MATH 117 - ELEMENTS OF
MATHEMATICS - Jacobian matrix
Question Bank - Set 9
Liberty University
Question 1
Question
Let f(x, y)=(x2+y, xy). Find the Jacobian matrix of fat the point (3,2).
Solution
Step 1: Find the partial derivatives of f(x, y) with respect to xand y.
∂f
∂x = (2x, y) and ∂f
∂y = (1, x)
Step 2: Evaluate the partial derivatives at the point (3,2).
∂f
∂x = (6,2) and ∂f
∂y = (1,3)
Step 3: Assemble the partial derivatives into the Jacobian matrix of fat
(3,2).
Jf(3,2) = 6 2
1 3
Therefore, the Jacobian matrix of fat the point (3,2) is 6 2
1 3.
Question 2
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, yz). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Computing ∂f1∂x,∂f1
∂y , and ∂f1
∂z
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
∂f1
∂z = 0 (since f1does not depend on z)
Step 2: Computing ∂f2∂x,∂f2
∂y , and ∂f2
∂z
∂f2
∂x =∂
∂x (yz)=0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 3: Constructing the Jacobian matrix The Jacobian matrix Jfof
fis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x1 0
0z y
Question 3
Question
Let f:R2→R2be the transformation given by f(x, y) = (3x2y, x3). Find the
Jacobian matrix of fat the point (1,2).
Solution
To find the Jacobian matrix of fat the point (1,2), we need to compute the
partial derivatives of fwith respect to xand yat that point.
Step 1: Compute ∂f1∂x and ∂f1
∂y
∂f1
∂x =∂
∂x (3x2y) = 6xy
∂f1
∂y =∂
∂y (3x2y)=3x2
2
Step 2: Compute ∂f2∂x and ∂f2
∂y
∂f2
∂x =∂
∂x (x3)=3x2
∂f2
∂y =∂
∂y (x3)=0
Step 3: Construct the Jacobian matrix The Jacobian matrix of fat
the point (1,2) is given by:
Jf= ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=6xy 3x2
3x20
Step 4: Evaluate the Jacobian matrix at (1,2) Substitute x= 1 and
y= 2 into the Jacobian matrix:
Jf(1,2) = 6(1)(2) 3(1)2
3(1)20=12 3
3 0
Therefore, the Jacobian matrix of fat the point (1,2) is 12 3
3 0.
Question 4
Question
Let f:R3→R2be a transformation defined by f(x, y, z)=(x2+y, xyz).
Compute the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of fis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !
where f1and f2are the components of the transformation f.
Step 2: Compute the partial derivatives of f:
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Evaluate the partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 0
3
∂f2
∂x (1,2,3) = 2(3) = 6,∂f2
∂y (1,2,3) = 1(3) = 3,∂f2
∂z (1,2,3) = 1(2) = 2
Step 4: Assemble the Jacobian matrix at the point (1,2,3):
Jf(1,2,3) = 210
632
Question 5
Question
Let f:R3→R2be a vector-valued function defined by f(x, y, z)=(xy +z, x2−
yz). 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 =y
2x,∂f
∂y =x
−z,∂f
∂z =1
−y
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2
2,∂f
∂y (1,2,3) = 1
−3,∂f
∂z (1,2,3) = 1
−2
Step 3: Assemble these partial derivatives into the Jacobian matrix of fat
(1,2,3).
Jf(1,2,3) = 2 1 1
2−3−2
Question 6
Question
Let f(x, y, z)=2x2y+y2z+xyz. Compute the Jacobian matrix of fat the
point (1,2,3).
Solution
To compute the Jacobian matrix of fat the point (1,2,3), we need to find the
partial derivatives of fwith respect to each variable and evaluate them at that
point.
Step 1: Compute ∂f
∂x .
∂f
∂x = 4xy +yz
4
Step 2: Evaluate ∂f
∂x at (1,2,3).
∂f
∂x (1,2,3) = 4(1)(2) + (2)(3) = 8 + 6 = 14
Step 3: Compute ∂f
∂y .
∂f
∂y = 2x2+ 2yz +xz
Step 4: Evaluate ∂f
∂y at (1,2,3).
∂f
∂y (1,2,3) = 2(1)2+ 2(2)(3) + (1)(3) = 2 + 12 + 3 = 17
Step 5: Compute ∂f
∂z .
∂f
∂z =y2+xy
Step 6: Evaluate ∂f
∂z at (1,2,3).
∂f
∂z (1,2,3) = (2)2+ (1)(2) = 4 + 2 = 6
Step 7: Assemble the Jacobian matrix Jfat the point (1,2,3).
Jf(1,2,3) = ∂f
∂x
∂f
∂y
∂f
∂z =14 17 6
Question 7
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+y, y2+z). Find
the Jacobian matrix of fat the point (1,2,−1).
Solution
To find the Jacobian matrix of fat the point (1,2,−1), we need to compute the
partial derivatives of fwith respect to each input variable and evaluate them
at the given point.
Step 1: Compute the partial derivatives of f: Let u=x2+yand v=y2+z.
Then, f1=∂u
∂x = 2x,f2=∂u
∂y = 1, f3=∂v
∂y = 2y,f4=∂v
∂z = 1.
Step 2: Evaluate the partial derivatives at (1,2,−1): At (1,2,−1), we have
f1(1,2,−1) = 2(1) = 2, f2(1,2,−1) = 1, f3(1,2,−1) = 2(2) = 4, f4(1,2,−1) =
1.
5
Step 3: Assemble the Jacobian matrix: The Jacobian matrix of fat
(1,2,−1) is given by
Jf(1,2,−1) = f1f2
f3f4=2 1
4 1.
Therefore, the Jacobian matrix of fat the point (1,2,−1) is 2 1
4 1.
Question 8
Question
Let f:R3→R2be defined by f(x, y, z) = (2x+y, x −3z). Determine the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we first need to compute the partial derivatives
of fwith respect to each variable.
Step 1: Compute ∂f1∂x
f1(x, y, z)=2x+y
∂f1
∂x =∂(2x+y)
∂x = 2
Step 2: Compute ∂f1∂y
f1(x, y, z)=2x+y
∂f1
∂y =∂(2x+y)
∂y = 1
Step 3: Compute ∂f1∂z Since f1does not depend on z,∂f1
∂z = 0.
Step 4: Compute ∂f2∂x
f2(x, y, z) = x−3z
∂f2
∂x =∂(x−3z)
∂x = 1
Step 5: Compute ∂f2∂y Since f2does not depend on y,∂f2
∂y = 0.
Step 6: Compute ∂f2∂z
f2(x, y, z) = x−3z
∂f2
∂z =∂(x−3z)
∂z =−3
Therefore, the Jacobian matrix of the function fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2 1 0
1 0 −3
6
Question 9
Question
Let f:R⊯→R⊭be a function defined by f(x, y, z)=(x2+yz, 3x−z). Calculate
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y ,and ∂f1
∂z for the first component function f1(x, y, z) =
x2+yz.
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Find ∂f2
∂x ,∂f2
∂y ,and ∂f2
∂z for the second component function f2(x, y, z) =
3x−z.
∂f2
∂x = 3
∂f2
∂y = 0
∂f2
∂z =−1
Step 3: Assemble the Jacobian matrix Jfusing the partial derivatives
calculated above.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
3 0 −1
Therefore, the Jacobian matrix of the function fis:
Jf=2x z y
3 0 −1
Question 10
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+y2, xz). Compute the Jacobian
matrix of fat the point (1,2,−1).
7
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable.
Given f(x, y, z)=(x2+y2, xz), we find the partial derivatives of fas follows:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
Substitute x= 1, y= 2, z=−1 into the partial derivatives:
∂f1
∂x (1,2,−1) = 2(1) = 2,∂f1
∂y (1,2,−1) = 2(2) = 4,∂f1
∂z (1,2,−1) = 0
∂f2
∂x (1,2,−1) = −1,∂f2
∂y (1,2,−1) = 0,∂f2
∂z (1,2,−1) = 1
Step 3: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,−1) is given by:
Jf(1,2,−1) = ∂f1
∂x (1,2,−1) ∂f1
∂y (1,2,−1) ∂f1
∂z (1,2,−1)
∂f2
∂x (1,2,−1) ∂f2
∂y (1,2,−1) ∂f2
∂z (1,2,−1)!
=2 4 0
−101
Question 11
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+yz, xy +z, xz +y).
Determine 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 = (2x, y, z),
∂f
∂y = (z, x, 1) ,
∂f
∂z = (y, 1, x).
8
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1),2,3) = (2,2,3),
∂f
∂y (1,2,3)
= (3,1,1),
∂f
∂z (1,2,3)
= (2,1,1).
