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2x1 0
0 2y1
11
Question 16
Question
Let f:R2→R2be a function defined by f(x, y) = (xy, x2). Determine the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each input variable.
Step 1: Compute ∂f1
∂x ,∂f1
∂y ,∂f2
∂x , and ∂f2
∂y .
Given f(x, y)=(xy, x2), we have:
∂f1
∂x =∂(xy)
∂x =y
∂f1
∂y =∂(xy)
∂y =x
∂f2
∂x =∂(x2)
∂x = 2x
∂f2
∂y = 0 (since f2does not involve y)
Step 2: Assemble the partial derivatives into a 2 ×2 Jacobian matrix.
The Jacobian matrix Jfof fis given by:
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Plugging in the partial derivatives we computed:
Jf=y x
2x0
Therefore, the Jacobian matrix of the function fis y x
2x0.
Question 17
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2y, sin(xy)).
Compute the Jacobian matrix of f.
12
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet F(x, y, z)=(u, v) =
(x2y, sin(xy)). The partial derivatives of Fare:
∂u
∂x ,∂u
∂y ,∂u
∂z ,∂v
∂x ,∂v
∂y ,∂v
∂z
Compute each partial derivative:
∂u
∂x = 2xy
∂u
∂y =x2
∂u
∂z = 0
∂v
∂x =ycos(xy)
∂v
∂y =xcos(xy)
∂v
∂z = 0
Step 2: Create the Jacobian matrix The Jacobian matrix of fis given
by:
J= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
Substitute the computed partial derivatives into the matrix:
J=2xy x20
ycos(xy)xcos(xy) 0
Therefore, the Jacobian matrix of fis:
J=2xy x20
ycos(xy)xcos(xy) 0
Question 18
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y, xz). Calculate
the Jacobian matrix of f.
13
Solution
To find the Jacobian matrix of a function f:Rn→Rm, we need to compute
the matrix of partial derivatives of f.
Step 1: Compute the partial derivatives of f.The components of
fare f1(x, y, z) = 2x+yand f2(x, y, z) = xz. We now compute the partial
derivatives of f1and f2:
∂f1
∂x = 2,∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 2: Assemble the Jacobian matrix. The Jacobian matrix of fis a
2×3 matrix where the i-th row and j-th column entry is the partial derivative
of fiwith respect to the j-th variable. Therefore, the Jacobian matrix of fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2 1 0
z0x
Therefore, the Jacobian matrix of the function f(x, y, z) = (2x+y, xz) is
2 1 0
z0x.
Question 19
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+yz, x2+ 2y+z).
Find the Jacobian matrix of f.
Solution
Step 1: Write down the components of the function f:
f(x, y, z) = (2x+yz, x2+ 2y+z)
Step 2: Calculate the partial derivatives of fwith respect to x,y, and z:
∂f
∂x = ∂(2x+yz)
∂x
∂(x2+2y+z)
∂x !=2
2x
∂f
∂y = ∂(2x+yz)
∂y
∂(x2+2y+z)
∂y !=z
2
∂f
∂z = ∂(2x+yz)
∂z
∂(x2+2y+z)
∂z !=y
1
14
Step 3: Assemble the partial derivatives into a matrix to form the Jacobian
matrix of f:
Jf=2z y
2x2 1
Question 20
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Determine the
Jacobian matrix of f.
Solution
To determine the Jacobian matrix of f, we first need to calculate the partial
derivatives of each component function with respect to each input variable.
Step 1: Calculate ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function
f1(x, y, z) = x2y.
∂f1
∂x = 2xy , ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Calculate ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0 ,∂f2
∂y =z , ∂f2
∂z =y
Step 3: Assemble the partial derivatives into a 2 ×3 Jacobian matrix Jf
for f:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of the function f(x, y, z)=(x2y, yz) is
Jf=2xy x20
0z y
.
Question 21
Question
Let f(x, y, z) = (x2+y2, y2+z2, z2+x2) be a vector-valued function. Calculate
the Jacobian matrix of fat the point (1,2,3).
15
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂f1
∂x
∂f2
∂x
∂f3
∂x
=
2x
0
2z
∂f
∂y =
∂f1
∂y
∂f2
∂y
∂f3
∂y
=
2y
2y
0
∂f
∂z =
∂f1
∂z
∂f2
∂z
∂f3
∂z
=
0
2z
2x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) =
2
0
6
∂f
∂y (1,2,3) =
4
4
0
∂f
∂z (1,2,3) =
0
6
2
Step 3: Assemble the Jacobian matrix using the partial derivatives evaluated
at (1,2,3).
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
240
046
602
Question 22
Question
Find the Jacobian matrix of the transformation (x, y)=(rcos θ, r sin θ).
Solution
To find the Jacobian matrix of the transformation, we need to first express x
and yin terms of rand θ, and then find the partial derivatives with respect to
rand θ.
16
Step 1: Express xand yin terms of rand θ:
x=rcos θ
y=rsin θ
Step 2: Find the partial derivatives:
∂x
∂r = cos θ
∂x
∂θ =−rsin θ
∂y
∂r = sin θ
∂y
∂θ =rcos θ
Step 3: Form the Jacobian matrix: The Jacobian matrix is given by
∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ
Substitute the partial derivatives:
cos θ−rsin θ
sin θ r cos θ
Therefore, the Jacobian matrix for the transformation (x, y) = (rcos θ, r sin θ)
is cos θ−rsin θ
sin θ r cos θ
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. 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.
∂f
∂x = 2xy +z2,
∂f
∂y =x2+ 2yz,
∂f
∂z =y2+ 2zx.
17
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= 2(1)(2) + 32= 4 + 9 = 13,
∂f
∂y (1,2,3)
= 12+ 2(2)(3) = 1 + 12 = 13,
∂f
∂z (1,2,3)
= 2(2)(1) + 32= 4 + 9 = 13.
Step 3: Assemble the Jacobian matrix using the computed partial deriva-
tives.
Jf(1,2,3) = 13 13 13.
Therefore, the Jacobian matrix of fat the point (1,2,3) is 13 13 13.
Question 24
Question
Let f:R2→R2be a function defined by f(x, y)=(excos(y), exsin(y)). Find
the Jacobian matrix of fat the point (0,0).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y.
∂f1
∂x =∂
∂x (excos(y)) = excos(y)
∂f1
∂y =∂
∂y (excos(y)) = −exsin(y)
∂f2
∂x =∂
∂x (exsin(y)) = exsin(y)
∂f2
∂y =∂
∂y (exsin(y)) = excos(y)
Step 2: Evaluate the partial derivatives at the point (0,0).
