MATH 117 - ELEMENTS OF
MATHEMATICS - Jacobian matrix
Question Bank - Set 5
Liberty University
Question 1
Question
Let f:R3→R3be a function defined by f(x, y, z) = (xy, xz, yz). Calculate
the Jacobian matrix of fat the point (2,3,4).
Solution
Step 1: Write out the function fin components:
f(x, y, z)=(xy, xz, yz)
Step 2: Calculate the partial derivatives of fwith respect to x,y, and z:
∂f
∂x =
y
z
0
,∂f
∂y =
x
0
z
,∂f
∂z =
0
x
y
Step 3: Evaluate the partial derivatives at the point (2,3,4):
∂f
∂x =
3
4
0
,∂f
∂y =
2
0
4
,∂f
∂z =
0
2
3
Step 4: Assemble the Jacobian matrix:
Jf(2,3,4) =
320
402
043
Therefore, the Jacobian matrix of fat the point (2,3,4) is:
Jf(2,3,4) =
320
402
043
Question 2
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, y sin(z)).
Determine the Jacobian matrix of f.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y,
∂f2
∂x = 0,∂f2
∂y = sin(z),∂f2
∂z =ycos(z).
Step 2: Assemble the Jacobian matrix using the partial derivatives computed
in Step 1.
J(f) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
0 sin(z)ycos(z).
Therefore, the Jacobian matrix of fis 2x z y
0 sin(z)ycos(z).
Question 3
Question
Consider the transformation T:R3→R2given by T(x, y, z) = (x2+y2,2z).
Find the Jacobian matrix of T.
Solution
To find the Jacobian matrix of T, we need to compute the partial derivatives of
each component function with respect to each input variable.
Step 1: Partial Derivatives of the First Component Function Let
f(x, y, z) = x2+y2. Compute the partial derivatives of f:
∂f
∂x = 2x,
∂f
∂y = 2y,
∂f
∂z = 0.
2
Step 2: Partial Derivatives of the Second Component Function Let
g(x, y, z)=2z. Compute the partial derivatives of g:
∂g
∂x = 0,
∂g
∂y = 0,
∂g
∂z = 2.
Step 3: Constructing the Jacobian Matrix The Jacobian matrix of T
is given by:
JT="∂f
∂x
∂f
∂y
∂f
∂z
∂g
∂x
∂g
∂y
∂g
∂z #=2x2y0
0 0 2.
Question 4
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y2, yz). Find the
Jacobian matrix of fat the point (1,2,3).
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 at the point
(1,2,3).
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x , ∂f1
∂y = 2y , ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0 ,∂f2
∂y =z , ∂f2
∂z =y
3
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) = 2 ·2=4
∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 0
∂f2
∂y (1,2,3) = 3
∂f2
∂z (1,2,3) = 2
Step 4: Assemble the Jacobian matrix of fat the point (1,2,3).
Jacobian matrix of fat (1,2,3) = 240
032
Question 5
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+ 2y, xyz). 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 and organize them into a matrix.
Step 1: Compute the partial derivatives of f
For the first component function f1(x, y, z) = x2+ 2y:
∂f1
∂x = 2x, ∂f1
∂y = 2,∂f1
∂z = 0
For the second component function f2(x, y, z) = xyz:
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 2: Organize the partial derivatives into a matrix
The Jacobian matrix of fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2 0
yz xz xy
Therefore, the Jacobian matrix of fis 2x2 0
yz xz xy.
4
Question 6
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, z −x). Compute the
Jacobian matrix Jfof f.
Solution
To compute the Jacobian matrix Jfof f, we need to find the partial derivatives
of each component function of f.
Step 1: Find ∂f1
∂x ,∂f1
∂y ,and ∂f1
∂z . Let f1=x2+yz. Then,
∂f1
∂x = 2x, ∂f1
∂y =z, and ∂f1
∂z =y.
Step 2: Find ∂f2
∂x ,∂f2
∂y ,and ∂f2
∂z . Let f2=z−x. Then,
∂f2
∂x =−1,∂f2
∂y = 0,and ∂f2
∂z = 1.
Step 3: Construct the Jacobian matrix Jf. The Jacobian matrix Jfis given
by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
=2x z y
−1 0 1.
Therefore, the Jacobian matrix of fis
Jf=2x z y
−1 0 1.
Question 7
Question
Let f(x, y, z) = (3x2y+z3, xz, 2y2z) be a vector-valued function. Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
5
Step 1: Compute the partial derivatives of fwith respect to x.
∂fi
∂x =
∂(3x2y+z3)
∂x
∂(xz)
∂x
∂(2y2z)
∂x
Step 2:
∂fi
∂x =
6xy
z
0
Step 3: Compute the partial derivatives of fwith respect to y.
∂fi
∂y =
∂(3x2y+z3)
∂y
∂(xz)
∂y
∂(2y2z)
∂y
Step 4:
∂fi
∂y =
3x2
0
4yz
Step 5: Compute the partial derivatives of fwith respect to z.
∂fi
∂z =
∂(3x2y+z3)
∂z
∂(xz)
∂z
∂(2y2z)
∂z
Step 6:
∂fi
∂z =
0
x
2y2
Step 7: The Jacobian matrix of fis
Jf=
6xy 3x20
z0 4yz
0x2y2
Question 8
Question
Let f:R3→R2be a function defined by f(x, y, z) = (xy, x2z). Determine the
Jacobian matrix of f.
6
Solution
To find the Jacobian matrix of the function f, we need to compute the par-
tial derivatives of each component function with respect to each variable. The
Jacobian matrix of fwill be a 2 ×3 matrix, where the (i, j) entry is ∂fi
∂xj.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f
∂x (x, y, z) = ∂(xy)
∂x
∂(xy)
∂y
∂(xy)
∂z
∂(x2z)
∂x
∂(x2z)
∂y
∂(x2z)
∂z !
Step 2:
∂f
∂x =y x 0
2xz 0x2
Step 3: Compute the partial derivatives of fwith respect to y:
∂f
∂y (x, y, z) = ∂(xy)
∂x
∂(xy)
∂y
∂(xy)
∂z
∂(x2z)
∂x
∂(x2z)
∂y
∂(x2z)
∂z !
Step 4:
∂f
∂y =y x 0
2xz 0x2
Step 5: The Jacobian matrix of fis formed by arranging the derivative
matrices with respect to each variable into a single matrix:
Jf(x, y, z) = y x 0
2xz 0x2
Question 9
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 0
7
∂f2
∂x (1,2,3) = 0,∂f2
∂y (1,2,3) = 3,∂f2
∂z (1,2,3) = 2
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf(1,2,3) = 210
032
Question 10
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+yz, y2+xz). Find
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function and arrange them in a matrix.
Step 1: Find ∂f1∂x,∂f1
∂y ,∂f1
∂z
Given f(x, y, z) = (x2+yz, y2+xz), we have:
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Find ∂f2∂x,∂f2
∂y ,∂f2
∂z
Continuing with the same function f(x, y, z)=(x2+yz, y2+xz), we have:
∂f2
∂x =z
∂f2
∂y = 2y
∂f2
∂z =x
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 #
8
Substitute the partial derivatives we found into the matrix:
Jf=2x z y
z2y x
Question 11
Question
Let f(x, y) = x2+y2and g(x, y) = xy. Compute the Jacobian matrix of the
composition h(x, y) = f(g(x, y), g(y, x)).
Solution
To find the Jacobian matrix of the composition h(x, y) = f(g(x, y), g(y, x)), we
need to compute the partial derivatives of fand gfirst.
Step 1: Compute partial derivatives of fand gGiven f(x, y) = x2+y2
and g(x, y) = xy:
Partial derivatives of f:
∂f
∂x = 2xand ∂f
∂y = 2y
Partial derivatives of g:
∂g
∂x =yand ∂g
∂y =x
Step 2: Compute the composition h(x, y) The composition h(x, y) =
f(g(x, y), g(y, x)) is:
h(x, y) = f(xy, yx)
Step 3: Compute the Jacobian matrix of hThe Jacobian matrix of h
is given by:
Jh=h∂h
∂x
∂h
∂y i=h∂f
∂x
∂g
∂x +∂f
∂y
∂g
∂y
∂f
∂x
∂g
∂y +∂f
∂y
∂g
∂x i
Substitute the partial derivatives of fand ginto the Jacobian matrix for-
mula:
Jh=2xy + 2yx 2xy + 2yx=4xy 4yx
Therefore, the Jacobian matrix of the composition h(x, y) = f(g(x, y), g(y, x))
is:
Jh=4xy 4yx
9
Question 12
Question
Let f(x, y, z)=(xy, yz, xz) be a vector-valued function. Find the Jacobian
matrix of f.
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂(xy)
∂x
∂(yz)
∂x
∂(xz)
∂x
=
y
0
z
Step 2: Calculate the partial derivatives of fwith respect to y.
∂f
∂y =
∂(xy)
∂y
∂(yz)
∂y
∂(xz)
∂y
=
x
z
0
Step 3: Calculate the partial derivatives of fwith respect to z.
∂f
∂z =
∂(xy)
∂z
∂(yz)
∂z
∂(xz)
∂z
=
0
y
x
Step 4: Assemble the partial derivatives into a matrix to obtain the Jacobian
matrix of f.