Therefore, the Jacobian matrix of fat the point (1,2,3) is
Jf(1,2,3) =
223
311
211
.
Question 12
Question
Let f(x, y, z) =
x2+yz
y2+zx
z2+xy
. Find the Jacobian matrix of f.
Solution
Step 1: Compute the partial derivatives of f(x, y, z) with respect to x,y, and
z.
∂f
∂x =
2x
z
y
,∂f
∂y =
z
2y
x
,∂f
∂z =
y
x
2z
Step 2: Assemble the partial derivatives into a matrix to obtain the Jacobian
matrix of f.
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 13
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, xyz). Find the
Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x z y
yz xz xy
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y+yz, xz). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
fwith respect to each variable.
Step 1: Compute the partial derivatives of f.
∂f
∂x =∂
∂x (x2y+yz),∂
∂x (xz)
= (2xy, z)
Step 2: Compute the partial derivatives of f.
∂f
∂y =∂
∂y (x2y+yz),∂
∂y (xz)
= (x2+z, 0)
10
Step 3: Compute the partial derivatives of f.
∂f
∂z =∂
∂z (x2y+yz),∂
∂z (xz)
= (y, x)
Step 4: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf=
2xy z
x2+z0
y x
Question 15
Question
Let f:R2→R2be defined by f(x, y) = (x2+y2, xy). Calculate the Jacobian
matrix of fat the point (1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (x2+y2)
∂
∂x (xy)=2x
y
∂f
∂y = ∂
∂y (x2+y2)
∂
∂y (xy)!=2y
x
Step 2: Evaluate the partial derivatives at the point (1,2).
∂f
∂x (1,2) = 2·1
2=2
2
∂f
∂y (1,2) = 2·2
1=4
1
Step 3: Assemble the Jacobian matrix by combining the partial derivatives.
Jf(1,2) = 2 2
4 1
Therefore, the Jacobian matrix of fat the point (1,2) is 2 2
4 1.
Question 16
Question
Let f(x, y, z)=(x2+y2+z2, exyz, xyz). Find the Jacobian matrix of f.
11
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
yexyz
yz
,∂f
∂y =
2y
xexyz
xz
,∂f
∂z =
2z
xyexyz
xy
.
Step 2: Assemble the partial derivatives into the Jacobian matrix.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y2z
yexyz xexyz xyexyz
yz xz xy
.
Question 17
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
To find the Jacobian matrix of f(x, y, z), we need to compute the partial deriva-
tives of fwith respect to each variable and organize them into a matrix.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y+y2z+z2x)=2xy +z2
Step 2: Find ∂f ∂y
∂f
∂y =∂
∂y (x2y+y2z+z2x) = x2+ 2yz
Step 3: Find ∂f ∂z
∂f
∂z =∂
∂z (x2y+y2z+z2x) = y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3) The Ja-
cobian matrix is given by:
Jf(1,2,3) = h∂f
∂x
∂f
∂y
∂f
∂z i
Substitute the values x= 1, y= 2, z= 3 into the partial derivatives:
Jf(1,2,3) = 2(1)(2) + 3212+ 2(2)(3) 2(3)(1) + 2(1)(3)
12
Jf(1,2,3) = 4 + 9 1 + 12 6 + 6
Jf(1,2,3) = 13 13 12
Therefore, the Jacobian matrix of fat the point (1,2,3) is Jf(1,2,3) =
13 13 12.
Question 18
Question
Let f(x, y, z) = (2x2+y2,3xy, 4z) be a vector-valued function. Find the Jaco-
bian matrix of fat the point (1,−1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (2x2+y2),∂
∂x (3xy),∂
∂x (4z)
= (4x, 3y, 0),
∂f
∂y =∂
∂y (2x2+y2),∂
∂y (3xy),∂
∂y (4z)
= (2y, 3x, 0),
∂f
∂z =∂
∂z (2x2+y2),∂
∂z (3xy),∂
∂z (4z)
= (0,0,4).
Step 2: Evaluate the partial derivatives at the point (1,−1,2).
∂f
∂x (1,−1,2) = (4(1),3(−1),0) = (4,−3,0),
∂f
∂y (1,−1,2) = (2(−1),3(1),0) = (−2,3,0),
∂f
∂z (1,−1,2) = (0,0,4).
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
into a 3 ×3 matrix.
Jf(1,−1,2) =
4−3 0
−230
0 0 4
.
Therefore, the Jacobian matrix of fat the point (1,−1,2) is
4−3 0
−230
0 0 4
.
13
Question 19
Question
Let f:R2→R2be a differentiable function defined by f(x, y)=(x2+y, xy2).
Determine the Jacobian matrix of fat the point (1,2).
Solution
Step 1: Write the components of f. The function f(x, y)=(x2+y, xy2) has
components f1(x, y) = x2+yand f2(x, y) = xy2.
Step 2: Calculate the partial derivatives of f1and f2. The partial derivatives
of f1are:
∂
∂x (x2+y) = 2xand ∂
∂y (x2+y) = 1
The partial derivatives of f2are:
∂
∂x (xy2) = y2and ∂
∂y (xy2)=2xy
Step 3: Evaluate the Jacobian matrix at (1,2). The Jacobian matrix of fis:
Jf(x, y) = "∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute x= 1 and y= 2 into the partial derivatives:
Jf(1,2) = 2(1) 1
2(2) 1(2)=2 1
4 2
Therefore, the Jacobian matrix of fat the point (1,2) is 2 1
4 2.
Question 20
Question
Let f:R3→R3be the vector-valued function defined by f(x, y, z) = 2xz, x2−z2,3y.
Find the Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of fis given by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
14
Step 2: We first need to calculate the partial derivatives of fwith respect
to x,y, and z.
Step 3: We have
∂f1
∂x = 2z, ∂f1
∂y = 0,∂f1
∂z = 2x
Step 4: Similarly, we find
∂f2
∂x = 2x, ∂f2
∂y = 0,∂f2
∂z =−2z
Step 5: Lastly, we calculate
∂f3
∂x = 0,∂f3
∂y = 3,∂f3
∂z = 0
Step 6: Now we can construct the Jacobian matrix Jfusing these partial
derivatives:
Jf=
2z0 2x
2x0−2z
0 3 0
Therefore, the Jacobian matrix of fis
Jf=
2z0 2x
2x0−2z
0 3 0
Question 21
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2+y2, y2+z2, z2+x2).
Find the Jacobian matrix of fat the point P= (1,1,1).
Solution
Step 1: Compute the partial derivatives of fwith respect to x, y, and z.
∂f
∂x = (2x, 0,2z),∂f
∂y = (0,2y, 2x),∂f
∂z = (0,2z, 2y)
Step 2: Evaluate the partial derivatives at the point P= (1,1,1).
∂f
∂x = (2,0,2) ,∂f
∂y = (0,2,2) ,∂f
∂z = (0,2,2)
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,1,1) =
202
022
022
15
Therefore, the Jacobian matrix of fat the point P= (1,1,1) is
202
022
022
.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y2,3xz). 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 input variable.
Step 1: Calculate the partial derivatives of the first component function
f1(x, y, z)=2x+y2.
∂f1
∂x = 2
∂f1
∂y = 2y
∂f1
∂z = 0
Step 2: Calculate the partial derivatives of the second component function
f2(x, y, z)=3xz.
∂f2
∂x = 3z
∂f2
∂y = 0
∂f2
∂z = 3x
Step 3: Assemble the partial derivatives into a 2 ×3 Jacobian matrix.
Jacobian Matrix of f="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
=2 2y0
3z0 3x
Therefore, the Jacobian matrix of fis 2 2y0
3z0 3x.
16
Question 23
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+
yz, xy −z). Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f:R3→R2, we first find the partial
derivatives of the components of f.
Step 1: Find ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
Step 2: Find ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z =−1
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
Substitute the partial derivatives we found into the Jacobian matrix:
J(f) = 2x z y
y x −1
So, the Jacobian matrix of fis 2x z y
y x −1.
Question 24
Question
Let f:R3→R2be a vector-valued function defined by f(x, y, z) = x2+yz, z2−xz.
Compute the Jacobian matrix of fat the point (1,−1,2).
Solution
To compute the Jacobian matrix of fat the point (1,−1,2), we need to find the
partial derivatives of each component function and evaluate them at the given
point. Recall that the Jacobian matrix is defined as follows:
17
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
where f1and f2are the component functions. Let’s start by finding the
partial derivatives:
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =−z, ∂f2
∂y = 0,∂f2
∂z = 2z−x
Now, evaluate these partial derivatives at the point (1,−1,2):
Step 3: Evaluate the derivatives at (1,−1,2).