∂f1
∂x (0,0)
= 1
∂f1
∂y (0,0)
= 0
∂f2
∂x (0,0)
= 0
∂f2
∂y (0,0)
= 1
18
Step 3: Assemble the Jacobian matrix of fat (0,0).
Jf(0,0) = 1 0
0 1
Question 25
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+yz, y2+zx, z2+xy).
Find the Jacobian matrix of f.
Solution
Step 1: Write out the components of the function f.
f(x, y, z) = x2+yz, y2+zx, z2+xy
Step 2: Compute 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
∂f3
∂x =y, ∂f3
∂y =x, ∂f3
∂z = 2z
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in rows.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x z y
z2y x
y x 2z
Question 26
Question
Let f:R3→Rbe a vector-valued function defined by f(x, y, z) = (x2yz, ex+y, xyz2).
Determine the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
19
Step 1: Compute the partial derivatives of f:
∂f1
∂x =∂(x2yz)
∂x = 2xyz,
∂f1
∂y =∂(x2yz)
∂y =x2z,
∂f1
∂z =∂(x2yz)
∂z =x2y,
∂f2
∂x =∂(ex+y)
∂x =ex+y,
∂f2
∂y =∂(ex+y)
∂y =ex+y,
∂f2
∂z = 0,
∂f3
∂x =∂(xyz2)
∂x =yz2,
∂f3
∂y =∂(xyz2)
∂y =xz2,
∂f3
∂z =∂(xyz2)
∂z = 2xyz.
Step 2: Assemble the partial derivatives into a matrix:
J(f) =
2xyz x2z x2y
ex+yex+y0
yz2xz22xyz
.
Therefore, the Jacobian matrix of fis
J(f) =
2xyz x2z x2y
ex+yex+y0
yz2xz22xyz
.
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z) = x2+y2, yz. Find the
Jacobian matrix of f.
Solution
The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix of partial
derivatives. For the function f(x, y, z) = x2+y2, yz, we need to determine
the partial derivatives of each component of fwith respect to x,y, and z.
20
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component f1=x2+y2.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component f2=yz.
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 3: Write the Jacobian matrix Jfof fusing the computed partial
derivatives.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z) = (x2, yz). Compute the
Jacobian matrix of fat the point (2,−1,3).
Solution
Step 1: The Jacobian matrix of a mapping f:Rn→Rmis an m×nmatrix
where each entry is a partial derivative of the corresponding component function.
Step 2: Consider the component functions of f:
f1(x, y, z) = x2
f2(x, y, z) = yz
Step 3: Compute the partial derivatives of f1and f2with respect to x, y,
and z.∂f1
∂x = 2x
∂f1
∂y = 0
∂f1
∂z = 0
Step 4: Now, compute the partial derivatives of f2:
∂f2
∂x = 0
∂f2
∂y =z
21
∂f2
∂z =y
Step 5: The Jacobian matrix Jfof fis then:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y
Step 6: Evaluate the Jacobian matrix at the point (2,−1,3):
Jf(2,−1,3) = 2(2) 0 0
0 3 −1=4 0 0
0 3 −1
Therefore, the Jacobian matrix of fat the point (2,−1,3) is 4 0 0
0 3 −1.
Question 29
Question
Let f(x, y, z) = (2x+y2,3y−z, xz) be a vector-valued function. Calculate the
Jacobian matrix Jfof f.
Solution
To find the Jacobian matrix Jfof f, we need to compute the partial derivatives
of each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (2x+y2)=2
∂f1
∂y =∂
∂y (2x+y2)=2y
∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (3y−z)=0
∂f2
∂y =∂
∂y (3y−z)=3
∂f2
∂z =−1
22
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xz) = z
∂f3
∂y =∂
∂y (xz)=0
∂f3
∂z =x
Step 4: Assemble the partial derivatives into the Jacobian matrix Jf:
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2 2y0
0 3 −1
z0x
Question 30
Question
Let f(x, y)=(x2y+ 2xy3, x2+ 3y2) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,1).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f1
∂x =∂
∂x (x2y+ 2xy3)=2xy + 2y3
∂f1
∂y =∂
∂y (x2y+ 2xy3) = x2+ 6xy2
∂f2
∂x =∂
∂x (x2+ 3y2)=2x
∂f2
∂y =∂
∂y (x2+ 3y2)=6y
Step 2: Evaluate the partial derivatives at the point (1,1).
∂f1
∂x (1,1) = 2(1)(1) + 2(1)3= 4
∂f1
∂y (1,1) = (1)2+ 6(1)(1)2= 7
∂f2
∂x (1,1) = 2(1) = 2
∂f2
∂y (1,1) = 6(1) = 6
23
Solution
Step 1: Calculate 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 Jacobian matrix. The Jacobian matrix of fis given
by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2xy x20
0z y
Therefore, the Jacobian matrix of fis
Jf=2xy x20
0z y
.
Question 3
Question
Let f(x, y) = x2+ 2y
3x−y. Find the Jacobian matrix of fat the point (1,−1).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (x2+ 2y)
∂
∂x (3x−y)=2x
3
∂f
∂y = ∂
∂y (x2+ 2y)
∂
∂y (3x−y)!=2
−1
2
Step 2: Evaluate the partial derivatives at the point (1,−1).
∂f
∂x (1,−1) = 2(1)
3=2
3
∂f
∂y (1,−1) = 2
−1
Step 3: Assemble the Jacobian matrix at the given point.
Jf(1,−1) = ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=2 2
3−1
Therefore, the Jacobian matrix of fat the point (1,−1) is 2 2
3−1.
Question 4
Question
Let f:R2→R2be a function defined by f(x, y)=(x2y, xy2). Find the
Jacobian matrix of fat the point (2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f1
∂x = 2xy and ∂f2
∂x =y2
∂f1
∂y =x2and ∂f2
∂y = 2xy
Step 2: Evaluate the partial derivatives at the point (2,3).
∂f1
∂x (2,3) = 2(2)(3) = 12,∂f2
∂x (2,3) = (3)2= 9
∂f1
∂y (2,3) = (2)2= 4,∂f2
∂y (2,3) = 2(2)(3) = 12
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat point
(2,3) is
Jf(2,3) = "∂f1
∂x (2,3) ∂f1
∂y (2,3)
∂f2
∂x (2,3) ∂f2
∂y (2,3)#=12 4
9 12
Question 5
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y2, yz). Find the
Jacobian matrix of f.
3
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of f
with respect to each variable and arrange them in a matrix.
Step 1: Calculate the partial derivatives of f.
∂f1
∂x = 2x
∂f1
∂y = 2y
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 2: Assemble the partial derivatives into a matrix. The Jacobian matrix
of fis:
Jf=2x2y0
0z y
Question 6
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2y, yz, xy2). 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 each variable.