J=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
y x 0
0z y
z0x
Question 13
Question
Let f(x, y) = (x2+3y, 2x−y2) and g(u, v) = uv2. Compute the Jacobian matrix
of the composition g◦fat the point (1,2).
Solution
To find the Jacobian matrix of the composition g◦f, we first need to compute
g(f(x, y)) and then evaluate the partial derivatives of this composition function
at the point (1,2).
10
Step 1: Find g(f(x, y)) Let’s first find f(x, y):
f(x, y) = (x2+ 3y, 2x−y2)
Now, substitute xand ywith uand vrespectively to find g(f(x, y)):
g(f(x, y)) = g(x2+ 3y, 2x−y2)=(x2+ 3y)(2x−y2)2
Step 2: Compute the Jacobian matrix Now, let’s find the partial deriva-
tives of g(f(x, y)) with respect to xand y:
∂g(f(x, y))
∂x =∂
∂x ((x2+ 3y)(2x−y2)2)
Differentiating with respect to x, we get:
∂g(f(x, y))
∂x = 4x(2x−y2)2+ (x2+ 3y)4(2x−y2)
Similarly, we find the partial derivative with respect to y:
∂g(f(x, y))
∂y =∂
∂y ((x2+ 3y)(2x−y2)2)
Differentiating with respect to y, we get:
∂g(f(x, y))
∂y = (x2+ 3y)2(2x−y2)2+ 3(2x−y2)2
Evaluate these partial derivatives at the point (1,2) to find the Jacobian
matrix of g◦fat that point.
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+yz, xyz). Compute
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each input variable.
f(x, y, z)=(x2+yz, xyz)
Step 1: Compute the partial derivatives of f1(x, y, z) = x2+yz.
∂f1
∂x = 2x∂f1
∂y =z∂f1
∂z =y
11
Step 2: Compute the partial derivatives of f2(x, y, z) = xyz.
∂f2
∂x =yz ∂f2
∂y =xz ∂f2
∂z =xy
Step 3: Assemble the partial derivatives into a Jacobian matrix.
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
yz xz xy
Question 15
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2yz, xyz). Calculate
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the derivatives of each
component function with respect to each variable.
Step 1: Partial derivatives of the first component Let u=x2yz and
v=xyz. Then, the first component of fis uand we need to find ∂u
∂x ,∂u
∂y , and
∂u
∂z .
∂u
∂x = 2xyz
∂u
∂y =x2z
∂u
∂z =x2y
Step 2: Partial derivatives of the second component Similarly, let
w=xyz. Then, the second component of fis wand we need to find ∂w
∂x ,∂w
∂y ,
and ∂w
∂z .
∂w
∂x =yz
∂w
∂y =xz
∂w
∂z =xy
Step 3: Assembling the Jacobian matrix The Jacobian matrix of fis
given by
J(f) = "∂u
∂x
∂u
∂y
∂u
∂z
∂w
∂x
∂w
∂y
∂w
∂z #
12
Substitute the partial derivatives we found into the Jacobian matrix:
J(f) = 2xyz x2z x2y
yz xz xy
Question 16
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2−yz, 2y+z). Compute
the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we first need to find the partial derivatives
of f.
Step 1: Find ∂f
∂x ,∂f
∂y , and ∂f
∂z .
∂f
∂x =∂
∂x (x2−yz),∂
∂x (2y+z)
= (2x, 0)
∂f
∂y =∂
∂y (x2−yz),∂
∂y (2y+z)
= (−z, 2)
∂f
∂z =∂
∂z (x2−yz),∂
∂z (2y+z)
= (−y, 1)
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Substitute the partial derivatives we found into the Jacobian matrix:
Jf=2x0 0
−z2 0
Therefore, the Jacobian matrix of fis:
Jf=2x0 0
−z2 0
13
Question 17
Question
Find the Jacobian matrix of the transformation (x, y)7→ (x2+y, xy) at the
point (2,3).
Solution
Step 1: Compute the partial derivatives of the transformation. Let u=x2+y
and v=xy. Then, the transformation can be represented by u=f(x, y) and
v=g(x, y).
The partial derivatives with respect to xand yare:
∂f
∂x = 2xand ∂f
∂y = 1
∂g
∂x =yand ∂g
∂y =x
Step 2: Evaluate the partial derivatives at the point (2,3). At the point
(2,3) we have:
∂f
∂x = 2(2) = 4 and ∂f
∂y = 1
∂g
∂x = 3 and ∂g
∂y = 2
Step 3: Assemble the Jacobian matrix. The Jacobian matrix is given by:
"∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y #
Substitute the computed partial derivatives:
4 1
3 2
Therefore, the Jacobian matrix of the transformation at the point (2,3) is:
4 1
3 2
Question 18
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, yz). Find the
Jacobian matrix of f.
14
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix of
partial derivatives of the components of f. Let’s compute the Jacobian matrix
of the given function f(x, y, z) = (x2+y, yz).
Step 2: The components of fare f1(x, y, z) = x2+yand f2(x, y, z) = yz.
The Jacobian matrix Jis given by:
J="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 3: Calculating the partial derivatives:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 4: Now, substitute these partial derivatives into the Jacobian matrix:
J=2x1 0
0z y
Therefore, the Jacobian matrix of the function f(x, y, z)=(x2+y, yz) is
2x1 0
0z y
.
Question 19
Question
Let f(x, y, z)=(x2+y2+z2, xy, xz) be a vector-valued function. Find the
Jacobian matrix of f.
15
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 2z
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 0
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =z, ∂f3
∂y = 0,∂f3
∂z =x
Step 4: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis:
Jf(x, y, z) =
2x2y2z
y x 0
z0x
Question 20
Question
Given the following transformation:
x=rcos(θ)
y=rsin(θ)
find the Jacobian matrix Jof the transformation.
Solution
Step 1: Start by expressing the transformation as a vector-valued function
F(r, θ)=(x, y).
F(r, θ)=(rcos(θ), r sin(θ))
Step 2: Now, find the partial derivatives of xand ywith respect to rand θ:
∂x
∂r = cos(θ),∂x
∂θ =−rsin(θ)
∂y
∂r = sin(θ),∂y
∂θ =rcos(θ)
16
Step 3: Assemble these partial derivatives into the Jacobian matrix J.
J=∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ =cos(θ)−rsin(θ)
sin(θ)rcos(θ)
Therefore, the Jacobian matrix of the transformation is J=cos(θ)−rsin(θ)
sin(θ)rcos(θ).
Question 21
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, 3x−2y+z).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each variable.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 1: Find ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz) = 2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
Step 2: Find ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (3x−2y+z)=3
∂f2
∂y =∂
∂y (3x−2y+z) = −2
∂f2
∂z =∂
∂z (3x−2y+z)=1
Step 3: Assemble the Jacobian matrix Jf.
Jf=2x z y
3−2 1
Therefore, the Jacobian matrix of fis:
Jf=2x z y
3−2 1
17
Question 22
Question
Let f:R3→R2be a mapping given by f(x, y, z) = (xy +z, x2y). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function with respect to each input variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x =y
∂f1
∂y =x
∂f1
∂z = 1
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x = 2xy
∂f2
∂y =x2
∂f2
∂z = 0
Step 3: Construct the Jacobian matrix Jf:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=y x 1
2xy x20
Therefore, the Jacobian matrix of fis:
Jf=y x 1
2xy x20
Question 23
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, xyz). Determine
the Jacobian matrix of f.
18
Solution
Step 1: Write fin component form. The function fcan be written as f(x, y, z) =
x2+y
xyz .
Step 2: Calculate the partial derivatives of f. The partial derivatives of f
are: - ∂f1
∂x = 2x-∂f1
∂y = 1 - ∂f1
∂z = 0 - ∂f2
∂x =yz -∂f2
∂y =xz -∂f2
∂z =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 !.
Step 4: Substitute the partial derivatives into the Jacobian matrix. Sub-
stitute the partial derivatives calculated in Step 2 into the Jacobian matrix to
get
Jf=2x1 0
yz xz xy .
Therefore, the Jacobian matrix of the function fis 2x1 0
yz xz xy .
Question 24
Question
Let f:R2→R2be defined by f(x, y) = (3x2+ 2y, x2−y). Determine the
Jacobian matrix of fat the point (1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y:
∂f1
∂x = 6xand ∂f1
∂y = 2
∂f2
∂x = 2xand ∂f2
∂y =−1
Step 2: Evaluate the partial derivatives at the point (1,2):
∂f1
∂x (1,2) = 6(1) = 6,∂f1
∂y (1,2) = 2
∂f2
∂x (1,2) = 2(1) = 2,∂f2
∂y (1,2) = −1
Step 3: Assemble the Jacobian matrix of fat (1,2):
Jf(1,2) = 6 2
2−1
Therefore, the Jacobian matrix of fat the point (1,2) is 6 2
2−1.
19
Question 25
Question
Let f:R3→R2be 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: Find the partial derivatives Let f(x, y, z)=(u, v) where u=
x2+yand v=yz.