∂f1
∂x = 2(1) = 2,∂f1
∂y = 2,∂f1
∂z =−1
∂f2
∂x =−2,∂f2
∂y = 0,∂f2
∂z = 2(2) −1=3
Step 4: Construct the Jacobian matrix Jfat (1,−1,2).
Jf=2 2 −1
−2 0 3
Therefore, the Jacobian matrix of fat the point (1,−1,2) is:
Jf=2 2 −1
−2 0 3
Question 25
Question
Let f(x, y, z)=(x2+y, 2y, z2) be a vector-valued function. Compute the Jaco-
bian matrix of f.
Solution
Step 1: The Jacobian matrix of a vector-valued function is a matrix of its first-
order partial derivatives. In this case, we need to find ∂f
∂x ,∂f
∂y , and ∂f
∂z .
18
Step 2: Let’s start by finding ∂f
∂x . We have:
∂f
∂x =
∂
∂x (x2+y)
∂
∂x (2y)
∂
∂x (z2)
=
2x
0
0
Step 3: Next, let’s find ∂f
∂y . We have:
∂f
∂y =
∂
∂y (x2+y)
∂
∂y (2y)
∂
∂y (z2)
=
1
2
0
Step 4: Finally, let’s find ∂f
∂z . We have:
∂f
∂z =
∂
∂z (x2+y)
∂
∂z (2y)
∂
∂z (z2)
=
0
0
2z
Step 5: Putting all the partial derivatives together, we get the Jacobian
matrix of f:
Jf=
2x1 0
0 2 0
0 0 2z
Question 26
Question
Let f(x, y) = sin(xy)
ey2cos(x). Compute the Jacobian matrix of fat the point
(π, 0).
Solution
To find the Jacobian matrix of fat the point (π, 0), we need to compute the
partial derivatives of fwith respect to xand yat that point.
Step 1: Compute ∂f1∂x and ∂f2
∂x
∂f1
∂x =∂
∂x (sin(xy)) = ycos(xy)
∂f2
∂x =∂
∂x (ey2cos(x)) = −ey2sin(x)
Step 2: Compute ∂f1∂y and ∂f2
∂y
∂f1
∂y =∂
∂y (sin(xy)) = xcos(xy)
∂f2
∂y =∂
∂y (ey2cos(x)) = 2yey2cos(x)
19
Step 3: Evaluate the Jacobian matrix at the point (π, 0) The Jacobian
matrix Jfis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !
Substitute the point (π, 0) into the partial derivatives we found earlier:
Jf(π, 0) = 0π
−e0sin(π) 0=0π
0 0
Therefore, the Jacobian matrix of fat the point (π, 0) is 0π
0 0.
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z) = (xy +z, x2+y2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of fis given by
J(f) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !,
where f1and f2are the components of f.
Step 2: We calculate the partial derivatives of f:
∂f1
∂x =y, ∂f1
∂y =x, ∂f1
∂z = 1
∂f2
∂x = 2x, ∂f2
∂y = 2y, ∂f2
∂z = 0
Step 3: Evaluate these partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 1
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 4,∂f2
∂z (1,2,3) = 0
Step 4: Finally, the Jacobian matrix of fat (1,2,3) is
J(f)(1,2,3) =211
240.
20
Question 28
Question
Consider the following system of equations:
2x2−y2+z2= 1
x2+y2+z2= 9
3x+ 2y−z= 5
Find the Jacobian matrix of the system at the point (x, y, z) = (1,2,2).
Solution
Step 1: Write the equations in a clearer form using matrix notation: Let F(x)
represent the system of equations, where x= (x, y, z):
F(x) =
2x2−y2+z2−1
x2+y2+z2−9
3x+ 2y−z−5
Step 2: Calculate the Jacobian matrix Jof F: The Jacobian matrix Jis
given by:
J=
∂F1
∂x
∂F1
∂y
∂F1
∂z
∂F2
∂x
∂F2
∂y
∂F2
∂z
∂F3
∂x
∂F3
∂y
∂F3
∂z
Step 3: Calculate the partial derivatives of each component of F:
∂F1
∂x = 4x, ∂F1
∂y =−2y, ∂F1
∂z = 2z
∂F2
∂x = 2x, ∂F2
∂y = 2y, ∂F2
∂z = 2z
∂F3
∂x = 3,∂F3
∂y = 2,∂F3
∂z =−1
Step 4: Evaluate the Jacobian matrix at (x, y, z) = (1,2,2):
J=
4−4 4
2 4 4
3 2 −1
Question 29
Question
Let f:R3→R2be the mapping defined by f(x, y, z) = (x2+y2, yz). Find the
Jacobian matrix of f.
21
Solution
Step 1: The Jacobian matrix of a mapping f:Rn→Rmis an m×nmatrix
whose entries are the first partial derivatives of the components of f.
Step 2: In this case, f(x, y, z)=(x2+y2, yz). We need to find the partial
derivatives of fwith respect to x,y, and z.
Step 3: The first component of fis x2+y2. The partial derivatives of this
component are:
∂
∂x (x2+y2)=2xand ∂
∂y (x2+y2)=2y.
Step 4: The second component of fis yz. The partial derivatives of this
component are:
∂
∂y (yz) = zand ∂
∂z (yz) = y.
Step 5: Therefore, the Jacobian matrix of fis:
Jf=2x2y0
0z y.
Question 30
Question
Let f(x, y, z) =
x2y
xyz
z2
. Find the Jacobian matrix of fat the point P(1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2xy
yz
0
∂f
∂y =
x2
xz
0
∂f
∂z =
0
xy
2z
Step 2: Evaluate the partial derivatives at the point P(1,2,3).
∂f
∂x (1,2,3) =
4
6
0
22
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Computing ∂f1∂x,∂f1
∂y , and ∂f1
∂z
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
∂f1
∂z = 0 (since f1does not depend on z)
Step 2: Computing ∂f2∂x,∂f2
∂y , and ∂f2
∂z
∂f2
∂x =∂
∂x (yz)=0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 3: Constructing the Jacobian matrix The Jacobian matrix Jfof
fis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x1 0
0z y
Question 3
Question
Let f:R2→R2be the transformation given by f(x, y) = (3x2y, x3). Find the
Jacobian matrix of fat the point (1,2).
Solution
To find the Jacobian matrix of fat the point (1,2), we need to compute the
partial derivatives of fwith respect to xand yat that point.
Step 1: Compute ∂f1∂x and ∂f1
∂y
∂f1
∂x =∂
∂x (3x2y) = 6xy
∂f1
∂y =∂
∂y (3x2y)=3x2
2
Step 2: Compute ∂f2∂x and ∂f2
∂y
∂f2
∂x =∂
∂x (x3)=3x2
∂f2
∂y =∂
∂y (x3)=0
Step 3: Construct the Jacobian matrix The Jacobian matrix of fat
the point (1,2) is given by:
Jf= ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=6xy 3x2
3x20
Step 4: Evaluate the Jacobian matrix at (1,2) Substitute x= 1 and
y= 2 into the Jacobian matrix:
Jf(1,2) = 6(1)(2) 3(1)2
3(1)20=12 3
3 0
Therefore, the Jacobian matrix of fat the point (1,2) is 12 3
3 0.
Question 4
Question
Let f:R3→R2be a transformation defined by f(x, y, z)=(x2+y, xyz).
Compute the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of fis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !
where f1and f2are the components of the transformation f.
Step 2: Compute the partial derivatives of f:
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Evaluate the partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 0
3
∂f2
∂x (1,2,3) = 2(3) = 6,∂f2
∂y (1,2,3) = 1(3) = 3,∂f2
∂z (1,2,3) = 1(2) = 2
Step 4: Assemble the Jacobian matrix at the point (1,2,3):
Jf(1,2,3) = 210
632
Question 5
Question
Let f:R3→R2be a vector-valued function defined by f(x, y, z)=(xy +z, x2−
yz). 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 =y
2x,∂f
∂y =x
−z,∂f
∂z =1
−y
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2
2,∂f
∂y (1,2,3) = 1
−3,∂f
∂z (1,2,3) = 1
−2
Step 3: Assemble these partial derivatives into the Jacobian matrix of fat
(1,2,3).
Jf(1,2,3) = 2 1 1
2−3−2
Question 6
Question
Let f(x, y, z)=2x2y+y2z+xyz. Compute the Jacobian matrix of fat the
point (1,2,3).
Solution
To compute the Jacobian matrix of fat the point (1,2,3), we need to find the
partial derivatives of fwith respect to each variable and evaluate them at that
point.
Step 1: Compute ∂f
∂x .
∂f
∂x = 4xy +yz
4
Step 2: Evaluate ∂f
∂x at (1,2,3).
∂f
∂x (1,2,3) = 4(1)(2) + (2)(3) = 8 + 6 = 14
Step 3: Compute ∂f
∂y .