Step 1: Compute the partial derivatives of f:
∂
∂x f1(x, y, z) = ∂
∂x (x2y) = 2xy
∂
∂y f1(x, y, z) = ∂
∂y (x2y) = x2
∂
∂z f1(x, y, z) = 0
4
Step 2: Repeat the process for the other component functions of f:
∂
∂x f2(x, y, z) = ∂
∂x (yz) = 0
∂
∂y f2(x, y, z) = ∂
∂y (yz) = z
∂
∂z f2(x, y, z) = y
∂
∂x f3(x, y, z) = ∂
∂x (xy2) = y2
∂
∂y f3(x, y, z) = ∂
∂y (xy2) = 2xy
∂
∂z f3(x, y, z) = 0
Step 3: Assemble the partial derivatives into the Jacobian matrix:
Jf(x, y, z) =
2xy x20
0z y
y22xy 0
Therefore, the Jacobian matrix of fis given by Jf(x, y, z) =
2xy x20
0z y
y22xy 0
.
Question 7
Question
Let f(x, y) = (2x3y, x2y2). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f1
∂x and ∂f1
∂y for the first component function f1(x, y) =
2x3y.
∂f1
∂x = 6x2yand ∂f1
∂y = 2x3
Step 2: Compute ∂f2
∂x and ∂f2
∂y for the second component function f2(x, y) =
x2y2.
∂f2
∂x = 2xy2and ∂f2
∂y = 2x2y
5
Step 3: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis:
6x2y2x3
2xy22x2y
Question 8
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2). Find the Jacobian matrix of f.
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x0 0
∂f
∂y =0 2y0
∂f
∂z =002z
Step 2: Arrange the partial derivatives as columns of the Jacobian matrix.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Jf=
2x0 0
0 2y0
0 0 2z
Question 9
Question
Let f(x, y, z)=(x2+y2, xy +z, x +y+z). Find the Jacobian matrix of fat
the point (1,−1,2).
Solution
To find the Jacobian matrix of fat the point (1,−1,2), we need to compute
the partial derivatives of each component function with respect to each variable
at this point.
6
Step 1: Find ∂f1∂x,∂f1
∂y ,∂f1
∂z .
∂f1
∂x = 2x∂f1
∂y = 2y∂f1
∂z = 0
Step 2: Find ∂f2∂x,∂f2
∂y ,∂f2
∂z .
∂f2
∂x =y∂f2
∂y =x∂f2
∂z = 1
Step 3: Find ∂f3∂x,∂f3
∂y ,∂f3
∂z .
∂f3
∂x = 1 ∂f3
∂y = 1 ∂f3
∂z = 1
Step 4: Evaluate the Jacobian matrix at (1,−1,2).The Jacobian
matrix Jis given by
J=
2−2 0
−111
1 1 1
Thus, the Jacobian matrix of fat the point (1,−1,2) is:
2−2 0
−111
1 1 1
Question 10
Question
Let f:R3→R2be a mapping given by f(x, y, z)=(x2+yz, z2−xy). Calculate
the Jacobian matrix of f.
Solution
To calculate the Jacobian matrix of f, we need to find the partial derivatives of
the components of fwith respect to each variable.
Step 1: Partial derivatives of f
7
Let’s find the partial derivatives of each component function of f:
∂
∂x (x2+yz)=2x,
∂
∂y (x2+yz) = z,
∂
∂z (x2+yz) = y,
∂
∂x (z2−xy) = −y,
∂
∂y (z2−xy) = −x,
∂
∂z (z2−xy)=2z.
Step 2: Construct the Jacobian matrix
The Jacobian matrix of fis given by:
Jf(x, y, z) = "∂
∂x (x2+yz)∂
∂y (x2+yz)∂
∂z (x2+yz)
∂
∂x (z2−xy)∂
∂y (z2−xy)∂
∂z (z2−xy)#.
Plugging in the partial derivatives, we get:
Jf(x, y, z) = 2x z y
−y−x2z.
Therefore, the Jacobian matrix of the mapping fis:
Jf(x, y, z) = 2x z y
−y−x2z.
Question 11
Question
Find the Jacobian matrix of the following system of equations:
(x2+y= 3
2x−y2= 1
Solution
Step 1: To find the Jacobian matrix, we first need to compute the partial
derivatives of the given system of equations: Let f(x, y) = x2+y−3 and
g(x, y)=2x−y2−1, then we have fx= 2x,fy= 1, gx= 2, and gy=−2y.
Step 2: The Jacobian matrix of the system is given by:
J=fxfy
gxgy=2x1
2−2y
8
Step 3: Now, substitute x= 1 and y= 2 into the Jacobian matrix:
J(1,2) = 2(1) 1
2−2(2)=2 1
2−4
Therefore, the Jacobian matrix of the given system of equations at the point
(1,2) is 2 1
2−4.
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, ex+yz). Compute
the Jacobian matrix of fat the point (1,0,1).
Solution
To compute the Jacobian matrix of fat the point (1,0,1), we need to find the
partial derivatives of fwith respect to each variable at this point.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y),∂
∂x (ex+yz)
∂f
∂x = (2xy, ex)
Step 2: Find ∂f ∂y
∂f
∂y =∂
∂y (x2y),∂
∂y (ex+yz)
∂f
∂y = (x2, z)
Step 3: Find ∂f ∂z
∂f
∂z =∂
∂z (x2y),∂
∂z (ex+yz)
∂f
∂z = (0, y)
Step 4: Evaluate the Jacobian matrix at (1,0,1) Now, we evaluate the
Jacobian matrix at the point (1,0,1) by plugging in the values.
Jf(1,0,1) =
2(1)(0) e1
121
0 0
Jf(1,0,1) =
0e
1 1
0 0
9
Question 13
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+y2, xyz). Determine 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.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y2, xyz), we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Again, we have:
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Therefore, the Jacobian matrix of fis:
Jf(x, y, z) = 2x2y0
yz xz xy
Question 14
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2yz, xy2z, xyz2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: To find the Jacobian matrix of f, we first need to compute the partial
derivatives of fwith respect to x,y, and z.
Step 2: The partial derivatives are given by:
∂f
∂x =
2xyz
y2z
yz2
,∂f
∂y =
x2z
2xyz
xz2
,and ∂f
∂z =
x2y
xy2
2xyz
.
Step 3: Evaluate the partial derivatives at the point (1,2,3):
∂f
∂x (1,2,3) =
12
6
6
,∂f
∂y (1,2,3) =
6
12
3
,and ∂f
∂z (1,2,3) =
2
4
6
.
10
Step 4: The Jacobian matrix of fat (1,2,3) is the matrix formed by arrang-
ing the three partial derivatives evaluated at (1,2,3) as rows:
Jf(1,2,3) =
12 6 6
6 12 3
2 4 6
.