The partial derivatives are given by:
∂u
∂x =∂
∂x (x2+y) = 2x,
∂u
∂y =∂
∂y (x2+y) = 1,
∂u
∂z =∂
∂z (x2+y)=0,
∂v
∂x =∂
∂x (yz) = 0,
∂v
∂y =∂
∂y (yz) = z,
∂v
∂z =∂
∂z (yz) = y.
Step 2: Assemble the Jacobian matrix The Jacobian matrix is then
given by:
"∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #=2x1 0
0z y.
Therefore, the Jacobian matrix of fis 2x1 0
0z y.
Question 26
Question
Let f(x, y) = x2+y2
2xy be a vector-valued function. Find the Jacobian matrix
of fat the point (3,4).
20
Solution
Step 1: Compute the partial derivatives of f:
∂f
∂x =∂f1
∂x
∂f2
∂x =2x
2y,
∂f
∂y = ∂f1
∂y
∂f2
∂y !=2y
2x.
Step 2: Evaluate the partial derivatives at the point (3,4):
∂f
∂x (3,4) = 2·3
2·4=6
8,
∂f
∂y (3,4) = 2·4
2·3=8
6.
Step 3: Construct the Jacobian matrix:
Jf(3,4) = ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=6 8
8 6.
Question 27
Question
Let f:R3→R2be 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: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z)=(x2−y, yz), we can see that f1(x, y, z) = x2−y. Taking
the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y =−1,and ∂f1
∂z = 0.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with the second component function f2(x, y, z) = yz:
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Assemble the Jacobian matrix.
21
The Jacobian matrix of fis given by:
"∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x−1 0
0z y.
Therefore, the Jacobian matrix of fis 2x−1 0
0z y.
Question 28
Question
Let f(x, y, z) = (x2+y, 2y−z) be a vector-valued function. Compute the
Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of fis defined as the matrix of all first-order
partial derivatives of the vector-valued function f. Step 2: Compute the partial
derivatives of fwith respect to x,y, and z. Step 3: The partial derivatives of
fwith respect to x,y, and zare given by
∂f
∂x =2x
0,∂f
∂y =1
2,∂f
∂z =0
−1.
Step 4: Assemble the partial derivatives into the Jacobian matrix of f:
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x1 0
0 2 −1.
Therefore, the Jacobian matrix of fis 2x1 0
0 2 −1.
Question 29
Question
Let f(x, y, z) = x2+y2+z2and g(x, y, z) = ex+yz . Compute the Jacobian
matrix of the composition g◦fat the point (1,1,1).
Solution
To find the Jacobian matrix of the composition g◦f, we first need to find the
compositions of gand f.
Step 1: Find f(1,1,1) Substitute x= 1, y= 1, and z= 1 into f(x, y, z):
f(1,1,1) = 12+ 12+ 12= 3
22
Step 2: Find the gradient of fat (1,1,1) The gradient of fis given by:
∇f=
∂f
∂x
∂f
∂y
∂f
∂z
Compute the partial derivatives of f:
∂f
∂x = 2x, ∂f
∂y = 2y, ∂f
∂z = 2z
At (1,1,1), the gradient of fis:
∇f(1,1,1) =
2
2
2
Step 3: Find g(f(1,1,1)) Substitute f(1,1,1) = 3 into g(x, y, z):
g(3) = e3
Step 4: Find the gradient of gat e3The gradient of gis given by:
∇g=
∂g
∂x
∂g
∂y
∂g
∂z
Compute the partial derivatives of g:
∂g
∂x =ex+yz ,∂g
∂y =zex+yz ,∂g
∂z =yex+yz
At e3, the gradient of gis:
∇g(e3) =
e3
e3
e3
Step 5: Find the Jacobian matrix of g◦fat (1,1,1) By chain rule,
the Jacobian matrix of g◦fat (1,1,1) is given by:
∇g(e3)· ∇f(1,1,1) =
e3
e3
e3
·
2
2
2
=
2e3
2e3
2e3
Therefore, the Jacobian matrix of the composition g◦fat the point (1,1,1)
is
2e3
2e3
2e3
.
23
Question 30
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2yz, xy3z, xyz2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable:
∂f1
∂x = 2xyz, ∂f1
∂y =x2z, ∂f1
∂z =x2y,
∂f2
∂x =y3z, ∂f2
∂y = 3xy2z, ∂f2
∂z =xy3,
∂f3
∂x =yz2,∂f3
∂y =xz2,∂f3
∂z = 2xyz.
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3)
= 2(1)(2)(3) = 12,∂f1
∂y (1,2,3)
= (1)2(3) = 3,∂f1
∂z (1,2,3)
= (1)2(2) = 2,
∂f2
∂x (1,2,3)
= (2)3(3) = 24,∂f2
∂y (1,2,3)
= 3(1)(2)2= 12,∂f2
∂z (1,2,3)
= (1)(2)3= 8,
∂f3
∂x (1,2,3)
= (2)(3)2= 18,∂f3
∂y (1,2,3)
= (1)(3)2= 9,∂f3
∂z (1,2,3)
= 2(1)(2)(3) = 12.
Step 3: Assemble the Jacobian matrix of fat (1,2,3):
Jf(1,2,3) =
12 3 2
24 12 8
18 9 12
.
Therefore, the Jacobian matrix of fat the point (1,2,3) is
12 3 2
24 12 8
18 9 12
.
24
Question 2
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, y sin(z)).
Determine the Jacobian matrix of f.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y,
∂f2
∂x = 0,∂f2
∂y = sin(z),∂f2
∂z =ycos(z).
Step 2: Assemble the Jacobian matrix using the partial derivatives computed
in Step 1.
J(f) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
0 sin(z)ycos(z).
Therefore, the Jacobian matrix of fis 2x z y
0 sin(z)ycos(z).
Question 3
Question
Consider the transformation T:R3→R2given by T(x, y, z) = (x2+y2,2z).
Find the Jacobian matrix of T.
Solution
To find the Jacobian matrix of T, we need to compute the partial derivatives of
each component function with respect to each input variable.
Step 1: Partial Derivatives of the First Component Function Let
f(x, y, z) = x2+y2. Compute the partial derivatives of f:
∂f
∂x = 2x,
∂f
∂y = 2y,
∂f
∂z = 0.
2
Step 2: Partial Derivatives of the Second Component Function Let
g(x, y, z)=2z. Compute the partial derivatives of g:
∂g
∂x = 0,
∂g
∂y = 0,
∂g
∂z = 2.
Step 3: Constructing the Jacobian Matrix The Jacobian matrix of T
is given by:
JT="∂f
∂x
∂f
∂y
∂f
∂z
∂g
∂x
∂g
∂y
∂g
∂z #=2x2y0
0 0 2.
Question 4
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y2, yz). Find the
Jacobian matrix of fat the point (1,2,3).
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 at the point
(1,2,3).
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x , ∂f1
∂y = 2y , ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0 ,∂f2
∂y =z , ∂f2
∂z =y
3
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) = 2 ·2=4
∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 0
∂f2
∂y (1,2,3) = 3
∂f2
∂z (1,2,3) = 2
Step 4: Assemble the Jacobian matrix of fat the point (1,2,3).
Jacobian matrix of fat (1,2,3) = 240
032
Question 5
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+ 2y, xyz). 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 and organize them into a matrix.
Step 1: Compute the partial derivatives of f
For the first component function f1(x, y, z) = x2+ 2y:
∂f1
∂x = 2x, ∂f1
∂y = 2,∂f1
∂z = 0
For the second component function f2(x, y, z) = xyz:
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 2: Organize the partial derivatives into a matrix
The Jacobian matrix of fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2 0
yz xz xy
Therefore, the Jacobian matrix of fis 2x2 0
yz xz xy.
4
Question 6
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, z −x). Compute the
Jacobian matrix Jfof f.
Solution
To compute the Jacobian matrix Jfof f, we need to find the partial derivatives
of each component function of f.
Step 1: Find ∂f1
∂x ,∂f1
∂y ,and ∂f1
∂z . Let f1=x2+yz. Then,
∂f1
∂x = 2x, ∂f1
∂y =z, and ∂f1
∂z =y.
Step 2: Find ∂f2
∂x ,∂f2
∂y ,and ∂f2
∂z . Let f2=z−x. Then,
∂f2
∂x =−1,∂f2
∂y = 0,and ∂f2
∂z = 1.
Step 3: Construct the Jacobian matrix Jf. The Jacobian matrix Jfis given
by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
=2x z y
−1 0 1.
Therefore, the Jacobian matrix of fis
Jf=2x z y
−1 0 1.
Question 7
Question
Let f(x, y, z) = (3x2y+z3, xz, 2y2z) be a vector-valued function. Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
5
Step 1: Compute the partial derivatives of fwith respect to x.
∂fi
∂x =
∂(3x2y+z3)
∂x
∂(xz)
∂x
∂(2y2z)
∂x
Step 2:
∂fi
∂x =
6xy
z
0
Step 3: Compute the partial derivatives of fwith respect to y.
∂fi
∂y =
∂(3x2y+z3)
∂y
∂(xz)
∂y
∂(2y2z)
∂y
Step 4:
∂fi
∂y =
3x2
0
4yz
Step 5: Compute the partial derivatives of fwith respect to z.