∂f
∂y = 2x2+ 2yz +xz
Step 4: Evaluate ∂f
∂y at (1,2,3).
∂f
∂y (1,2,3) = 2(1)2+ 2(2)(3) + (1)(3) = 2 + 12 + 3 = 17
Step 5: Compute ∂f
∂z .
∂f
∂z =y2+xy
Step 6: Evaluate ∂f
∂z at (1,2,3).
∂f
∂z (1,2,3) = (2)2+ (1)(2) = 4 + 2 = 6
Step 7: Assemble the Jacobian matrix Jfat the point (1,2,3).
Jf(1,2,3) = ∂f
∂x
∂f
∂y
∂f
∂z =14 17 6
Question 7
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+y, y2+z). Find
the Jacobian matrix of fat the point (1,2,−1).
Solution
To find the Jacobian matrix of fat the point (1,2,−1), we need to compute the
partial derivatives of fwith respect to each input variable and evaluate them
at the given point.
Step 1: Compute the partial derivatives of f: Let u=x2+yand v=y2+z.
Then, f1=∂u
∂x = 2x,f2=∂u
∂y = 1, f3=∂v
∂y = 2y,f4=∂v
∂z = 1.
Step 2: Evaluate the partial derivatives at (1,2,−1): At (1,2,−1), we have
f1(1,2,−1) = 2(1) = 2, f2(1,2,−1) = 1, f3(1,2,−1) = 2(2) = 4, f4(1,2,−1) =
1.
5
Step 3: Assemble the Jacobian matrix: The Jacobian matrix of fat
(1,2,−1) is given by
Jf(1,2,−1) = f1f2
f3f4=2 1
4 1.
Therefore, the Jacobian matrix of fat the point (1,2,−1) is 2 1
4 1.
Question 8
Question
Let f:R3→R2be defined by f(x, y, z) = (2x+y, x −3z). Determine the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we first need to compute the partial derivatives
of fwith respect to each variable.
Step 1: Compute ∂f1∂x
f1(x, y, z)=2x+y
∂f1
∂x =∂(2x+y)
∂x = 2
Step 2: Compute ∂f1∂y
f1(x, y, z)=2x+y
∂f1
∂y =∂(2x+y)
∂y = 1
Step 3: Compute ∂f1∂z Since f1does not depend on z,∂f1
∂z = 0.
Step 4: Compute ∂f2∂x
f2(x, y, z) = x−3z
∂f2
∂x =∂(x−3z)
∂x = 1
Step 5: Compute ∂f2∂y Since f2does not depend on y,∂f2
∂y = 0.
Step 6: Compute ∂f2∂z
f2(x, y, z) = x−3z
∂f2
∂z =∂(x−3z)
∂z =−3
Therefore, the Jacobian matrix of the function fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2 1 0
1 0 −3
6
Question 9
Question
Let f:R⊯→R⊭be a function defined by f(x, y, z)=(x2+yz, 3x−z). Calculate
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y ,and ∂f1
∂z for the first component function f1(x, y, z) =
x2+yz.
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Find ∂f2
∂x ,∂f2
∂y ,and ∂f2
∂z for the second component function f2(x, y, z) =
3x−z.
∂f2
∂x = 3
∂f2
∂y = 0
∂f2
∂z =−1
Step 3: Assemble the Jacobian matrix Jfusing the partial derivatives
calculated above.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
3 0 −1
Therefore, the Jacobian matrix of the function fis:
Jf=2x z y
3 0 −1
Question 10
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+y2, xz). Compute the Jacobian
matrix of fat the point (1,2,−1).
7
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable.
Given f(x, y, z)=(x2+y2, xz), we find the partial derivatives of fas follows:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
Substitute x= 1, y= 2, z=−1 into the partial derivatives:
∂f1
∂x (1,2,−1) = 2(1) = 2,∂f1
∂y (1,2,−1) = 2(2) = 4,∂f1
∂z (1,2,−1) = 0
∂f2
∂x (1,2,−1) = −1,∂f2
∂y (1,2,−1) = 0,∂f2
∂z (1,2,−1) = 1
Step 3: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,−1) is given by:
Jf(1,2,−1) = ∂f1
∂x (1,2,−1) ∂f1
∂y (1,2,−1) ∂f1
∂z (1,2,−1)
∂f2
∂x (1,2,−1) ∂f2
∂y (1,2,−1) ∂f2
∂z (1,2,−1)!
=2 4 0
−101
Question 11
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+yz, xy +z, xz +y).
Determine 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 = (2x, y, z),
∂f
∂y = (z, x, 1) ,
∂f
∂z = (y, 1, x).
8
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1),2,3) = (2,2,3),
∂f
∂y (1,2,3)
= (3,1,1),
∂f
∂z (1,2,3)
= (2,1,1).
Therefore, the Jacobian matrix of fat the point (1,2,3) is
Jf(1,2,3) =
223
311
211
.
Question 12
Question
Let f(x, y, z) =
x2+yz
y2+zx
z2+xy
. Find the Jacobian matrix of f.
Solution
Step 1: Compute the partial derivatives of f(x, y, z) with respect to x,y, and
z.
∂f
∂x =
2x
z
y
,∂f
∂y =
z
2y
x
,∂f
∂z =
y
x
2z
Step 2: Assemble the partial derivatives into a matrix to obtain the Jacobian
matrix of f.
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 13
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, xyz). Find the
Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x z y
yz xz xy
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y+yz, xz). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
fwith respect to each variable.
Step 1: Compute the partial derivatives of f.
∂f
∂x =∂
∂x (x2y+yz),∂
∂x (xz)
= (2xy, z)
Step 2: Compute the partial derivatives of f.
∂f
∂y =∂
∂y (x2y+yz),∂
∂y (xz)
= (x2+z, 0)
10
Step 3: Compute the partial derivatives of f.
∂f
∂z =∂
∂z (x2y+yz),∂
∂z (xz)
= (y, x)
Step 4: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf=
2xy z
x2+z0
y x
Question 15
Question
Let f:R2→R2be defined by f(x, y) = (x2+y2, xy). Calculate the Jacobian
matrix of fat the point (1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (x2+y2)
∂
∂x (xy)=2x
y
∂f
∂y = ∂
∂y (x2+y2)
∂
∂y (xy)!=2y
x
Step 2: Evaluate the partial derivatives at the point (1,2).
∂f
∂x (1,2) = 2·1
2=2
2
∂f
∂y (1,2) = 2·2
1=4
1
Step 3: Assemble the Jacobian matrix by combining the partial derivatives.
Jf(1,2) = 2 2
4 1
Therefore, the Jacobian matrix of fat the point (1,2) is 2 2
4 1.
Question 16
Question
Let f(x, y, z)=(x2+y2+z2, exyz, xyz). Find the Jacobian matrix of f.
11
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
yexyz
yz
,∂f
∂y =
2y
xexyz
xz
,∂f
∂z =
2z
xyexyz
xy
.
Step 2: Assemble the partial derivatives into the Jacobian matrix.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y2z
yexyz xexyz xyexyz
yz xz xy
.
Question 17
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
To find the Jacobian matrix of f(x, y, z), we need to compute the partial deriva-
tives of fwith respect to each variable and organize them into a matrix.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y+y2z+z2x)=2xy +z2
Step 2: Find ∂f ∂y
∂f
∂y =∂
∂y (x2y+y2z+z2x) = x2+ 2yz
Step 3: Find ∂f ∂z
∂f
∂z =∂
∂z (x2y+y2z+z2x) = y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3) The Ja-
cobian matrix is given by:
Jf(1,2,3) = h∂f
∂x
∂f
∂y
∂f
∂z i
Substitute the values x= 1, y= 2, z= 3 into the partial derivatives:
Jf(1,2,3) = 2(1)(2) + 3212+ 2(2)(3) 2(3)(1) + 2(1)(3)
12
Jf(1,2,3) = 4 + 9 1 + 12 6 + 6
Jf(1,2,3) = 13 13 12
Therefore, the Jacobian matrix of fat the point (1,2,3) is Jf(1,2,3) =
13 13 12.
Question 18
Question
Let f(x, y, z) = (2x2+y2,3xy, 4z) be a vector-valued function. Find the Jaco-
bian matrix of fat the point (1,−1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (2x2+y2),∂
∂x (3xy),∂
∂x (4z)
= (4x, 3y, 0),
∂f
∂y =∂
∂y (2x2+y2),∂
∂y (3xy),∂
∂y (4z)
= (2y, 3x, 0),
∂f
∂z =∂
∂z (2x2+y2),∂
∂z (3xy),∂
∂z (4z)
= (0,0,4).