Question 15
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, y2+z). Compute the
Jacobian matrix of f.
Solution
Step 1: Write out the components of the function f.
f(x, y, z)=(x2+y, y2+z)
Step 2: Determine the partial derivatives of each component of fwith respect
to x,y, and z. For the first component:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
For the second component:
∂f2
∂x = 0
∂f2
∂y = 2y
∂f2
∂z = 1
Step 3: Assemble the Jacobian matrix using the partial derivatives calculated
above.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2x1 0
0 2y1
11
Question 16
Question
Let f:R2→R2be a function defined by f(x, y) = (xy, x2). Determine the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each input variable.
Step 1: Compute ∂f1
∂x ,∂f1
∂y ,∂f2
∂x , and ∂f2
∂y .
Given f(x, y)=(xy, x2), we have:
∂f1
∂x =∂(xy)
∂x =y
∂f1
∂y =∂(xy)
∂y =x
∂f2
∂x =∂(x2)
∂x = 2x
∂f2
∂y = 0 (since f2does not involve y)
Step 2: Assemble the partial derivatives into a 2 ×2 Jacobian matrix.
The Jacobian matrix Jfof fis given by:
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Plugging in the partial derivatives we computed:
Jf=y x
2x0
Therefore, the Jacobian matrix of the function fis y x
2x0.
Question 17
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2y, sin(xy)).
Compute the Jacobian matrix of f.
12
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet F(x, y, z)=(u, v) =
(x2y, sin(xy)). The partial derivatives of Fare:
∂u
∂x ,∂u
∂y ,∂u
∂z ,∂v
∂x ,∂v
∂y ,∂v
∂z
Compute each partial derivative:
∂u
∂x = 2xy
∂u
∂y =x2
∂u
∂z = 0
∂v
∂x =ycos(xy)
∂v
∂y =xcos(xy)
∂v
∂z = 0
Step 2: Create the Jacobian matrix The Jacobian matrix of fis given
by:
J= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
Substitute the computed partial derivatives into the matrix:
J=2xy x20
ycos(xy)xcos(xy) 0
Therefore, the Jacobian matrix of fis:
J=2xy x20
ycos(xy)xcos(xy) 0
Question 18
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y, xz). Calculate
the Jacobian matrix of f.
13
Solution
To find the Jacobian matrix of a function f:Rn→Rm, we need to compute
the matrix of partial derivatives of f.
Step 1: Compute the partial derivatives of f.The components of
fare f1(x, y, z) = 2x+yand f2(x, y, z) = xz. We now compute the partial
derivatives of f1and f2:
∂f1
∂x = 2,∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 2: Assemble the Jacobian matrix. The Jacobian matrix of fis a
2×3 matrix where the i-th row and j-th column entry is the partial derivative
of fiwith respect to the j-th variable. Therefore, the Jacobian matrix of fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2 1 0
z0x
Therefore, the Jacobian matrix of the function f(x, y, z) = (2x+y, xz) is
2 1 0
z0x.
Question 19
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+yz, x2+ 2y+z).
Find the Jacobian matrix of f.
Solution
Step 1: Write down the components of the function f:
f(x, y, z) = (2x+yz, x2+ 2y+z)
Step 2: Calculate the partial derivatives of fwith respect to x,y, and z:
∂f
∂x = ∂(2x+yz)
∂x
∂(x2+2y+z)
∂x !=2
2x
∂f
∂y = ∂(2x+yz)
∂y
∂(x2+2y+z)
∂y !=z
2
∂f
∂z = ∂(2x+yz)
∂z
∂(x2+2y+z)
∂z !=y
1
14
Step 3: Assemble the partial derivatives into a matrix to form the Jacobian
matrix of f:
Jf=2z y
2x2 1
Question 20
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Determine the
Jacobian matrix of f.
Solution
To determine the Jacobian matrix of f, we first need to calculate the partial
derivatives of each component function with respect to each input variable.
Step 1: Calculate ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function
f1(x, y, z) = x2y.
∂f1
∂x = 2xy , ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Calculate ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0 ,∂f2
∂y =z , ∂f2
∂z =y
Step 3: Assemble the partial derivatives into a 2 ×3 Jacobian matrix Jf
for f:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of the function f(x, y, z)=(x2y, yz) is
Jf=2xy x20
0z y
.
Question 21
Question
Let f(x, y, z) = (x2+y2, y2+z2, z2+x2) be a vector-valued function. Calculate
the Jacobian matrix of fat the point (1,2,3).
15
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂f1
∂x
∂f2
∂x
∂f3
∂x
=
2x
0
2z
∂f
∂y =
∂f1
∂y
∂f2
∂y
∂f3
∂y
=
2y
2y
0
∂f
∂z =
∂f1
∂z
∂f2
∂z
∂f3
∂z
=
0
2z
2x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) =
2
0
6
∂f
∂y (1,2,3) =
4
4
0
∂f
∂z (1,2,3) =
0
6
2
Step 3: Assemble the Jacobian matrix using the partial derivatives evaluated
at (1,2,3).
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
240
046
602
Question 22
Question
Find the Jacobian matrix of the transformation (x, y)=(rcos θ, r sin θ).
Solution
To find the Jacobian matrix of the transformation, we need to first express x
and yin terms of rand θ, and then find the partial derivatives with respect to
rand θ.
16
Step 1: Express xand yin terms of rand θ:
x=rcos θ
y=rsin θ
Step 2: Find the partial derivatives:
∂x
∂r = cos θ
∂x
∂θ =−rsin θ
∂y
∂r = sin θ
∂y
∂θ =rcos θ
Step 3: Form the Jacobian matrix: The Jacobian matrix is given by
∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ
Substitute the partial derivatives:
cos θ−rsin θ
sin θ r cos θ
Therefore, the Jacobian matrix for the transformation (x, y) = (rcos θ, r sin θ)
is cos θ−rsin θ
sin θ r cos θ
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. 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.
∂f
∂x = 2xy +z2,
∂f
∂y =x2+ 2yz,
∂f
∂z =y2+ 2zx.
17
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= 2(1)(2) + 32= 4 + 9 = 13,
∂f
∂y (1,2,3)
= 12+ 2(2)(3) = 1 + 12 = 13,
∂f
∂z (1,2,3)
= 2(2)(1) + 32= 4 + 9 = 13.
Step 3: Assemble the Jacobian matrix using the computed partial deriva-
tives.
Jf(1,2,3) = 13 13 13.
Therefore, the Jacobian matrix of fat the point (1,2,3) is 13 13 13.
Question 24
Question
Let f:R2→R2be a function defined by f(x, y)=(excos(y), exsin(y)). Find
the Jacobian matrix of fat the point (0,0).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y.