∂fi
∂z =
∂(3x2y+z3)
∂z
∂(xz)
∂z
∂(2y2z)
∂z
Step 6:
∂fi
∂z =
0
x
2y2
Step 7: The Jacobian matrix of fis
Jf=
6xy 3x20
z0 4yz
0x2y2
Question 8
Question
Let f:R3→R2be a function defined by f(x, y, z) = (xy, x2z). Determine the
Jacobian matrix of f.
6
Solution
To find the Jacobian matrix of the function f, we need to compute the par-
tial derivatives of each component function with respect to each variable. The
Jacobian matrix of fwill be a 2 ×3 matrix, where the (i, j) entry is ∂fi
∂xj.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f
∂x (x, y, z) = ∂(xy)
∂x
∂(xy)
∂y
∂(xy)
∂z
∂(x2z)
∂x
∂(x2z)
∂y
∂(x2z)
∂z !
Step 2:
∂f
∂x =y x 0
2xz 0x2
Step 3: Compute the partial derivatives of fwith respect to y:
∂f
∂y (x, y, z) = ∂(xy)
∂x
∂(xy)
∂y
∂(xy)
∂z
∂(x2z)
∂x
∂(x2z)
∂y
∂(x2z)
∂z !
Step 4:
∂f
∂y =y x 0
2xz 0x2
Step 5: The Jacobian matrix of fis formed by arranging the derivative
matrices with respect to each variable into a single matrix:
Jf(x, y, z) = y x 0
2xz 0x2
Question 9
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 0
7
∂f2
∂x (1,2,3) = 0,∂f2
∂y (1,2,3) = 3,∂f2
∂z (1,2,3) = 2
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf(1,2,3) = 210
032
Question 10
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+yz, y2+xz). Find
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function and arrange them in a matrix.
Step 1: Find ∂f1∂x,∂f1
∂y ,∂f1
∂z
Given f(x, y, z) = (x2+yz, y2+xz), we have:
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Find ∂f2∂x,∂f2
∂y ,∂f2
∂z
Continuing with the same function f(x, y, z)=(x2+yz, y2+xz), we have:
∂f2
∂x =z
∂f2
∂y = 2y
∂f2
∂z =x
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 #
8
Substitute the partial derivatives we found into the matrix:
Jf=2x z y
z2y x
Question 11
Question
Let f(x, y) = x2+y2and g(x, y) = xy. Compute the Jacobian matrix of the
composition h(x, y) = f(g(x, y), g(y, x)).
Solution
To find the Jacobian matrix of the composition h(x, y) = f(g(x, y), g(y, x)), we
need to compute the partial derivatives of fand gfirst.
Step 1: Compute partial derivatives of fand gGiven f(x, y) = x2+y2
and g(x, y) = xy:
Partial derivatives of f:
∂f
∂x = 2xand ∂f
∂y = 2y
Partial derivatives of g:
∂g
∂x =yand ∂g
∂y =x
Step 2: Compute the composition h(x, y) The composition h(x, y) =
f(g(x, y), g(y, x)) is:
h(x, y) = f(xy, yx)
Step 3: Compute the Jacobian matrix of hThe Jacobian matrix of h
is given by:
Jh=h∂h
∂x
∂h
∂y i=h∂f
∂x
∂g
∂x +∂f
∂y
∂g
∂y
∂f
∂x
∂g
∂y +∂f
∂y
∂g
∂x i
Substitute the partial derivatives of fand ginto the Jacobian matrix for-
mula:
Jh=2xy + 2yx 2xy + 2yx=4xy 4yx
Therefore, the Jacobian matrix of the composition h(x, y) = f(g(x, y), g(y, x))
is:
Jh=4xy 4yx
9
Question 12
Question
Let f(x, y, z)=(xy, yz, xz) be a vector-valued function. Find the Jacobian
matrix of f.
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂(xy)
∂x
∂(yz)
∂x
∂(xz)
∂x
=
y
0
z
Step 2: Calculate the partial derivatives of fwith respect to y.
∂f
∂y =
∂(xy)
∂y
∂(yz)
∂y
∂(xz)
∂y
=
x
z
0
Step 3: Calculate the partial derivatives of fwith respect to z.
∂f
∂z =
∂(xy)
∂z
∂(yz)
∂z
∂(xz)
∂z
=
0
y
x
Step 4: Assemble the partial derivatives into a matrix to obtain the Jacobian
matrix of f.
J=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
y x 0
0z y
z0x
Question 13
Question
Let f(x, y) = (x2+3y, 2x−y2) and g(u, v) = uv2. Compute the Jacobian matrix
of the composition g◦fat the point (1,2).
Solution
To find the Jacobian matrix of the composition g◦f, we first need to compute
g(f(x, y)) and then evaluate the partial derivatives of this composition function
at the point (1,2).
10
Step 1: Find g(f(x, y)) Let’s first find f(x, y):
f(x, y) = (x2+ 3y, 2x−y2)
Now, substitute xand ywith uand vrespectively to find g(f(x, y)):
g(f(x, y)) = g(x2+ 3y, 2x−y2)=(x2+ 3y)(2x−y2)2
Step 2: Compute the Jacobian matrix Now, let’s find the partial deriva-
tives of g(f(x, y)) with respect to xand y:
∂g(f(x, y))
∂x =∂
∂x ((x2+ 3y)(2x−y2)2)
Differentiating with respect to x, we get:
∂g(f(x, y))
∂x = 4x(2x−y2)2+ (x2+ 3y)4(2x−y2)
Similarly, we find the partial derivative with respect to y:
∂g(f(x, y))
∂y =∂
∂y ((x2+ 3y)(2x−y2)2)
Differentiating with respect to y, we get:
∂g(f(x, y))
∂y = (x2+ 3y)2(2x−y2)2+ 3(2x−y2)2
Evaluate these partial derivatives at the point (1,2) to find the Jacobian
matrix of g◦fat that point.
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+yz, xyz). Compute
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each input variable.
f(x, y, z)=(x2+yz, xyz)
Step 1: Compute the partial derivatives of f1(x, y, z) = x2+yz.
∂f1
∂x = 2x∂f1
∂y =z∂f1
∂z =y
11
Step 2: Compute the partial derivatives of f2(x, y, z) = xyz.
∂f2
∂x =yz ∂f2
∂y =xz ∂f2
∂z =xy
Step 3: Assemble the partial derivatives into a Jacobian matrix.
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
yz xz xy
Question 15
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2yz, xyz). Calculate
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the derivatives of each
component function with respect to each variable.
Step 1: Partial derivatives of the first component Let u=x2yz and
v=xyz. Then, the first component of fis uand we need to find ∂u
∂x ,∂u
∂y , and
∂u
∂z .
∂u
∂x = 2xyz
∂u
∂y =x2z
∂u
∂z =x2y
Step 2: Partial derivatives of the second component Similarly, let
w=xyz. Then, the second component of fis wand we need to find ∂w
∂x ,∂w
∂y ,
and ∂w
∂z .
∂w
∂x =yz
∂w
∂y =xz
∂w
∂z =xy
Step 3: Assembling the Jacobian matrix The Jacobian matrix of fis
given by
J(f) = "∂u
∂x
∂u
∂y
∂u
∂z
∂w
∂x
∂w
∂y
∂w
∂z #
12
Substitute the partial derivatives we found into the Jacobian matrix:
J(f) = 2xyz x2z x2y
yz xz xy
Question 16
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2−yz, 2y+z). Compute
the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we first need to find the partial derivatives
of f.
Step 1: Find ∂f
∂x ,∂f
∂y , and ∂f
∂z .
∂f
∂x =∂
∂x (x2−yz),∂
∂x (2y+z)
= (2x, 0)
∂f
∂y =∂
∂y (x2−yz),∂
∂y (2y+z)
= (−z, 2)
∂f
∂z =∂
∂z (x2−yz),∂
∂z (2y+z)
= (−y, 1)
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Substitute the partial derivatives we found into the Jacobian matrix:
Jf=2x0 0
−z2 0
Therefore, the Jacobian matrix of fis:
Jf=2x0 0
−z2 0
13
Question 17
Question
Find the Jacobian matrix of the transformation (x, y)7→ (x2+y, xy) at the
point (2,3).
Solution
Step 1: Compute the partial derivatives of the transformation. Let u=x2+y
and v=xy. Then, the transformation can be represented by u=f(x, y) and
v=g(x, y).
The partial derivatives with respect to xand yare:
∂f
∂x = 2xand ∂f
∂y = 1
∂g
∂x =yand ∂g
∂y =x
Step 2: Evaluate the partial derivatives at the point (2,3). At the point
(2,3) we have:
∂f
∂x = 2(2) = 4 and ∂f
∂y = 1
∂g
∂x = 3 and ∂g
∂y = 2
Step 3: Assemble the Jacobian matrix. The Jacobian matrix is given by:
"∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y #
Substitute the computed partial derivatives:
4 1
3 2
Therefore, the Jacobian matrix of the transformation at the point (2,3) is:
4 1
3 2
Question 18
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, yz). Find the
Jacobian matrix of f.
14
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix of
partial derivatives of the components of f. Let’s compute the Jacobian matrix
of the given function f(x, y, z) = (x2+y, yz).