Step 2: Evaluate the partial derivatives at the point (1,−1,2).
∂f
∂x (1,−1,2) = (4(1),3(−1),0) = (4,−3,0),
∂f
∂y (1,−1,2) = (2(−1),3(1),0) = (−2,3,0),
∂f
∂z (1,−1,2) = (0,0,4).
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
into a 3 ×3 matrix.
Jf(1,−1,2) =
4−3 0
−230
0 0 4
.
Therefore, the Jacobian matrix of fat the point (1,−1,2) is
4−3 0
−230
0 0 4
.
13
Question 19
Question
Let f:R2→R2be a differentiable function defined by f(x, y)=(x2+y, xy2).
Determine the Jacobian matrix of fat the point (1,2).
Solution
Step 1: Write the components of f. The function f(x, y)=(x2+y, xy2) has
components f1(x, y) = x2+yand f2(x, y) = xy2.
Step 2: Calculate the partial derivatives of f1and f2. The partial derivatives
of f1are:
∂
∂x (x2+y) = 2xand ∂
∂y (x2+y) = 1
The partial derivatives of f2are:
∂
∂x (xy2) = y2and ∂
∂y (xy2)=2xy
Step 3: Evaluate the Jacobian matrix at (1,2). The Jacobian matrix of fis:
Jf(x, y) = "∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute x= 1 and y= 2 into the partial derivatives:
Jf(1,2) = 2(1) 1
2(2) 1(2)=2 1
4 2
Therefore, the Jacobian matrix of fat the point (1,2) is 2 1
4 2.
Question 20
Question
Let f:R3→R3be the vector-valued function defined by f(x, y, z) = 2xz, x2−z2,3y.
Find the Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of fis given by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
14
Step 2: We first need to calculate the partial derivatives of fwith respect
to x,y, and z.
Step 3: We have
∂f1
∂x = 2z, ∂f1
∂y = 0,∂f1
∂z = 2x
Step 4: Similarly, we find
∂f2
∂x = 2x, ∂f2
∂y = 0,∂f2
∂z =−2z
Step 5: Lastly, we calculate
∂f3
∂x = 0,∂f3
∂y = 3,∂f3
∂z = 0
Step 6: Now we can construct the Jacobian matrix Jfusing these partial
derivatives:
Jf=
2z0 2x
2x0−2z
0 3 0
Therefore, the Jacobian matrix of fis
Jf=
2z0 2x
2x0−2z
0 3 0
Question 21
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2+y2, y2+z2, z2+x2).
Find the Jacobian matrix of fat the point P= (1,1,1).
Solution
Step 1: Compute the partial derivatives of fwith respect to x, y, and z.
∂f
∂x = (2x, 0,2z),∂f
∂y = (0,2y, 2x),∂f
∂z = (0,2z, 2y)
Step 2: Evaluate the partial derivatives at the point P= (1,1,1).
∂f
∂x = (2,0,2) ,∂f
∂y = (0,2,2) ,∂f
∂z = (0,2,2)
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,1,1) =
202
022
022
15
Therefore, the Jacobian matrix of fat the point P= (1,1,1) is
202
022
022
.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y2,3xz). 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 input variable.
Step 1: Calculate the partial derivatives of the first component function
f1(x, y, z)=2x+y2.
∂f1
∂x = 2
∂f1
∂y = 2y
∂f1
∂z = 0
Step 2: Calculate the partial derivatives of the second component function
f2(x, y, z)=3xz.
∂f2
∂x = 3z
∂f2
∂y = 0
∂f2
∂z = 3x
Step 3: Assemble the partial derivatives into a 2 ×3 Jacobian matrix.
Jacobian Matrix of f="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
=2 2y0
3z0 3x
Therefore, the Jacobian matrix of fis 2 2y0
3z0 3x.
16
Question 23
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+
yz, xy −z). Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f:R3→R2, we first find the partial
derivatives of the components of f.
Step 1: Find ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
Step 2: Find ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z =−1
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
Substitute the partial derivatives we found into the Jacobian matrix:
J(f) = 2x z y
y x −1
So, the Jacobian matrix of fis 2x z y
y x −1.
Question 24
Question
Let f:R3→R2be a vector-valued function defined by f(x, y, z) = x2+yz, z2−xz.
Compute the Jacobian matrix of fat the point (1,−1,2).
Solution
To compute the Jacobian matrix of fat the point (1,−1,2), we need to find the
partial derivatives of each component function and evaluate them at the given
point. Recall that the Jacobian matrix is defined as follows:
17
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
where f1and f2are the component functions. Let’s start by finding the
partial derivatives:
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =−z, ∂f2
∂y = 0,∂f2
∂z = 2z−x
Now, evaluate these partial derivatives at the point (1,−1,2):
Step 3: Evaluate the derivatives at (1,−1,2).
∂f1
∂x = 2(1) = 2,∂f1
∂y = 2,∂f1
∂z =−1
∂f2
∂x =−2,∂f2
∂y = 0,∂f2
∂z = 2(2) −1=3
Step 4: Construct the Jacobian matrix Jfat (1,−1,2).
Jf=2 2 −1
−2 0 3
Therefore, the Jacobian matrix of fat the point (1,−1,2) is:
Jf=2 2 −1
−2 0 3
Question 25
Question
Let f(x, y, z)=(x2+y, 2y, z2) be a vector-valued function. Compute the Jaco-
bian matrix of f.
Solution
Step 1: The Jacobian matrix of a vector-valued function is a matrix of its first-
order partial derivatives. In this case, we need to find ∂f
∂x ,∂f
∂y , and ∂f
∂z .
18
Step 2: Let’s start by finding ∂f
∂x . We have:
∂f
∂x =
∂
∂x (x2+y)
∂
∂x (2y)
∂
∂x (z2)
=
2x
0
0
Step 3: Next, let’s find ∂f
∂y . We have:
∂f
∂y =
∂
∂y (x2+y)
∂
∂y (2y)
∂
∂y (z2)
=
1
2
0
Step 4: Finally, let’s find ∂f
∂z . We have:
∂f
∂z =
∂
∂z (x2+y)
∂
∂z (2y)
∂
∂z (z2)
=
0
0
2z
Step 5: Putting all the partial derivatives together, we get the Jacobian
matrix of f:
Jf=
2x1 0
0 2 0
0 0 2z
Question 26
Question
Let f(x, y) = sin(xy)
ey2cos(x). Compute the Jacobian matrix of fat the point
(π, 0).
Solution
To find the Jacobian matrix of fat the point (π, 0), we need to compute the
partial derivatives of fwith respect to xand yat that point.
Step 1: Compute ∂f1∂x and ∂f2
∂x
∂f1
∂x =∂
∂x (sin(xy)) = ycos(xy)
∂f2
∂x =∂
∂x (ey2cos(x)) = −ey2sin(x)
Step 2: Compute ∂f1∂y and ∂f2
∂y
∂f1
∂y =∂
∂y (sin(xy)) = xcos(xy)
∂f2
∂y =∂
∂y (ey2cos(x)) = 2yey2cos(x)
19
Step 3: Evaluate the Jacobian matrix at the point (π, 0) The Jacobian
matrix Jfis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !
Substitute the point (π, 0) into the partial derivatives we found earlier:
Jf(π, 0) = 0π
−e0sin(π) 0=0π
0 0
Therefore, the Jacobian matrix of fat the point (π, 0) is 0π
0 0.
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z) = (xy +z, x2+y2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of fis given by
J(f) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !,
where f1and f2are the components of f.
Step 2: We calculate the partial derivatives of f:
∂f1
∂x =y, ∂f1
∂y =x, ∂f1
∂z = 1
∂f2
∂x = 2x, ∂f2
∂y = 2y, ∂f2
∂z = 0
Step 3: Evaluate these partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 1
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 4,∂f2
∂z (1,2,3) = 0
Step 4: Finally, the Jacobian matrix of fat (1,2,3) is
J(f)(1,2,3) =211
240.
20
Question 28
Question
Consider the following system of equations:
2x2−y2+z2= 1
x2+y2+z2= 9
3x+ 2y−z= 5
Find the Jacobian matrix of the system at the point (x, y, z) = (1,2,2).