∂f1
∂x =∂
∂x (excos(y)) = excos(y)
∂f1
∂y =∂
∂y (excos(y)) = −exsin(y)
∂f2
∂x =∂
∂x (exsin(y)) = exsin(y)
∂f2
∂y =∂
∂y (exsin(y)) = excos(y)
Step 2: Evaluate the partial derivatives at the point (0,0).
∂f1
∂x (0,0)
= 1
∂f1
∂y (0,0)
= 0
∂f2
∂x (0,0)
= 0
∂f2
∂y (0,0)
= 1
18
Step 3: Assemble the Jacobian matrix of fat (0,0).
Jf(0,0) = 1 0
0 1
Question 25
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+yz, y2+zx, z2+xy).
Find the Jacobian matrix of f.
Solution
Step 1: Write out the components of the function f.
f(x, y, z) = x2+yz, y2+zx, z2+xy
Step 2: Compute 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
∂f3
∂x =y, ∂f3
∂y =x, ∂f3
∂z = 2z
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in rows.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x z y
z2y x
y x 2z
Question 26
Question
Let f:R3→Rbe a vector-valued function defined by f(x, y, z) = (x2yz, ex+y, xyz2).
Determine the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
19
Step 1: Compute the partial derivatives of f:
∂f1
∂x =∂(x2yz)
∂x = 2xyz,
∂f1
∂y =∂(x2yz)
∂y =x2z,
∂f1
∂z =∂(x2yz)
∂z =x2y,
∂f2
∂x =∂(ex+y)
∂x =ex+y,
∂f2
∂y =∂(ex+y)
∂y =ex+y,
∂f2
∂z = 0,
∂f3
∂x =∂(xyz2)
∂x =yz2,
∂f3
∂y =∂(xyz2)
∂y =xz2,
∂f3
∂z =∂(xyz2)
∂z = 2xyz.
Step 2: Assemble the partial derivatives into a matrix:
J(f) =
2xyz x2z x2y
ex+yex+y0
yz2xz22xyz
.
Therefore, the Jacobian matrix of fis
J(f) =
2xyz x2z x2y
ex+yex+y0
yz2xz22xyz
.
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z) = x2+y2, yz. Find the
Jacobian matrix of f.
Solution
The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix of partial
derivatives. For the function f(x, y, z) = x2+y2, yz, we need to determine
the partial derivatives of each component of fwith respect to x,y, and z.
20
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component f1=x2+y2.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component f2=yz.
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 3: Write the Jacobian matrix Jfof fusing the computed partial
derivatives.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z) = (x2, yz). Compute the
Jacobian matrix of fat the point (2,−1,3).
Solution
Step 1: The Jacobian matrix of a mapping f:Rn→Rmis an m×nmatrix
where each entry is a partial derivative of the corresponding component function.
Step 2: Consider the component functions of f:
f1(x, y, z) = x2
f2(x, y, z) = yz
Step 3: Compute the partial derivatives of f1and f2with respect to x, y,
and z.∂f1
∂x = 2x
∂f1
∂y = 0
∂f1
∂z = 0
Step 4: Now, compute the partial derivatives of f2:
∂f2
∂x = 0
∂f2
∂y =z
21
∂f2
∂z =y
Step 5: The Jacobian matrix Jfof fis then:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y
Step 6: Evaluate the Jacobian matrix at the point (2,−1,3):
Jf(2,−1,3) = 2(2) 0 0
0 3 −1=4 0 0
0 3 −1
Therefore, the Jacobian matrix of fat the point (2,−1,3) is 4 0 0
0 3 −1.
Question 29
Question
Let f(x, y, z) = (2x+y2,3y−z, xz) be a vector-valued function. Calculate the
Jacobian matrix Jfof f.
Solution
To find the Jacobian matrix Jfof f, we need to compute the partial derivatives
of each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (2x+y2)=2
∂f1
∂y =∂
∂y (2x+y2)=2y
∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (3y−z)=0
∂f2
∂y =∂
∂y (3y−z)=3
∂f2
∂z =−1
22
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xz) = z
∂f3
∂y =∂
∂y (xz)=0
∂f3
∂z =x
Step 4: Assemble the partial derivatives into the Jacobian matrix Jf:
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2 2y0
0 3 −1
z0x
Question 30
Question
Let f(x, y)=(x2y+ 2xy3, x2+ 3y2) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,1).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f1
∂x =∂
∂x (x2y+ 2xy3)=2xy + 2y3
∂f1
∂y =∂
∂y (x2y+ 2xy3) = x2+ 6xy2
∂f2
∂x =∂
∂x (x2+ 3y2)=2x
∂f2
∂y =∂
∂y (x2+ 3y2)=6y
Step 2: Evaluate the partial derivatives at the point (1,1).
∂f1
∂x (1,1) = 2(1)(1) + 2(1)3= 4
∂f1
∂y (1,1) = (1)2+ 6(1)(1)2= 7
∂f2
∂x (1,1) = 2(1) = 2
∂f2
∂y (1,1) = 6(1) = 6
23
Solution
Step 1: Calculate 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 Jacobian matrix. The Jacobian matrix of fis given
by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2xy x20
0z y
Therefore, the Jacobian matrix of fis
Jf=2xy x20
0z y
.
Question 3
Question
Let f(x, y) = x2+ 2y
3x−y. Find the Jacobian matrix of fat the point (1,−1).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (x2+ 2y)
∂
∂x (3x−y)=2x
3
∂f
∂y = ∂
∂y (x2+ 2y)
∂
∂y (3x−y)!=2
−1
2
Step 2: Evaluate the partial derivatives at the point (1,−1).
∂f
∂x (1,−1) = 2(1)
3=2
3
∂f
∂y (1,−1) = 2
−1
Step 3: Assemble the Jacobian matrix at the given point.
Jf(1,−1) = ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=2 2
3−1
Therefore, the Jacobian matrix of fat the point (1,−1) is 2 2
3−1.
Question 4
Question
Let f:R2→R2be a function defined by f(x, y)=(x2y, xy2). Find the
Jacobian matrix of fat the point (2,3).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f1
∂x = 2xy and ∂f2
∂x =y2
∂f1
∂y =x2and ∂f2
∂y = 2xy
Step 2: Evaluate the partial derivatives at the point (2,3).
∂f1
∂x (2,3) = 2(2)(3) = 12,∂f2
∂x (2,3) = (3)2= 9
∂f1
∂y (2,3) = (2)2= 4,∂f2
∂y (2,3) = 2(2)(3) = 12
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fat point
(2,3) is
Jf(2,3) = "∂f1
∂x (2,3) ∂f1
∂y (2,3)
∂f2
∂x (2,3) ∂f2
∂y (2,3)#=12 4
9 12
Question 5
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y2, yz). Find the
Jacobian matrix of f.