Step 2: The components of fare f1(x, y, z) = x2+yand f2(x, y, z) = yz.
The Jacobian matrix Jis given by:
J="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 3: Calculating the partial derivatives:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 4: Now, substitute these partial derivatives into the Jacobian matrix:
J=2x1 0
0z y
Therefore, the Jacobian matrix of the function f(x, y, z)=(x2+y, yz) is
2x1 0
0z y
.
Question 19
Question
Let f(x, y, z)=(x2+y2+z2, xy, xz) be a vector-valued function. Find the
Jacobian matrix of f.
15
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 2z
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 0
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =z, ∂f3
∂y = 0,∂f3
∂z =x
Step 4: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis:
Jf(x, y, z) =
2x2y2z
y x 0
z0x
Question 20
Question
Given the following transformation:
x=rcos(θ)
y=rsin(θ)
find the Jacobian matrix Jof the transformation.
Solution
Step 1: Start by expressing the transformation as a vector-valued function
F(r, θ)=(x, y).
F(r, θ)=(rcos(θ), r sin(θ))
Step 2: Now, find the partial derivatives of xand ywith respect to rand θ:
∂x
∂r = cos(θ),∂x
∂θ =−rsin(θ)
∂y
∂r = sin(θ),∂y
∂θ =rcos(θ)
16
Step 3: Assemble these partial derivatives into the Jacobian matrix J.
J=∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ =cos(θ)−rsin(θ)
sin(θ)rcos(θ)
Therefore, the Jacobian matrix of the transformation is J=cos(θ)−rsin(θ)
sin(θ)rcos(θ).
Question 21
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, 3x−2y+z).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each variable.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 1: Find ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz) = 2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
Step 2: Find ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (3x−2y+z)=3
∂f2
∂y =∂
∂y (3x−2y+z) = −2
∂f2
∂z =∂
∂z (3x−2y+z)=1
Step 3: Assemble the Jacobian matrix Jf.
Jf=2x z y
3−2 1
Therefore, the Jacobian matrix of fis:
Jf=2x z y
3−2 1
17
Question 22
Question
Let f:R3→R2be a mapping given by f(x, y, z) = (xy +z, x2y). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function with respect to each input variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x =y
∂f1
∂y =x
∂f1
∂z = 1
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x = 2xy
∂f2
∂y =x2
∂f2
∂z = 0
Step 3: Construct the Jacobian matrix Jf:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=y x 1
2xy x20
Therefore, the Jacobian matrix of fis:
Jf=y x 1
2xy x20
Question 23
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, xyz). Determine
the Jacobian matrix of f.
18
Solution
Step 1: Write fin component form. The function fcan be written as f(x, y, z) =
x2+y
xyz .
Step 2: Calculate the partial derivatives of f. The partial derivatives of f
are: - ∂f1
∂x = 2x-∂f1
∂y = 1 - ∂f1
∂z = 0 - ∂f2
∂x =yz -∂f2
∂y =xz -∂f2
∂z =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 !.
Step 4: Substitute the partial derivatives into the Jacobian matrix. Sub-
stitute the partial derivatives calculated in Step 2 into the Jacobian matrix to
get
Jf=2x1 0
yz xz xy .
Therefore, the Jacobian matrix of the function fis 2x1 0
yz xz xy .
Question 24
Question
Let f:R2→R2be defined by f(x, y) = (3x2+ 2y, x2−y). Determine the
Jacobian matrix of fat the point (1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y:
∂f1
∂x = 6xand ∂f1
∂y = 2
∂f2
∂x = 2xand ∂f2
∂y =−1
Step 2: Evaluate the partial derivatives at the point (1,2):
∂f1
∂x (1,2) = 6(1) = 6,∂f1
∂y (1,2) = 2
∂f2
∂x (1,2) = 2(1) = 2,∂f2
∂y (1,2) = −1
Step 3: Assemble the Jacobian matrix of fat (1,2):
Jf(1,2) = 6 2
2−1
Therefore, the Jacobian matrix of fat the point (1,2) is 6 2
2−1.
19
Question 25
Question
Let f:R3→R2be 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: Find the partial derivatives Let f(x, y, z)=(u, v) where u=
x2+yand v=yz.
The partial derivatives are given by:
∂u
∂x =∂
∂x (x2+y) = 2x,
∂u
∂y =∂
∂y (x2+y) = 1,
∂u
∂z =∂
∂z (x2+y)=0,
∂v
∂x =∂
∂x (yz) = 0,
∂v
∂y =∂
∂y (yz) = z,
∂v
∂z =∂
∂z (yz) = y.
Step 2: Assemble the Jacobian matrix The Jacobian matrix is then
given by:
"∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #=2x1 0
0z y.
Therefore, the Jacobian matrix of fis 2x1 0
0z y.
Question 26
Question
Let f(x, y) = x2+y2
2xy be a vector-valued function. Find the Jacobian matrix
of fat the point (3,4).
20
Solution
Step 1: Compute the partial derivatives of f:
∂f
∂x =∂f1
∂x
∂f2
∂x =2x
2y,
∂f
∂y = ∂f1
∂y
∂f2
∂y !=2y
2x.
Step 2: Evaluate the partial derivatives at the point (3,4):
∂f
∂x (3,4) = 2·3
2·4=6
8,
∂f
∂y (3,4) = 2·4
2·3=8
6.
Step 3: Construct the Jacobian matrix:
Jf(3,4) = ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=6 8
8 6.
Question 27
Question
Let f:R3→R2be 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: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z)=(x2−y, yz), we can see that f1(x, y, z) = x2−y. Taking
the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y =−1,and ∂f1
∂z = 0.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with the second component function f2(x, y, z) = yz:
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Assemble the Jacobian matrix.
21
The Jacobian matrix of fis given by:
"∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x−1 0
0z y.
Therefore, the Jacobian matrix of fis 2x−1 0
0z y.
Question 28
Question
Let f(x, y, z) = (x2+y, 2y−z) be a vector-valued function. Compute the
Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of fis defined as the matrix of all first-order
partial derivatives of the vector-valued function f. Step 2: Compute the partial
derivatives of fwith respect to x,y, and z. Step 3: The partial derivatives of
fwith respect to x,y, and zare given by
∂f
∂x =2x
0,∂f
∂y =1
2,∂f
∂z =0
−1.
Step 4: Assemble the partial derivatives into the Jacobian matrix of f:
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x1 0
0 2 −1.
Therefore, the Jacobian matrix of fis 2x1 0
0 2 −1.
Question 29
Question
Let f(x, y, z) = x2+y2+z2and g(x, y, z) = ex+yz . Compute the Jacobian
matrix of the composition g◦fat the point (1,1,1).
Solution
To find the Jacobian matrix of the composition g◦f, we first need to find the
compositions of gand f.
Step 1: Find f(1,1,1) Substitute x= 1, y= 1, and z= 1 into f(x, y, z):
f(1,1,1) = 12+ 12+ 12= 3
22
Step 2: Find the gradient of fat (1,1,1) The gradient of fis given by:
∇f=
∂f
∂x
∂f
∂y
∂f
∂z
Compute the partial derivatives of f:
∂f
∂x = 2x, ∂f
∂y = 2y, ∂f
∂z = 2z
At (1,1,1), the gradient of fis:
∇f(1,1,1) =
2
2
2
Step 3: Find g(f(1,1,1)) Substitute f(1,1,1) = 3 into g(x, y, z):
g(3) = e3
Step 4: Find the gradient of gat e3The gradient of gis given by:
∇g=
∂g
∂x
∂g
∂y
∂g
∂z
Compute the partial derivatives of g:
∂g
∂x =ex+yz ,∂g
∂y =zex+yz ,∂g
∂z =yex+yz
At e3, the gradient of gis:
∇g(e3) =
e3
e3
e3
Step 5: Find the Jacobian matrix of g◦fat (1,1,1) By chain rule,
the Jacobian matrix of g◦fat (1,1,1) is given by:
∇g(e3)· ∇f(1,1,1) =
e3
e3
e3
·
2
2
2
=
2e3
2e3
2e3
Therefore, the Jacobian matrix of the composition g◦fat the point (1,1,1)
is
2e3
2e3
2e3
.
23
Question 30
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2yz, xy3z, xyz2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable:
∂f1
∂x = 2xyz, ∂f1
∂y =x2z, ∂f1
∂z =x2y,
∂f2
∂x =y3z, ∂f2
∂y = 3xy2z, ∂f2
∂z =xy3,
∂f3
∂x =yz2,∂f3
∂y =xz2,∂f3
∂z = 2xyz.
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3)
= 2(1)(2)(3) = 12,∂f1
∂y (1,2,3)
= (1)2(3) = 3,∂f1
∂z (1,2,3)
= (1)2(2) = 2,
∂f2
∂x (1,2,3)
= (2)3(3) = 24,∂f2
∂y (1,2,3)
= 3(1)(2)2= 12,∂f2
∂z (1,2,3)
= (1)(2)3= 8,
∂f3
∂x (1,2,3)
= (2)(3)2= 18,∂f3
∂y (1,2,3)
= (1)(3)2= 9,∂f3
∂z (1,2,3)
= 2(1)(2)(3) = 12.