Solution
Step 1: Write the equations in a clearer form using matrix notation: Let F(x)
represent the system of equations, where x= (x, y, z):
F(x) =
2x2−y2+z2−1
x2+y2+z2−9
3x+ 2y−z−5
Step 2: Calculate the Jacobian matrix Jof F: The Jacobian matrix Jis
given by:
J=
∂F1
∂x
∂F1
∂y
∂F1
∂z
∂F2
∂x
∂F2
∂y
∂F2
∂z
∂F3
∂x
∂F3
∂y
∂F3
∂z
Step 3: Calculate the partial derivatives of each component of F:
∂F1
∂x = 4x, ∂F1
∂y =−2y, ∂F1
∂z = 2z
∂F2
∂x = 2x, ∂F2
∂y = 2y, ∂F2
∂z = 2z
∂F3
∂x = 3,∂F3
∂y = 2,∂F3
∂z =−1
Step 4: Evaluate the Jacobian matrix at (x, y, z) = (1,2,2):
J=
4−4 4
2 4 4
3 2 −1
Question 29
Question
Let f:R3→R2be the mapping defined by f(x, y, z) = (x2+y2, yz). Find the
Jacobian matrix of f.
21
Solution
Step 1: The Jacobian matrix of a mapping f:Rn→Rmis an m×nmatrix
whose entries are the first partial derivatives of the components of f.
Step 2: In this case, f(x, y, z)=(x2+y2, yz). We need to find the partial
derivatives of fwith respect to x,y, and z.
Step 3: The first component of fis x2+y2. The partial derivatives of this
component are:
∂
∂x (x2+y2)=2xand ∂
∂y (x2+y2)=2y.
Step 4: The second component of fis yz. The partial derivatives of this
component are:
∂
∂y (yz) = zand ∂
∂z (yz) = y.
Step 5: Therefore, the Jacobian matrix of fis:
Jf=2x2y0
0z y.
Question 30
Question
Let f(x, y, z) =
x2y
xyz
z2
. Find the Jacobian matrix of fat the point P(1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2xy
yz
0
∂f
∂y =
x2
xz
0
∂f
∂z =
0
xy
2z
Step 2: Evaluate the partial derivatives at the point P(1,2,3).
∂f
∂x (1,2,3) =
4
6
0
22
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Computing ∂f1∂x,∂f1
∂y , and ∂f1
∂z
∂f1
∂x =∂
∂x (x2+y)=2x
∂f1
∂y =∂
∂y (x2+y)=1
∂f1
∂z = 0 (since f1does not depend on z)
Step 2: Computing ∂f2∂x,∂f2
∂y , and ∂f2
∂z
∂f2
∂x =∂
∂x (yz)=0
∂f2
∂y =∂
∂y (yz) = z
∂f2
∂z =∂
∂z (yz) = y
Step 3: Constructing the Jacobian matrix The Jacobian matrix Jfof
fis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x1 0
0z y
Question 3
Question
Let f:R2→R2be the transformation given by f(x, y) = (3x2y, x3). Find the
Jacobian matrix of fat the point (1,2).
Solution
To find the Jacobian matrix of fat the point (1,2), we need to compute the
partial derivatives of fwith respect to xand yat that point.
Step 1: Compute ∂f1∂x and ∂f1
∂y
∂f1
∂x =∂
∂x (3x2y) = 6xy
∂f1
∂y =∂
∂y (3x2y)=3x2
2
Step 2: Compute ∂f2∂x and ∂f2
∂y
∂f2
∂x =∂
∂x (x3)=3x2
∂f2
∂y =∂
∂y (x3)=0
Step 3: Construct the Jacobian matrix The Jacobian matrix of fat
the point (1,2) is given by:
Jf= ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=6xy 3x2
3x20
Step 4: Evaluate the Jacobian matrix at (1,2) Substitute x= 1 and
y= 2 into the Jacobian matrix:
Jf(1,2) = 6(1)(2) 3(1)2
3(1)20=12 3
3 0
Therefore, the Jacobian matrix of fat the point (1,2) is 12 3
3 0.
Question 4
Question
Let f:R3→R2be a transformation defined by f(x, y, z)=(x2+y, xyz).
Compute the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of fis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !
where f1and f2are the components of the transformation f.
Step 2: Compute the partial derivatives of f:
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 3: Evaluate the partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 0
3
∂f2
∂x (1,2,3) = 2(3) = 6,∂f2
∂y (1,2,3) = 1(3) = 3,∂f2
∂z (1,2,3) = 1(2) = 2
Step 4: Assemble the Jacobian matrix at the point (1,2,3):
Jf(1,2,3) = 210
632
Question 5
Question
Let f:R3→R2be a vector-valued function defined by f(x, y, z)=(xy +z, x2−
yz). 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 =y
2x,∂f
∂y =x
−z,∂f
∂z =1
−y
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3) = 2
2,∂f
∂y (1,2,3) = 1
−3,∂f
∂z (1,2,3) = 1
−2
Step 3: Assemble these partial derivatives into the Jacobian matrix of fat
(1,2,3).
Jf(1,2,3) = 2 1 1
2−3−2
Question 6
Question
Let f(x, y, z)=2x2y+y2z+xyz. Compute the Jacobian matrix of fat the
point (1,2,3).
Solution
To compute the Jacobian matrix of fat the point (1,2,3), we need to find the
partial derivatives of fwith respect to each variable and evaluate them at that
point.
Step 1: Compute ∂f
∂x .
∂f
∂x = 4xy +yz
4
Step 2: Evaluate ∂f
∂x at (1,2,3).
∂f
∂x (1,2,3) = 4(1)(2) + (2)(3) = 8 + 6 = 14
Step 3: Compute ∂f
∂y .
∂f
∂y = 2x2+ 2yz +xz
Step 4: Evaluate ∂f
∂y at (1,2,3).
∂f
∂y (1,2,3) = 2(1)2+ 2(2)(3) + (1)(3) = 2 + 12 + 3 = 17
Step 5: Compute ∂f
∂z .
∂f
∂z =y2+xy
Step 6: Evaluate ∂f
∂z at (1,2,3).
∂f
∂z (1,2,3) = (2)2+ (1)(2) = 4 + 2 = 6
Step 7: Assemble the Jacobian matrix Jfat the point (1,2,3).
Jf(1,2,3) = ∂f
∂x
∂f
∂y
∂f
∂z =14 17 6
Question 7
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+y, y2+z). Find
the Jacobian matrix of fat the point (1,2,−1).
Solution
To find the Jacobian matrix of fat the point (1,2,−1), we need to compute the
partial derivatives of fwith respect to each input variable and evaluate them
at the given point.
Step 1: Compute the partial derivatives of f: Let u=x2+yand v=y2+z.
Then, f1=∂u
∂x = 2x,f2=∂u
∂y = 1, f3=∂v
∂y = 2y,f4=∂v
∂z = 1.
Step 2: Evaluate the partial derivatives at (1,2,−1): At (1,2,−1), we have
f1(1,2,−1) = 2(1) = 2, f2(1,2,−1) = 1, f3(1,2,−1) = 2(2) = 4, f4(1,2,−1) =
1.
5
Step 3: Assemble the Jacobian matrix: The Jacobian matrix of fat
(1,2,−1) is given by
Jf(1,2,−1) = f1f2
f3f4=2 1
4 1.
Therefore, the Jacobian matrix of fat the point (1,2,−1) is 2 1
4 1.
Question 8
Question
Let f:R3→R2be defined by f(x, y, z) = (2x+y, x −3z). Determine the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we first need to compute the partial derivatives
of fwith respect to each variable.
Step 1: Compute ∂f1∂x
f1(x, y, z)=2x+y
∂f1
∂x =∂(2x+y)
∂x = 2
Step 2: Compute ∂f1∂y
f1(x, y, z)=2x+y
∂f1
∂y =∂(2x+y)
∂y = 1
Step 3: Compute ∂f1∂z Since f1does not depend on z,∂f1
∂z = 0.
Step 4: Compute ∂f2∂x
f2(x, y, z) = x−3z
∂f2
∂x =∂(x−3z)
∂x = 1
Step 5: Compute ∂f2∂y Since f2does not depend on y,∂f2
∂y = 0.
Step 6: Compute ∂f2∂z
f2(x, y, z) = x−3z
∂f2
∂z =∂(x−3z)
∂z =−3
Therefore, the Jacobian matrix of the function fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2 1 0
1 0 −3
6
Question 9
Question
Let f:R⊯→R⊭be a function defined by f(x, y, z)=(x2+yz, 3x−z). Calculate
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y ,and ∂f1
∂z for the first component function f1(x, y, z) =
x2+yz.
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Find ∂f2
∂x ,∂f2
∂y ,and ∂f2
∂z for the second component function f2(x, y, z) =
3x−z.
∂f2
∂x = 3
∂f2
∂y = 0
∂f2
∂z =−1
Step 3: Assemble the Jacobian matrix Jfusing the partial derivatives
calculated above.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
3 0 −1
Therefore, the Jacobian matrix of the function fis:
Jf=2x z y
3 0 −1
Question 10
Question
Let f:R3→R2be defined by f(x, y, z)=(x2+y2, xz). Compute the Jacobian
matrix of fat the point (1,2,−1).