3
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of f
with respect to each variable and arrange them in a matrix.
Step 1: Calculate the partial derivatives of f.
∂f1
∂x = 2x
∂f1
∂y = 2y
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 2: Assemble the partial derivatives into a matrix. The Jacobian matrix
of fis:
Jf=2x2y0
0z y
Question 6
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2y, yz, xy2). 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 each variable.
Step 1: Compute the partial derivatives of f:
∂
∂x f1(x, y, z) = ∂
∂x (x2y) = 2xy
∂
∂y f1(x, y, z) = ∂
∂y (x2y) = x2
∂
∂z f1(x, y, z) = 0
4
Step 2: Repeat the process for the other component functions of f:
∂
∂x f2(x, y, z) = ∂
∂x (yz) = 0
∂
∂y f2(x, y, z) = ∂
∂y (yz) = z
∂
∂z f2(x, y, z) = y
∂
∂x f3(x, y, z) = ∂
∂x (xy2) = y2
∂
∂y f3(x, y, z) = ∂
∂y (xy2) = 2xy
∂
∂z f3(x, y, z) = 0
Step 3: Assemble the partial derivatives into the Jacobian matrix:
Jf(x, y, z) =
2xy x20
0z y
y22xy 0
Therefore, the Jacobian matrix of fis given by Jf(x, y, z) =
2xy x20
0z y
y22xy 0
.
Question 7
Question
Let f(x, y) = (2x3y, x2y2). Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
Step 1: Compute ∂f1
∂x and ∂f1
∂y for the first component function f1(x, y) =
2x3y.
∂f1
∂x = 6x2yand ∂f1
∂y = 2x3
Step 2: Compute ∂f2
∂x and ∂f2
∂y for the second component function f2(x, y) =
x2y2.
∂f2
∂x = 2xy2and ∂f2
∂y = 2x2y
5
Step 3: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis:
6x2y2x3
2xy22x2y
Question 8
Question
Let f(x, y, z)=(x2+y2, y2+z2, z2+x2). Find the Jacobian matrix of f.
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =2x0 0
∂f
∂y =0 2y0
∂f
∂z =002z
Step 2: Arrange the partial derivatives as columns of the Jacobian matrix.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
Jf=
2x0 0
0 2y0
0 0 2z
Question 9
Question
Let f(x, y, z)=(x2+y2, xy +z, x +y+z). Find the Jacobian matrix of fat
the point (1,−1,2).
Solution
To find the Jacobian matrix of fat the point (1,−1,2), we need to compute
the partial derivatives of each component function with respect to each variable
at this point.
6
Step 1: Find ∂f1∂x,∂f1
∂y ,∂f1
∂z .
∂f1
∂x = 2x∂f1
∂y = 2y∂f1
∂z = 0
Step 2: Find ∂f2∂x,∂f2
∂y ,∂f2
∂z .
∂f2
∂x =y∂f2
∂y =x∂f2
∂z = 1
Step 3: Find ∂f3∂x,∂f3
∂y ,∂f3
∂z .
∂f3
∂x = 1 ∂f3
∂y = 1 ∂f3
∂z = 1
Step 4: Evaluate the Jacobian matrix at (1,−1,2).The Jacobian
matrix Jis given by
J=
2−2 0
−111
1 1 1
Thus, the Jacobian matrix of fat the point (1,−1,2) is:
2−2 0
−111
1 1 1
Question 10
Question
Let f:R3→R2be a mapping given by f(x, y, z)=(x2+yz, z2−xy). Calculate
the Jacobian matrix of f.
Solution
To calculate the Jacobian matrix of f, we need to find the partial derivatives of
the components of fwith respect to each variable.
Step 1: Partial derivatives of f
7
Let’s find the partial derivatives of each component function of f:
∂
∂x (x2+yz)=2x,
∂
∂y (x2+yz) = z,
∂
∂z (x2+yz) = y,
∂
∂x (z2−xy) = −y,
∂
∂y (z2−xy) = −x,
∂
∂z (z2−xy)=2z.
Step 2: Construct the Jacobian matrix
The Jacobian matrix of fis given by:
Jf(x, y, z) = "∂
∂x (x2+yz)∂
∂y (x2+yz)∂
∂z (x2+yz)
∂
∂x (z2−xy)∂
∂y (z2−xy)∂
∂z (z2−xy)#.
Plugging in the partial derivatives, we get:
Jf(x, y, z) = 2x z y
−y−x2z.
Therefore, the Jacobian matrix of the mapping fis:
Jf(x, y, z) = 2x z y
−y−x2z.
Question 11
Question
Find the Jacobian matrix of the following system of equations:
(x2+y= 3
2x−y2= 1
Solution
Step 1: To find the Jacobian matrix, we first need to compute the partial
derivatives of the given system of equations: Let f(x, y) = x2+y−3 and
g(x, y)=2x−y2−1, then we have fx= 2x,fy= 1, gx= 2, and gy=−2y.
Step 2: The Jacobian matrix of the system is given by:
J=fxfy
gxgy=2x1
2−2y
8
Step 3: Now, substitute x= 1 and y= 2 into the Jacobian matrix:
J(1,2) = 2(1) 1
2−2(2)=2 1
2−4
Therefore, the Jacobian matrix of the given system of equations at the point
(1,2) is 2 1
2−4.
Question 12
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, ex+yz). Compute
the Jacobian matrix of fat the point (1,0,1).
Solution
To compute the Jacobian matrix of fat the point (1,0,1), we need to find the
partial derivatives of fwith respect to each variable at this point.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y),∂
∂x (ex+yz)
∂f
∂x = (2xy, ex)
Step 2: Find ∂f ∂y
∂f
∂y =∂
∂y (x2y),∂
∂y (ex+yz)
∂f
∂y = (x2, z)
Step 3: Find ∂f ∂z
∂f
∂z =∂
∂z (x2y),∂
∂z (ex+yz)
∂f
∂z = (0, y)
Step 4: Evaluate the Jacobian matrix at (1,0,1) Now, we evaluate the
Jacobian matrix at the point (1,0,1) by plugging in the values.
Jf(1,0,1) =
2(1)(0) e1
121
0 0
Jf(1,0,1) =
0e
1 1
0 0
9
Question 13
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+y2, xyz). Determine 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.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z) = (x2+y2, xyz), we have:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Again, we have:
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Therefore, the Jacobian matrix of fis:
Jf(x, y, z) = 2x2y0
yz xz xy
Question 14
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2yz, xy2z, xyz2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: To find the Jacobian matrix of f, we first need to compute the partial
derivatives of fwith respect to x,y, and z.