Step 3: Assemble the Jacobian matrix of fat (1,2,3):
Jf(1,2,3) =
12 3 2
24 12 8
18 9 12
.
Therefore, the Jacobian matrix of fat the point (1,2,3) is
12 3 2
24 12 8
18 9 12
.
24
Question 2
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, y sin(z)).
Determine the Jacobian matrix of f.
Solution
Step 1: Compute the partial derivatives of fwith respect to x,y, and z.
∂f1
∂x = 2x, ∂f1
∂y =z, ∂f1
∂z =y,
∂f2
∂x = 0,∂f2
∂y = sin(z),∂f2
∂z =ycos(z).
Step 2: Assemble the Jacobian matrix using the partial derivatives computed
in Step 1.
J(f) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
0 sin(z)ycos(z).
Therefore, the Jacobian matrix of fis 2x z y
0 sin(z)ycos(z).
Question 3
Question
Consider the transformation T:R3→R2given by T(x, y, z) = (x2+y2,2z).
Find the Jacobian matrix of T.
Solution
To find the Jacobian matrix of T, we need to compute the partial derivatives of
each component function with respect to each input variable.
Step 1: Partial Derivatives of the First Component Function Let
f(x, y, z) = x2+y2. Compute the partial derivatives of f:
∂f
∂x = 2x,
∂f
∂y = 2y,
∂f
∂z = 0.
2
Step 2: Partial Derivatives of the Second Component Function Let
g(x, y, z)=2z. Compute the partial derivatives of g:
∂g
∂x = 0,
∂g
∂y = 0,
∂g
∂z = 2.
Step 3: Constructing the Jacobian Matrix The Jacobian matrix of T
is given by:
JT="∂f
∂x
∂f
∂y
∂f
∂z
∂g
∂x
∂g
∂y
∂g
∂z #=2x2y0
0 0 2.
Question 4
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+y2, yz). Find the
Jacobian matrix of fat the point (1,2,3).
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 at the point
(1,2,3).
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x , ∂f1
∂y = 2y , ∂f1
∂z = 0
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x = 0 ,∂f2
∂y =z , ∂f2
∂z =y
3
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) = 2 ·2=4
∂f1
∂z (1,2,3) = 0
∂f2
∂x (1,2,3) = 0
∂f2
∂y (1,2,3) = 3
∂f2
∂z (1,2,3) = 2
Step 4: Assemble the Jacobian matrix of fat the point (1,2,3).
Jacobian matrix of fat (1,2,3) = 240
032
Question 5
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2+ 2y, xyz). 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 and organize them into a matrix.
Step 1: Compute the partial derivatives of f
For the first component function f1(x, y, z) = x2+ 2y:
∂f1
∂x = 2x, ∂f1
∂y = 2,∂f1
∂z = 0
For the second component function f2(x, y, z) = xyz:
∂f2
∂x =yz, ∂f2
∂y =xz, ∂f2
∂z =xy
Step 2: Organize the partial derivatives into a matrix
The Jacobian matrix of fis:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x2 0
yz xz xy
Therefore, the Jacobian matrix of fis 2x2 0
yz xz xy.
4
Question 6
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+yz, z −x). Compute the
Jacobian matrix Jfof f.
Solution
To compute the Jacobian matrix Jfof f, we need to find the partial derivatives
of each component function of f.
Step 1: Find ∂f1
∂x ,∂f1
∂y ,and ∂f1
∂z . Let f1=x2+yz. Then,
∂f1
∂x = 2x, ∂f1
∂y =z, and ∂f1
∂z =y.
Step 2: Find ∂f2
∂x ,∂f2
∂y ,and ∂f2
∂z . Let f2=z−x. Then,
∂f2
∂x =−1,∂f2
∂y = 0,and ∂f2
∂z = 1.
Step 3: Construct the Jacobian matrix Jf. The Jacobian matrix Jfis given
by
Jf=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
=2x z y
−1 0 1.
Therefore, the Jacobian matrix of fis
Jf=2x z y
−1 0 1.
Question 7
Question
Let f(x, y, z) = (3x2y+z3, xz, 2y2z) be a vector-valued function. Determine
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each variable.
5
Step 1: Compute the partial derivatives of fwith respect to x.
∂fi
∂x =
∂(3x2y+z3)
∂x
∂(xz)
∂x
∂(2y2z)
∂x
Step 2:
∂fi
∂x =
6xy
z
0
Step 3: Compute the partial derivatives of fwith respect to y.
∂fi
∂y =
∂(3x2y+z3)
∂y
∂(xz)
∂y
∂(2y2z)
∂y
Step 4:
∂fi
∂y =
3x2
0
4yz
Step 5: Compute the partial derivatives of fwith respect to z.
∂fi
∂z =
∂(3x2y+z3)
∂z
∂(xz)
∂z
∂(2y2z)
∂z
Step 6:
∂fi
∂z =
0
x
2y2
Step 7: The Jacobian matrix of fis
Jf=
6xy 3x20
z0 4yz
0x2y2
Question 8
Question
Let f:R3→R2be a function defined by f(x, y, z) = (xy, x2z). Determine the
Jacobian matrix of f.
6
Solution
To find the Jacobian matrix of the function f, we need to compute the par-
tial derivatives of each component function with respect to each variable. The
Jacobian matrix of fwill be a 2 ×3 matrix, where the (i, j) entry is ∂fi
∂xj.
Step 1: Compute the partial derivatives of fwith respect to x:
∂f
∂x (x, y, z) = ∂(xy)
∂x
∂(xy)
∂y
∂(xy)
∂z
∂(x2z)
∂x
∂(x2z)
∂y
∂(x2z)
∂z !
Step 2:
∂f
∂x =y x 0
2xz 0x2
Step 3: Compute the partial derivatives of fwith respect to y:
∂f
∂y (x, y, z) = ∂(xy)
∂x
∂(xy)
∂y
∂(xy)
∂z
∂(x2z)
∂x
∂(x2z)
∂y
∂(x2z)
∂z !
Step 4:
∂f
∂y =y x 0
2xz 0x2
Step 5: The Jacobian matrix of fis formed by arranging the derivative
matrices with respect to each variable into a single matrix:
Jf(x, y, z) = y x 0
2xz 0x2
Question 9
Question
Let f:R3→R2be defined by f(x, y, z) = (x2+y, yz). Find the Jacobian
matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable.
∂f1
∂x = 2x, ∂f1
∂y = 1,∂f1
∂z = 0
∂f2
∂x = 0,∂f2
∂y =z, ∂f2
∂z =y
Step 2: Evaluate the partial derivatives at the point (1,2,3).
∂f1
∂x (1,2,3) = 2(1) = 2,∂f1
∂y (1,2,3) = 1,∂f1
∂z (1,2,3) = 0
7
∂f2
∂x (1,2,3) = 0,∂f2
∂y (1,2,3) = 3,∂f2
∂z (1,2,3) = 2
Step 3: Assemble the Jacobian matrix of fat (1,2,3).
Jf(1,2,3) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf(1,2,3) = 210
032
Question 10
Question
Let f:R3→R2be a mapping defined by f(x, y, z)=(x2+yz, y2+xz). Find
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function and arrange them in a matrix.
Step 1: Find ∂f1∂x,∂f1
∂y ,∂f1
∂z
Given f(x, y, z) = (x2+yz, y2+xz), we have:
∂f1
∂x = 2x
∂f1
∂y =z
∂f1
∂z =y
Step 2: Find ∂f2∂x,∂f2
∂y ,∂f2
∂z
Continuing with the same function f(x, y, z)=(x2+yz, y2+xz), we have:
∂f2
∂x =z
∂f2
∂y = 2y
∂f2
∂z =x
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 #
8
Substitute the partial derivatives we found into the matrix:
Jf=2x z y
z2y x
Question 11
Question
Let f(x, y) = x2+y2and g(x, y) = xy. Compute the Jacobian matrix of the
composition h(x, y) = f(g(x, y), g(y, x)).
Solution
To find the Jacobian matrix of the composition h(x, y) = f(g(x, y), g(y, x)), we
need to compute the partial derivatives of fand gfirst.
Step 1: Compute partial derivatives of fand gGiven f(x, y) = x2+y2
and g(x, y) = xy:
Partial derivatives of f:
∂f
∂x = 2xand ∂f
∂y = 2y
Partial derivatives of g:
∂g
∂x =yand ∂g
∂y =x
Step 2: Compute the composition h(x, y) The composition h(x, y) =
f(g(x, y), g(y, x)) is:
h(x, y) = f(xy, yx)
Step 3: Compute the Jacobian matrix of hThe Jacobian matrix of h
is given by:
Jh=h∂h
∂x
∂h
∂y i=h∂f
∂x
∂g
∂x +∂f
∂y
∂g
∂y
∂f
∂x
∂g
∂y +∂f
∂y
∂g
∂x i
Substitute the partial derivatives of fand ginto the Jacobian matrix for-
mula:
Jh=2xy + 2yx 2xy + 2yx=4xy 4yx
Therefore, the Jacobian matrix of the composition h(x, y) = f(g(x, y), g(y, x))
is:
Jh=4xy 4yx
9
Question 12
Question
Let f(x, y, z)=(xy, yz, xz) be a vector-valued function. Find the Jacobian
matrix of f.