7
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable.
Given f(x, y, z)=(x2+y2, xz), we find the partial derivatives of fas follows:
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 0
∂f2
∂x =z, ∂f2
∂y = 0,∂f2
∂z =x
Step 2: Evaluate the partial derivatives at the point (1,2,−1).
Substitute x= 1, y= 2, z=−1 into the partial derivatives:
∂f1
∂x (1,2,−1) = 2(1) = 2,∂f1
∂y (1,2,−1) = 2(2) = 4,∂f1
∂z (1,2,−1) = 0
∂f2
∂x (1,2,−1) = −1,∂f2
∂y (1,2,−1) = 0,∂f2
∂z (1,2,−1) = 1
Step 3: Assemble the Jacobian matrix.
The Jacobian matrix of fat the point (1,2,−1) is given by:
Jf(1,2,−1) = ∂f1
∂x (1,2,−1) ∂f1
∂y (1,2,−1) ∂f1
∂z (1,2,−1)
∂f2
∂x (1,2,−1) ∂f2
∂y (1,2,−1) ∂f2
∂z (1,2,−1)!
=2 4 0
−101
Question 11
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2+yz, xy +z, xz +y).
Determine 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 = (2x, y, z),
∂f
∂y = (z, x, 1) ,
∂f
∂z = (y, 1, x).
8
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f
∂x (1,2,3)
= (2(1),2,3) = (2,2,3),
∂f
∂y (1,2,3)
= (3,1,1),
∂f
∂z (1,2,3)
= (2,1,1).
Therefore, the Jacobian matrix of fat the point (1,2,3) is
Jf(1,2,3) =
223
311
211
.
Question 12
Question
Let f(x, y, z) =
x2+yz
y2+zx
z2+xy
. Find the Jacobian matrix of f.
Solution
Step 1: Compute the partial derivatives of f(x, y, z) with respect to x,y, and
z.
∂f
∂x =
2x
z
y
,∂f
∂y =
z
2y
x
,∂f
∂z =
y
x
2z
Step 2: Assemble the partial derivatives into a matrix to obtain the Jacobian
matrix of f.
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 13
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, xyz). Find the
Jacobian matrix of f.
9
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz)=2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (xyz) = yz
∂f2
∂y =∂
∂y (xyz) = xz
∂f2
∂z =∂
∂z (xyz) = xy
Step 3: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x z y
yz xz xy
Question 14
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2y+yz, xz). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
fwith respect to each variable.
Step 1: Compute the partial derivatives of f.
∂f
∂x =∂
∂x (x2y+yz),∂
∂x (xz)
= (2xy, z)
Step 2: Compute the partial derivatives of f.
∂f
∂y =∂
∂y (x2y+yz),∂
∂y (xz)
= (x2+z, 0)
10
Step 3: Compute the partial derivatives of f.
∂f
∂z =∂
∂z (x2y+yz),∂
∂z (xz)
= (y, x)
Step 4: Assemble the Jacobian matrix. The Jacobian matrix of fis given
by
Jf=
2xy z
x2+z0
y x
Question 15
Question
Let f:R2→R2be defined by f(x, y) = (x2+y2, xy). Calculate the Jacobian
matrix of fat the point (1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y.
∂f
∂x =∂
∂x (x2+y2)
∂
∂x (xy)=2x
y
∂f
∂y = ∂
∂y (x2+y2)
∂
∂y (xy)!=2y
x
Step 2: Evaluate the partial derivatives at the point (1,2).
∂f
∂x (1,2) = 2·1
2=2
2
∂f
∂y (1,2) = 2·2
1=4
1
Step 3: Assemble the Jacobian matrix by combining the partial derivatives.
Jf(1,2) = 2 2
4 1
Therefore, the Jacobian matrix of fat the point (1,2) is 2 2
4 1.
Question 16
Question
Let f(x, y, z)=(x2+y2+z2, exyz, xyz). Find the Jacobian matrix of f.
11
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2x
yexyz
yz
,∂f
∂y =
2y
xexyz
xz
,∂f
∂z =
2z
xyexyz
xy
.
Step 2: Assemble the partial derivatives into the Jacobian matrix.
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
2x2y2z
yexyz xexyz xyexyz
yz xz xy
.
Question 17
Question
Let f(x, y, z) = x2y+y2z+z2x. Find the Jacobian matrix of fat the point
(1,2,3).
Solution
To find the Jacobian matrix of f(x, y, z), we need to compute the partial deriva-
tives of fwith respect to each variable and organize them into a matrix.
Step 1: Find ∂f ∂x
∂f
∂x =∂
∂x (x2y+y2z+z2x)=2xy +z2
Step 2: Find ∂f ∂y
∂f
∂y =∂
∂y (x2y+y2z+z2x) = x2+ 2yz
Step 3: Find ∂f ∂z
∂f
∂z =∂
∂z (x2y+y2z+z2x) = y2+ 2zx
Step 4: Evaluate the Jacobian matrix at the point (1,2,3) The Ja-
cobian matrix is given by:
Jf(1,2,3) = h∂f
∂x
∂f
∂y
∂f
∂z i
Substitute the values x= 1, y= 2, z= 3 into the partial derivatives:
Jf(1,2,3) = 2(1)(2) + 3212+ 2(2)(3) 2(3)(1) + 2(1)(3)
12
Jf(1,2,3) = 4 + 9 1 + 12 6 + 6
Jf(1,2,3) = 13 13 12
Therefore, the Jacobian matrix of fat the point (1,2,3) is Jf(1,2,3) =
13 13 12.
Question 18
Question
Let f(x, y, z) = (2x2+y2,3xy, 4z) be a vector-valued function. Find the Jaco-
bian matrix of fat the point (1,−1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =∂
∂x (2x2+y2),∂
∂x (3xy),∂
∂x (4z)
= (4x, 3y, 0),
∂f
∂y =∂
∂y (2x2+y2),∂
∂y (3xy),∂
∂y (4z)
= (2y, 3x, 0),
∂f
∂z =∂
∂z (2x2+y2),∂
∂z (3xy),∂
∂z (4z)
= (0,0,4).
Step 2: Evaluate the partial derivatives at the point (1,−1,2).
∂f
∂x (1,−1,2) = (4(1),3(−1),0) = (4,−3,0),
∂f
∂y (1,−1,2) = (2(−1),3(1),0) = (−2,3,0),
∂f
∂z (1,−1,2) = (0,0,4).
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
into a 3 ×3 matrix.
Jf(1,−1,2) =
4−3 0
−230
0 0 4
.
Therefore, the Jacobian matrix of fat the point (1,−1,2) is
4−3 0
−230
0 0 4
.
13
Question 19
Question
Let f:R2→R2be a differentiable function defined by f(x, y)=(x2+y, xy2).
Determine the Jacobian matrix of fat the point (1,2).
Solution
Step 1: Write the components of f. The function f(x, y)=(x2+y, xy2) has
components f1(x, y) = x2+yand f2(x, y) = xy2.
Step 2: Calculate the partial derivatives of f1and f2. The partial derivatives
of f1are:
∂
∂x (x2+y) = 2xand ∂
∂y (x2+y) = 1
The partial derivatives of f2are:
∂
∂x (xy2) = y2and ∂
∂y (xy2)=2xy
Step 3: Evaluate the Jacobian matrix at (1,2). The Jacobian matrix of fis:
Jf(x, y) = "∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y #
Substitute x= 1 and y= 2 into the partial derivatives:
Jf(1,2) = 2(1) 1
2(2) 1(2)=2 1
4 2
Therefore, the Jacobian matrix of fat the point (1,2) is 2 1
4 2.
Question 20
Question
Let f:R3→R3be the vector-valued function defined by f(x, y, z) = 2xz, x2−z2,3y.
Find the Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of fis given by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
14
Step 2: We first need to calculate the partial derivatives of fwith respect
to x,y, and z.
Step 3: We have
∂f1
∂x = 2z, ∂f1
∂y = 0,∂f1
∂z = 2x
Step 4: Similarly, we find
∂f2
∂x = 2x, ∂f2
∂y = 0,∂f2
∂z =−2z
Step 5: Lastly, we calculate
∂f3
∂x = 0,∂f3
∂y = 3,∂f3
∂z = 0
Step 6: Now we can construct the Jacobian matrix Jfusing these partial
derivatives:
Jf=
2z0 2x
2x0−2z
0 3 0
Therefore, the Jacobian matrix of fis
Jf=
2z0 2x
2x0−2z
0 3 0
Question 21
Question
Let f:R3→R3be a function defined by f(x, y, z) = (x2+y2, y2+z2, z2+x2).
Find the Jacobian matrix of fat the point P= (1,1,1).