Step 2: The partial derivatives are given by:
∂f
∂x =
2xyz
y2z
yz2
,∂f
∂y =
x2z
2xyz
xz2
,and ∂f
∂z =
x2y
xy2
2xyz
.
Step 3: Evaluate the partial derivatives at the point (1,2,3):
∂f
∂x (1,2,3) =
12
6
6
,∂f
∂y (1,2,3) =
6
12
3
,and ∂f
∂z (1,2,3) =
2
4
6
.
10
Step 4: The Jacobian matrix of fat (1,2,3) is the matrix formed by arrang-
ing the three partial derivatives evaluated at (1,2,3) as rows:
Jf(1,2,3) =
12 6 6
6 12 3
2 4 6
.
Question 15
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, y2+z). Compute the
Jacobian matrix of f.
Solution
Step 1: Write out the components of the function f.
f(x, y, z)=(x2+y, y2+z)
Step 2: Determine the partial derivatives of each component of fwith respect
to x,y, and z. For the first component:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
For the second component:
∂f2
∂x = 0
∂f2
∂y = 2y
∂f2
∂z = 1
Step 3: Assemble the Jacobian matrix using the partial derivatives calculated
above.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2x1 0
0 2y1
11
Question 16
Question
Let f:R2→R2be a function defined by f(x, y) = (xy, x2). Determine the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each input variable.
Step 1: Compute ∂f1
∂x ,∂f1
∂y ,∂f2
∂x , and ∂f2
∂y .
Given f(x, y)=(xy, x2), we have:
∂f1
∂x =∂(xy)
∂x =y
∂f1
∂y =∂(xy)
∂y =x
∂f2
∂x =∂(x2)
∂x = 2x
∂f2
∂y = 0 (since f2does not involve y)
Step 2: Assemble the partial derivatives into a 2 ×2 Jacobian matrix.
The Jacobian matrix Jfof fis given by:
Jf="∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Plugging in the partial derivatives we computed:
Jf=y x
2x0
Therefore, the Jacobian matrix of the function fis y x
2x0.
Question 17
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2y, sin(xy)).
Compute the Jacobian matrix of f.
12
Solution
To compute the Jacobian matrix of f, we need to find the partial derivatives of
each component function.
Step 1: Compute the partial derivatives of fLet F(x, y, z)=(u, v) =
(x2y, sin(xy)). The partial derivatives of Fare:
∂u
∂x ,∂u
∂y ,∂u
∂z ,∂v
∂x ,∂v
∂y ,∂v
∂z
Compute each partial derivative:
∂u
∂x = 2xy
∂u
∂y =x2
∂u
∂z = 0
∂v
∂x =ycos(xy)
∂v
∂y =xcos(xy)
∂v
∂z = 0
Step 2: Create the Jacobian matrix The Jacobian matrix of fis given
by:
J= ∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z !
Substitute the computed partial derivatives into the matrix:
J=2xy x20
ycos(xy)xcos(xy) 0
Therefore, the Jacobian matrix of fis:
J=2xy x20
ycos(xy)xcos(xy) 0
Question 18
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y, xz). Calculate
the Jacobian matrix of f.
13
Solution
To find the Jacobian matrix of a function f:Rn→Rm, we need to compute
the matrix of partial derivatives of f.
Step 1: Compute the partial derivatives of f.The components of
fare f1(x, y, z) = 2x+yand f2(x, y, z) = xz. We now compute the partial
derivatives of f1and f2:
∂f1
∂x = 2,∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 2: Assemble the Jacobian matrix. The Jacobian matrix of fis a
2×3 matrix where the i-th row and j-th column entry is the partial derivative
of fiwith respect to the j-th variable. Therefore, the Jacobian matrix of fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2 1 0
z0x
Therefore, the Jacobian matrix of the function f(x, y, z) = (2x+y, xz) is
2 1 0
z0x.
Question 19
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+yz, x2+ 2y+z).
Find the Jacobian matrix of f.
Solution
Step 1: Write down the components of the function f:
f(x, y, z) = (2x+yz, x2+ 2y+z)
Step 2: Calculate the partial derivatives of fwith respect to x,y, and z:
∂f
∂x = ∂(2x+yz)
∂x
∂(x2+2y+z)
∂x !=2
2x
∂f
∂y = ∂(2x+yz)
∂y
∂(x2+2y+z)
∂y !=z
2
∂f
∂z = ∂(2x+yz)
∂z
∂(x2+2y+z)
∂z !=y
1
14
Step 3: Assemble the partial derivatives into a matrix to form the Jacobian
matrix of f:
Jf=2z y
2x2 1
Question 20
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2y, yz). Determine the
Jacobian matrix of f.
Solution
To determine the Jacobian matrix of f, we first need to calculate the partial
derivatives of each component function with respect to each input variable.
Step 1: Calculate ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component function
f1(x, y, z) = x2y.
∂f1
∂x = 2xy , ∂f1
∂y =x2,∂f1
∂z = 0
Step 2: Calculate ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component function
f2(x, y, z) = yz.
∂f2
∂x = 0 ,∂f2
∂y =z , ∂f2
∂z =y
Step 3: Assemble the partial derivatives into a 2 ×3 Jacobian matrix Jf
for f:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=2xy x20
0z y
Therefore, the Jacobian matrix of the function f(x, y, z)=(x2y, yz) is
Jf=2xy x20
0z y
.
Question 21
Question
Let f(x, y, z) = (x2+y2, y2+z2, z2+x2) be a vector-valued function. Calculate
the Jacobian matrix of fat the point (1,2,3).
15
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂f1
∂x
∂f2
∂x
∂f3
∂x
=
2x
0
2z
∂f
∂y =
∂f1
∂y
∂f2
∂y
∂f3
∂y
=
2y
2y
0
∂f
∂z =
∂f1
∂z
∂f2
∂z
∂f3
∂z
=
0
2z
2x
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) =
2
0
6
∂f
∂y (1,2,3) =
4
4
0
∂f
∂z (1,2,3) =
0
6
2
Step 3: Assemble the Jacobian matrix using the partial derivatives evaluated
at (1,2,3).
Jf(1,2,3) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
240
046
602
Question 22
Question
Find the Jacobian matrix of the transformation (x, y)=(rcos θ, r sin θ).
Solution
To find the Jacobian matrix of the transformation, we need to first express x
and yin terms of rand θ, and then find the partial derivatives with respect to
rand θ.