Solution
Step 1: Calculate the partial derivatives of fwith respect to x,y, and z.
∂f
∂x =
∂(xy)
∂x
∂(yz)
∂x
∂(xz)
∂x
=
y
0
z
Step 2: Calculate the partial derivatives of fwith respect to y.
∂f
∂y =
∂(xy)
∂y
∂(yz)
∂y
∂(xz)
∂y
=
x
z
0
Step 3: Calculate the partial derivatives of fwith respect to z.
∂f
∂z =
∂(xy)
∂z
∂(yz)
∂z
∂(xz)
∂z
=
0
y
x
Step 4: Assemble the partial derivatives into a matrix to obtain the Jacobian
matrix of f.
J=
∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z
∂f3
∂x
∂f3
∂y
∂f3
∂z
=
y x 0
0z y
z0x
Question 13
Question
Let f(x, y) = (x2+3y, 2x−y2) and g(u, v) = uv2. Compute the Jacobian matrix
of the composition g◦fat the point (1,2).
Solution
To find the Jacobian matrix of the composition g◦f, we first need to compute
g(f(x, y)) and then evaluate the partial derivatives of this composition function
at the point (1,2).
10
Step 1: Find g(f(x, y)) Let’s first find f(x, y):
f(x, y) = (x2+ 3y, 2x−y2)
Now, substitute xand ywith uand vrespectively to find g(f(x, y)):
g(f(x, y)) = g(x2+ 3y, 2x−y2)=(x2+ 3y)(2x−y2)2
Step 2: Compute the Jacobian matrix Now, let’s find the partial deriva-
tives of g(f(x, y)) with respect to xand y:
∂g(f(x, y))
∂x =∂
∂x ((x2+ 3y)(2x−y2)2)
Differentiating with respect to x, we get:
∂g(f(x, y))
∂x = 4x(2x−y2)2+ (x2+ 3y)4(2x−y2)
Similarly, we find the partial derivative with respect to y:
∂g(f(x, y))
∂y =∂
∂y ((x2+ 3y)(2x−y2)2)
Differentiating with respect to y, we get:
∂g(f(x, y))
∂y = (x2+ 3y)2(2x−y2)2+ 3(2x−y2)2
Evaluate these partial derivatives at the point (1,2) to find the Jacobian
matrix of g◦fat that point.
Question 14
Question
Let f:R3→R2be the function defined by f(x, y, z)=(x2+yz, xyz). Compute
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component function with respect to each input variable.
f(x, y, z)=(x2+yz, xyz)
Step 1: Compute the partial derivatives of f1(x, y, z) = x2+yz.
∂f1
∂x = 2x∂f1
∂y =z∂f1
∂z =y
11
Step 2: Compute the partial derivatives of f2(x, y, z) = xyz.
∂f2
∂x =yz ∂f2
∂y =xz ∂f2
∂z =xy
Step 3: Assemble the partial derivatives into a Jacobian matrix.
Jf(x, y, z) = "∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x z y
yz xz xy
Question 15
Question
Let f:R3→R2be a function defined by f(x, y, z)=(x2yz, xyz). Calculate
the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the derivatives of each
component function with respect to each variable.
Step 1: Partial derivatives of the first component Let u=x2yz and
v=xyz. Then, the first component of fis uand we need to find ∂u
∂x ,∂u
∂y , and
∂u
∂z .
∂u
∂x = 2xyz
∂u
∂y =x2z
∂u
∂z =x2y
Step 2: Partial derivatives of the second component Similarly, let
w=xyz. Then, the second component of fis wand we need to find ∂w
∂x ,∂w
∂y ,
and ∂w
∂z .
∂w
∂x =yz
∂w
∂y =xz
∂w
∂z =xy
Step 3: Assembling the Jacobian matrix The Jacobian matrix of fis
given by
J(f) = "∂u
∂x
∂u
∂y
∂u
∂z
∂w
∂x
∂w
∂y
∂w
∂z #
12
Substitute the partial derivatives we found into the Jacobian matrix:
J(f) = 2xyz x2z x2y
yz xz xy
Question 16
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2−yz, 2y+z). Compute
the Jacobian matrix of f.
Solution
To compute the Jacobian matrix of f, we first need to find the partial derivatives
of f.
Step 1: Find ∂f
∂x ,∂f
∂y , and ∂f
∂z .
∂f
∂x =∂
∂x (x2−yz),∂
∂x (2y+z)
= (2x, 0)
∂f
∂y =∂
∂y (x2−yz),∂
∂y (2y+z)
= (−z, 2)
∂f
∂z =∂
∂z (x2−yz),∂
∂z (2y+z)
= (−y, 1)
Step 2: Construct the Jacobian matrix. The Jacobian matrix of fis given
by:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Substitute the partial derivatives we found into the Jacobian matrix:
Jf=2x0 0
−z2 0
Therefore, the Jacobian matrix of fis:
Jf=2x0 0
−z2 0
13
Question 17
Question
Find the Jacobian matrix of the transformation (x, y)7→ (x2+y, xy) at the
point (2,3).
Solution
Step 1: Compute the partial derivatives of the transformation. Let u=x2+y
and v=xy. Then, the transformation can be represented by u=f(x, y) and
v=g(x, y).
The partial derivatives with respect to xand yare:
∂f
∂x = 2xand ∂f
∂y = 1
∂g
∂x =yand ∂g
∂y =x
Step 2: Evaluate the partial derivatives at the point (2,3). At the point
(2,3) we have:
∂f
∂x = 2(2) = 4 and ∂f
∂y = 1
∂g
∂x = 3 and ∂g
∂y = 2
Step 3: Assemble the Jacobian matrix. The Jacobian matrix is given by:
"∂f
∂x
∂f
∂y
∂g
∂x
∂g
∂y #
Substitute the computed partial derivatives:
4 1
3 2
Therefore, the Jacobian matrix of the transformation at the point (2,3) is:
4 1
3 2
Question 18
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, yz). Find the
Jacobian matrix of f.
14
Solution
Step 1: The Jacobian matrix of a function f:Rn→Rmis an m×nmatrix of
partial derivatives of the components of f. Let’s compute the Jacobian matrix
of the given function f(x, y, z) = (x2+y, yz).
Step 2: The components of fare f1(x, y, z) = x2+yand f2(x, y, z) = yz.
The Jacobian matrix Jis given by:
J="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 3: Calculating the partial derivatives:
∂f1
∂x = 2x
∂f1
∂y = 1
∂f1
∂z = 0
∂f2
∂x = 0
∂f2
∂y =z
∂f2
∂z =y
Step 4: Now, substitute these partial derivatives into the Jacobian matrix:
J=2x1 0
0z y
Therefore, the Jacobian matrix of the function f(x, y, z)=(x2+y, yz) is
2x1 0
0z y
.
Question 19
Question
Let f(x, y, z)=(x2+y2+z2, xy, xz) be a vector-valued function. Find the
Jacobian matrix of f.
15
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to x,y, and z.
Step 1: Compute ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x = 2x, ∂f1
∂y = 2y, ∂f1
∂z = 2z
Step 2: Compute ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =y, ∂f2
∂y =x, ∂f2
∂z = 0
Step 3: Compute ∂f3
∂x ,∂f3
∂y , and ∂f3
∂z .
∂f3
∂x =z, ∂f3
∂y = 0,∂f3
∂z =x
Step 4: Assemble the partial derivatives into the Jacobian matrix. The
Jacobian matrix of fis:
Jf(x, y, z) =
2x2y2z
y x 0
z0x
Question 20
Question
Given the following transformation:
x=rcos(θ)
y=rsin(θ)
find the Jacobian matrix Jof the transformation.
Solution
Step 1: Start by expressing the transformation as a vector-valued function
F(r, θ)=(x, y).
F(r, θ)=(rcos(θ), r sin(θ))
Step 2: Now, find the partial derivatives of xand ywith respect to rand θ:
∂x
∂r = cos(θ),∂x
∂θ =−rsin(θ)
∂y
∂r = sin(θ),∂y
∂θ =rcos(θ)
16
Step 3: Assemble these partial derivatives into the Jacobian matrix J.
J=∂x
∂r
∂x
∂θ
∂y
∂r
∂y
∂θ =cos(θ)−rsin(θ)
sin(θ)rcos(θ)
Therefore, the Jacobian matrix of the transformation is J=cos(θ)−rsin(θ)
sin(θ)rcos(θ).
Question 21
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+yz, 3x−2y+z).
Find the Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to compute the partial derivatives of
each component of fwith respect to each variable.
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Step 1: Find ∂f1∂x,∂f1
∂y , and ∂f1
∂z .
∂f1
∂x =∂
∂x (x2+yz) = 2x
∂f1
∂y =∂
∂y (x2+yz) = z
∂f1
∂z =∂
∂z (x2+yz) = y
Step 2: Find ∂f2∂x,∂f2
∂y , and ∂f2
∂z .