Solution
Step 1: Compute the partial derivatives of fwith respect to x, y, and z.
∂f
∂x = (2x, 0,2z),∂f
∂y = (0,2y, 2x),∂f
∂z = (0,2z, 2y)
Step 2: Evaluate the partial derivatives at the point P= (1,1,1).
∂f
∂x = (2,0,2) ,∂f
∂y = (0,2,2) ,∂f
∂z = (0,2,2)
Step 3: Assemble the Jacobian matrix by arranging the partial derivatives
as rows.
Jf(1,1,1) =
202
022
022
15
Therefore, the Jacobian matrix of fat the point P= (1,1,1) is
202
022
022
.
Question 22
Question
Let f:R3→R2be a function defined by f(x, y, z) = (2x+y2,3xz). 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 input variable.
Step 1: Calculate the partial derivatives of the first component function
f1(x, y, z)=2x+y2.
∂f1
∂x = 2
∂f1
∂y = 2y
∂f1
∂z = 0
Step 2: Calculate the partial derivatives of the second component function
f2(x, y, z)=3xz.
∂f2
∂x = 3z
∂f2
∂y = 0
∂f2
∂z = 3x
Step 3: Assemble the partial derivatives into a 2 ×3 Jacobian matrix.
Jacobian Matrix of f="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
=2 2y0
3z0 3x
Therefore, the Jacobian matrix of fis 2 2y0
3z0 3x.
16
Question 23
Question
Let f:R3→R2be a differentiable function defined by f(x, y, z)=(x2+
yz, xy −z). Compute the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f:R3→R2, we first find the partial
derivatives of the components of f.
Step 1: Find ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
Step 2: Find ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z =−1
Step 3: Construct the Jacobian matrix. The Jacobian matrix of fis
given by
J(f) =
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
Substitute the partial derivatives we found into the Jacobian matrix:
J(f) = 2x z y
y x −1
So, the Jacobian matrix of fis 2x z y
y x −1.
Question 24
Question
Let f:R3→R2be a vector-valued function defined by f(x, y, z) = x2+yz, z2−xz.
Compute the Jacobian matrix of fat the point (1,−1,2).
Solution
To compute the Jacobian matrix of fat the point (1,−1,2), we need to find the
partial derivatives of each component function and evaluate them at the given
point. Recall that the Jacobian matrix is defined as follows:
17
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
where f1and f2are the component functions. Let’s start by finding the
partial derivatives:
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =−z, ∂f2
∂y = 0,∂f2
∂z = 2z−x
Now, evaluate these partial derivatives at the point (1,−1,2):
Step 3: Evaluate the derivatives at (1,−1,2).
∂f1
∂x = 2(1) = 2,∂f1
∂y = 2,∂f1
∂z =−1
∂f2
∂x =−2,∂f2
∂y = 0,∂f2
∂z = 2(2) −1=3
Step 4: Construct the Jacobian matrix Jfat (1,−1,2).
Jf=2 2 −1
−2 0 3
Therefore, the Jacobian matrix of fat the point (1,−1,2) is:
Jf=2 2 −1
−2 0 3
Question 25
Question
Let f(x, y, z)=(x2+y, 2y, z2) be a vector-valued function. Compute the Jaco-
bian matrix of f.
Solution
Step 1: The Jacobian matrix of a vector-valued function is a matrix of its first-
order partial derivatives. In this case, we need to find ∂f
∂x ,∂f
∂y , and ∂f
∂z .
18
Step 2: Let’s start by finding ∂f
∂x . We have:
∂f
∂x =
∂
∂x (x2+y)
∂
∂x (2y)
∂
∂x (z2)
=
2x
0
0
Step 3: Next, let’s find ∂f
∂y . We have:
∂f
∂y =
∂
∂y (x2+y)
∂
∂y (2y)
∂
∂y (z2)
=
1
2
0
Step 4: Finally, let’s find ∂f
∂z . We have:
∂f
∂z =
∂
∂z (x2+y)
∂
∂z (2y)
∂
∂z (z2)
=
0
0
2z
Step 5: Putting all the partial derivatives together, we get the Jacobian
matrix of f:
Jf=
2x1 0
0 2 0
0 0 2z
Question 26
Question
Let f(x, y) = sin(xy)
ey2cos(x). Compute the Jacobian matrix of fat the point
(π, 0).
Solution
To find the Jacobian matrix of fat the point (π, 0), we need to compute the
partial derivatives of fwith respect to xand yat that point.
Step 1: Compute ∂f1∂x and ∂f2
∂x
∂f1
∂x =∂
∂x (sin(xy)) = ycos(xy)
∂f2
∂x =∂
∂x (ey2cos(x)) = −ey2sin(x)
Step 2: Compute ∂f1∂y and ∂f2
∂y
∂f1
∂y =∂
∂y (sin(xy)) = xcos(xy)
∂f2
∂y =∂
∂y (ey2cos(x)) = 2yey2cos(x)
19
Step 3: Evaluate the Jacobian matrix at the point (π, 0) The Jacobian
matrix Jfis given by
Jf= ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !
Substitute the point (π, 0) into the partial derivatives we found earlier:
Jf(π, 0) = 0π
−e0sin(π) 0=0π
0 0
Therefore, the Jacobian matrix of fat the point (π, 0) is 0π
0 0.
Question 27
Question
Let f:R3→R2be a function defined by f(x, y, z) = (xy +z, x2+y2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: The Jacobian matrix of fis given by
J(f) = ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !,
where f1and f2are the components of f.
Step 2: We calculate the partial derivatives of f:
∂f1
∂x =y, ∂f1
∂y =x, ∂f1
∂z = 1
∂f2
∂x = 2x, ∂f2
∂y = 2y, ∂f2
∂z = 0
Step 3: Evaluate these partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 1
∂f2
∂x (1,2,3) = 2,∂f2
∂y (1,2,3) = 4,∂f2
∂z (1,2,3) = 0
Step 4: Finally, the Jacobian matrix of fat (1,2,3) is
J(f)(1,2,3) =211
240.
20
Question 28
Question
Consider the following system of equations:
2x2−y2+z2= 1
x2+y2+z2= 9
3x+ 2y−z= 5
Find the Jacobian matrix of the system at the point (x, y, z) = (1,2,2).
Solution
Step 1: Write the equations in a clearer form using matrix notation: Let F(x)
represent the system of equations, where x= (x, y, z):
F(x) =
2x2−y2+z2−1
x2+y2+z2−9
3x+ 2y−z−5
Step 2: Calculate the Jacobian matrix Jof F: The Jacobian matrix Jis
given by:
J=
∂F1
∂x
∂F1
∂y
∂F1
∂z
∂F2
∂x
∂F2
∂y
∂F2
∂z
∂F3
∂x
∂F3
∂y
∂F3
∂z
Step 3: Calculate the partial derivatives of each component of F:
∂F1
∂x = 4x, ∂F1
∂y =−2y, ∂F1
∂z = 2z
∂F2
∂x = 2x, ∂F2
∂y = 2y, ∂F2
∂z = 2z
∂F3
∂x = 3,∂F3
∂y = 2,∂F3
∂z =−1
Step 4: Evaluate the Jacobian matrix at (x, y, z) = (1,2,2):
J=
4−4 4
2 4 4
3 2 −1
Question 29
Question
Let f:R3→R2be the mapping defined by f(x, y, z) = (x2+y2, yz). Find the
Jacobian matrix of f.
21
Solution
Step 1: The Jacobian matrix of a mapping f:Rn→Rmis an m×nmatrix
whose entries are the first partial derivatives of the components of f.
Step 2: In this case, f(x, y, z)=(x2+y2, yz). We need to find the partial
derivatives of fwith respect to x,y, and z.
Step 3: The first component of fis x2+y2. The partial derivatives of this
component are:
∂
∂x (x2+y2)=2xand ∂
∂y (x2+y2)=2y.
Step 4: The second component of fis yz. The partial derivatives of this
component are:
∂
∂y (yz) = zand ∂
∂z (yz) = y.
Step 5: Therefore, the Jacobian matrix of fis:
Jf=2x2y0
0z y.
Question 30
Question
Let f(x, y, z) =
x2y
xyz
z2
. Find the Jacobian matrix of fat the point P(1,2,3).
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
2xy
yz
0
∂f
∂y =
x2
xz
0
∂f
∂z =
0
xy
2z
Step 2: Evaluate the partial derivatives at the point P(1,2,3).
∂f
∂x (1,2,3) =
4
6
0
22
∂f
∂y (1,2,3) =
1
3
0
∂f
∂z (1,2,3) =
0
2
6
Step 3: Construct the Jacobian matrix by arranging the partial derivatives
in rows.
Jf(1,2,3) =
460
130
026
23