16
Step 1: Express xand yin terms of rand θ:
x=rcos θ
y=rsin θ
Step 2: Find the partial derivatives:
∂x
∂r = cos θ
∂x
∂θ =−rsin θ
∂y
∂r = sin θ
∂y
∂θ =rcos θ
Step 3: Form the Jacobian matrix: The Jacobian matrix is given by
∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ
Substitute the partial derivatives:
cos θ−rsin θ
sin θ r cos θ
Therefore, the Jacobian matrix for the transformation (x, y) = (rcos θ, r sin θ)
is cos θ−rsin θ
sin θ r cos θ
Question 23
Question
Let f(x, y, z) = x2y+y2z+z2x. 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.
∂f
∂x = 2xy +z2,
∂f
∂y =x2+ 2yz,
∂f
∂z =y2+ 2zx.
17
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= 2(1)(2) + 32= 4 + 9 = 13,
∂f
∂y (1,2,3)
= 12+ 2(2)(3) = 1 + 12 = 13,
∂f
∂z (1,2,3)
= 2(2)(1) + 32= 4 + 9 = 13.
Step 3: Assemble the Jacobian matrix using the computed partial deriva-
tives.
Jf(1,2,3) = 13 13 13.
Therefore, the Jacobian matrix of fat the point (1,2,3) is 13 13 13.
Question 24
Question
Let f:R2→R2be a function defined by f(x, y)=(excos(y), exsin(y)). Find
the Jacobian matrix of fat the point (0,0).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y.
∂f1
∂x =∂
∂x (excos(y)) = excos(y)
∂f1
∂y =∂
∂y (excos(y)) = −exsin(y)
∂f2
∂x =∂
∂x (exsin(y)) = exsin(y)
∂f2
∂y =∂
∂y (exsin(y)) = excos(y)
Step 2: Evaluate the partial derivatives at the point (0,0).
∂f1
∂x (0,0)
= 1
∂f1
∂y (0,0)
= 0
∂f2
∂x (0,0)
= 0
∂f2
∂y (0,0)
= 1
18
Step 3: Assemble the Jacobian matrix of fat (0,0).
Jf(0,0) = 1 0
0 1
Question 25
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+yz, y2+zx, z2+xy).
Find the Jacobian matrix of f.
Solution
Step 1: Write out the components of the function f.
f(x, y, z) = x2+yz, y2+zx, z2+xy
Step 2: Compute 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
∂f3
∂x =y, ∂f3
∂y =x, ∂f3
∂z = 2z
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
in rows.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x z y
z2y x
y x 2z
Question 26
Question
Let f:R3→Rbe a vector-valued function defined by f(x, y, z) = (x2yz, ex+y, xyz2).
Determine the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
19
Step 1: Compute the partial derivatives of f:
∂f1
∂x =∂(x2yz)
∂x = 2xyz,
∂f1
∂y =∂(x2yz)
∂y =x2z,
∂f1
∂z =∂(x2yz)
∂z =x2y,
∂f2
∂x =∂(ex+y)
∂x =ex+y,
∂f2
∂y =∂(ex+y)
∂y =ex+y,
∂f2
∂z = 0,
∂f3
∂x =∂(xyz2)
∂x =yz2,
∂f3
∂y =∂(xyz2)
∂y =xz2,
∂f3
∂z =∂(xyz2)
∂z = 2xyz.
Step 2: Assemble the partial derivatives into a matrix:
J(f) =
2xyz x2z x2y
ex+yex+y0
yz2xz22xyz
.
Therefore, the Jacobian matrix of fis
J(f) =
2xyz x2z x2y
ex+yex+y0
yz2xz22xyz
.
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z) = x2+y2, yz. Find the
Jacobian matrix of f.
Solution
The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix of partial
derivatives. For the function f(x, y, z) = x2+y2, yz, we need to determine
the partial derivatives of each component of fwith respect to x,y, and z.
20
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z for the first component f1=x2+y2.
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z for the second component f2=yz.
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 3: Write the Jacobian matrix Jfof fusing the computed partial
derivatives.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2y0
0z y
Question 28
Question
Let f:R3→R2be a mapping defined by f(x, y, z) = (x2, yz). Compute the
Jacobian matrix of fat the point (2,−1,3).
Solution
Step 1: The Jacobian matrix of a mapping f:Rn→Rmis an m×nmatrix
where each entry is a partial derivative of the corresponding component function.
Step 2: Consider the component functions of f:
f1(x, y, z) = x2
f2(x, y, z) = yz
Step 3: Compute the partial derivatives of f1and f2with respect to x, y,
and z.∂f1
∂x = 2x
∂f1
∂y = 0
∂f1
∂z = 0
Step 4: Now, compute the partial derivatives of f2:
∂f2
∂x = 0
∂f2
∂y =z
21
∂f2
∂z =y
Step 5: The Jacobian matrix Jfof fis then:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x0 0
0z y
Step 6: Evaluate the Jacobian matrix at the point (2,−1,3):
Jf(2,−1,3) = 2(2) 0 0
0 3 −1=4 0 0
0 3 −1
Therefore, the Jacobian matrix of fat the point (2,−1,3) is 4 0 0
0 3 −1.
Question 29
Question
Let f(x, y, z) = (2x+y2,3y−z, xz) be a vector-valued function. Calculate the
Jacobian matrix Jfof f.
Solution
To find the Jacobian matrix Jfof f, we need to compute the partial derivatives
of each component function with respect to each variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (2x+y2)=2
∂f1
∂y =∂
∂y (2x+y2)=2y
∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (3y−z)=0
∂f2
∂y =∂
∂y (3y−z)=3
∂f2
∂z =−1
22
Step 3: Find ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =∂
∂x (xz) = z
∂f3
∂y =∂
∂y (xz)=0
∂f3
∂z =x
Step 4: Assemble the partial derivatives into the Jacobian matrix Jf:
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2 2y0
0 3 −1
z0x
Question 30
Question
Let f(x, y)=(x2y+ 2xy3, x2+ 3y2) be a vector-valued function. Find the
Jacobian matrix of fat the point (1,1).
Solution
Step 1: Find the partial derivatives of fwith respect to xand y.
∂f1
∂x =∂
∂x (x2y+ 2xy3)=2xy + 2y3
∂f1
∂y =∂
∂y (x2y+ 2xy3) = x2+ 6xy2
∂f2
∂x =∂
∂x (x2+ 3y2)=2x
∂f2
∂y =∂
∂y (x2+ 3y2)=6y
Step 2: Evaluate the partial derivatives at the point (1,1).
∂f1
∂x (1,1) = 2(1)(1) + 2(1)3= 4
∂f1
∂y (1,1) = (1)2+ 6(1)(1)2= 7
∂f2
∂x (1,1) = 2(1) = 2
∂f2
∂y (1,1) = 6(1) = 6
23
Step 3: Construct the Jacobian matrix of fat (1,1). The Jacobian matrix
of fat (1,1) is given by
Jf(1,1) = 4 7
2 6
24