∂f2
∂x =∂
∂x (3x−2y+z)=3
∂f2
∂y =∂
∂y (3x−2y+z) = −2
∂f2
∂z =∂
∂z (3x−2y+z)=1
Step 3: Assemble the Jacobian matrix Jf.
Jf=2x z y
3−2 1
Therefore, the Jacobian matrix of fis:
Jf=2x z y
3−2 1
17
Question 22
Question
Let f:R3→R2be a mapping given by f(x, y, z) = (xy +z, x2y). Find the
Jacobian matrix of f.
Solution
To find the Jacobian matrix of f, we need to find the partial derivatives of each
component function with respect to each input variable.
Step 1: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z :
∂f1
∂x =y
∂f1
∂y =x
∂f1
∂z = 1
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z :
∂f2
∂x = 2xy
∂f2
∂y =x2
∂f2
∂z = 0
Step 3: Construct the Jacobian matrix Jf:
Jf="∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #
Jf=y x 1
2xy x20
Therefore, the Jacobian matrix of fis:
Jf=y x 1
2xy x20
Question 23
Question
Let f:R3→R2be a function defined by f(x, y, z) = (x2+y, xyz). Determine
the Jacobian matrix of f.
18
Solution
Step 1: Write fin component form. The function fcan be written as f(x, y, z) =
x2+y
xyz .
Step 2: Calculate the partial derivatives of f. The partial derivatives of f
are: - ∂f1
∂x = 2x-∂f1
∂y = 1 - ∂f1
∂z = 0 - ∂f2
∂x =yz -∂f2
∂y =xz -∂f2
∂z =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 !.
Step 4: Substitute the partial derivatives into the Jacobian matrix. Sub-
stitute the partial derivatives calculated in Step 2 into the Jacobian matrix to
get
Jf=2x1 0
yz xz xy .
Therefore, the Jacobian matrix of the function fis 2x1 0
yz xz xy .
Question 24
Question
Let f:R2→R2be defined by f(x, y) = (3x2+ 2y, x2−y). Determine the
Jacobian matrix of fat the point (1,2).
Solution
Step 1: Compute the partial derivatives of fwith respect to xand y:
∂f1
∂x = 6xand ∂f1
∂y = 2
∂f2
∂x = 2xand ∂f2
∂y =−1
Step 2: Evaluate the partial derivatives at the point (1,2):
∂f1
∂x (1,2) = 6(1) = 6,∂f1
∂y (1,2) = 2
∂f2
∂x (1,2) = 2(1) = 2,∂f2
∂y (1,2) = −1
Step 3: Assemble the Jacobian matrix of fat (1,2):
Jf(1,2) = 6 2
2−1
Therefore, the Jacobian matrix of fat the point (1,2) is 6 2
2−1.
19
Question 25
Question
Let f:R3→R2be 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: Find the partial derivatives Let f(x, y, z)=(u, v) where u=
x2+yand v=yz.
The partial derivatives are given by:
∂u
∂x =∂
∂x (x2+y) = 2x,
∂u
∂y =∂
∂y (x2+y) = 1,
∂u
∂z =∂
∂z (x2+y)=0,
∂v
∂x =∂
∂x (yz) = 0,
∂v
∂y =∂
∂y (yz) = z,
∂v
∂z =∂
∂z (yz) = y.
Step 2: Assemble the Jacobian matrix The Jacobian matrix is then
given by:
"∂u
∂x
∂u
∂y
∂u
∂z
∂v
∂x
∂v
∂y
∂v
∂z #=2x1 0
0z y.
Therefore, the Jacobian matrix of fis 2x1 0
0z y.
Question 26
Question
Let f(x, y) = x2+y2
2xy be a vector-valued function. Find the Jacobian matrix
of fat the point (3,4).
20
Solution
Step 1: Compute the partial derivatives of f:
∂f
∂x =∂f1
∂x
∂f2
∂x =2x
2y,
∂f
∂y = ∂f1
∂y
∂f2
∂y !=2y
2x.
Step 2: Evaluate the partial derivatives at the point (3,4):
∂f
∂x (3,4) = 2·3
2·4=6
8,
∂f
∂y (3,4) = 2·4
2·3=8
6.
Step 3: Construct the Jacobian matrix:
Jf(3,4) = ∂f1
∂x
∂f1
∂y
∂f2
∂x
∂f2
∂y !=6 8
8 6.
Question 27
Question
Let f:R3→R2be 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: Find ∂f1
∂x ,∂f1
∂y , and ∂f1
∂z .
Given f(x, y, z)=(x2−y, yz), we can see that f1(x, y, z) = x2−y. Taking
the partial derivatives:
∂f1
∂x = 2x, ∂f1
∂y =−1,and ∂f1
∂z = 0.
Step 2: Find ∂f2
∂x ,∂f2
∂y , and ∂f2
∂z .
Continuing with the second component function f2(x, y, z) = yz:
∂f2
∂x = 0,∂f2
∂y =z, and ∂f2
∂z =y.
Step 3: Assemble the Jacobian matrix.
21
The Jacobian matrix of fis given by:
"∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z #=2x−1 0
0z y.
Therefore, the Jacobian matrix of fis 2x−1 0
0z y.
Question 28
Question
Let f(x, y, z) = (x2+y, 2y−z) be a vector-valued function. Compute the
Jacobian matrix of f.
Solution
Step 1: The Jacobian matrix of fis defined as the matrix of all first-order
partial derivatives of the vector-valued function f. Step 2: Compute the partial
derivatives of fwith respect to x,y, and z. Step 3: The partial derivatives of
fwith respect to x,y, and zare given by
∂f
∂x =2x
0,∂f
∂y =1
2,∂f
∂z =0
−1.
Step 4: Assemble the partial derivatives into the Jacobian matrix of f:
Jf= ∂f1
∂x
∂f1
∂y
∂f1
∂z
∂f2
∂x
∂f2
∂y
∂f2
∂z !=2x1 0
0 2 −1.
Therefore, the Jacobian matrix of fis 2x1 0
0 2 −1.
Question 29
Question
Let f(x, y, z) = x2+y2+z2and g(x, y, z) = ex+yz . Compute the Jacobian
matrix of the composition g◦fat the point (1,1,1).
Solution
To find the Jacobian matrix of the composition g◦f, we first need to find the
compositions of gand f.
Step 1: Find f(1,1,1) Substitute x= 1, y= 1, and z= 1 into f(x, y, z):
f(1,1,1) = 12+ 12+ 12= 3
22
Step 2: Find the gradient of fat (1,1,1) The gradient of fis given by:
∇f=
∂f
∂x
∂f
∂y
∂f
∂z
Compute the partial derivatives of f:
∂f
∂x = 2x, ∂f
∂y = 2y, ∂f
∂z = 2z
At (1,1,1), the gradient of fis:
∇f(1,1,1) =
2
2
2
Step 3: Find g(f(1,1,1)) Substitute f(1,1,1) = 3 into g(x, y, z):
g(3) = e3
Step 4: Find the gradient of gat e3The gradient of gis given by:
∇g=
∂g
∂x
∂g
∂y
∂g
∂z
Compute the partial derivatives of g:
∂g
∂x =ex+yz ,∂g
∂y =zex+yz ,∂g
∂z =yex+yz
At e3, the gradient of gis:
∇g(e3) =
e3
e3
e3
Step 5: Find the Jacobian matrix of g◦fat (1,1,1) By chain rule,
the Jacobian matrix of g◦fat (1,1,1) is given by:
∇g(e3)· ∇f(1,1,1) =
e3
e3
e3
·
2
2
2
=
2e3
2e3
2e3
Therefore, the Jacobian matrix of the composition g◦fat the point (1,1,1)
is
2e3
2e3
2e3
.
23
Question 30
Question
Let f:R3→R3be a function defined by f(x, y, z)=(x2yz, xy3z, xyz2). Find
the Jacobian matrix of fat the point (1,2,3).
Solution
Step 1: Compute the partial derivatives of fwith respect to each variable:
∂f1
∂x = 2xyz, ∂f1
∂y =x2z, ∂f1
∂z =x2y,
∂f2
∂x =y3z, ∂f2
∂y = 3xy2z, ∂f2
∂z =xy3,
∂f3
∂x =yz2,∂f3
∂y =xz2,∂f3
∂z = 2xyz.
Step 2: Evaluate the partial derivatives at the point (1,2,3):
∂f1
∂x (1,2,3)
= 2(1)(2)(3) = 12,∂f1
∂y (1,2,3)
= (1)2(3) = 3,∂f1
∂z (1,2,3)
= (1)2(2) = 2,
∂f2
∂x (1,2,3)
= (2)3(3) = 24,∂f2
∂y (1,2,3)
= 3(1)(2)2= 12,∂f2
∂z (1,2,3)
= (1)(2)3= 8,
∂f3
∂x (1,2,3)
= (2)(3)2= 18,∂f3
∂y (1,2,3)
= (1)(3)2= 9,∂f3
∂z (1,2,3)
= 2(1)(2)(3) = 12.
Step 3: Assemble the Jacobian matrix of fat (1,2,3):
Jf(1,2,3) =
12 3 2
24 12 8
18 9 12
.
Therefore, the Jacobian matrix of fat the point (1,2,3) is
12 3 2
24 12 8
18 9 12
.
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