Multivariable Calculus Practice Final Exam with Complete Solutions, Exams of Mathematics

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MATH V 253

MULTIVARIABLE

CALCULUS —

PRACTICE FINAL A

LATEST VERSION

WITH COMPLETE

SOLUTIONS

Study Ace Smart

APRIL 6, 2025

Multivariable Calculus — Practice Final A — 150

minutes

• The test consists of 11 pages and 9 questions worth a total of 90 marks.
• Electronic devices of any kind (including calculators, cell phones, etc.) are

forbidden.

• This is a closed book exam. No calculators, notes, or books are allowed.
• No work on this page will be marked.
• Fill in the information below before turning to the questions.

1. Let ρ ( x,y,z ) =

p x

2

• y

2

• z

2

. Find the constants a,b such that

3. A function f ( x,y,z ) is known to have f (2 , 1 , 1) = 5, and its derivatives

at that point are fx (2 , 1 , 1) = 1, fy (2 , 1 , 1) = 2, fz (2 , 1 , 1) = 3.

(a) Make a unit vector from (2 , 1 , 1) towards (3 , 2 , 2). Calculate the rate of

change of f ( x,y,z ) at (2 , 1 , 1) in this direction.

(b) Use linear approximation to approximate f (1_._ 9 , 1 , 1_._ 2).

Solution: Next, we do the linear approximation. f (1_._ 9 , 1 , 1_._ 2) ≈

f (2 , 1 , 1) + fx (2 − 1_._ 9) + fy (1 − 1) + fz (1_._ 2 − 1)

This gives

f (1_._ 9 , 1 , 1_._ 2) ≈ 5 + (1)(− 0_._ 1) + 0 + 3(0_._ 2) = 5_._ 5

(c) Let g ( x,y,z ) = x + z. Find all unit vector directions where the rate of change

of f ( x,y,z ) and g ( x,y,z ) are both zero at the point (2 , 1 , 1).

Solution: We differentiate once.

d

dt

f ( x ( t ) ,y ( t ))= f (^) x x ˙ + f (^) y y ˙

=(4 t ) f (^) x +3 t

2 f (^) y.

We differentiate twice.

d

2

dt

2

f ( x ( t ) ,y ( t ))=4 f (^) x +(4 t )( f (^) xx x ˙ + f (^) xy y ˙ )+6 tf (^) y +3 t

2 ( f (^) yx x ˙ + f (^) yy y ˙ )

At t = 1 , we obtain ˙ x = 4 , ˙ y = 3 and the second derivative evaluates to

which becomes

The answer is 28.

Solution: The gradient of f at the point in consideration is ∇^ f =

h 1 , 2 , 3 i. The unit vector in the given direction is ~

u

√^1 3

h 1 , 1 , 1 i.

Therefore

D ~

u

f = h 1 , 2 , 3 i·^

h 1 , 1 , 1 i (^) =

Solution:

Finally, we compute the direction where D~uf = D~ug = 0. Note ∇ g = h 1 , 0 , 1 i.

We need to solve

h 1 , 2 , 3 i · ~u = 0 , h 1 , 0 , 1 i · ~u = 0_._

For this, we can take a cross product.

h 1 , 2 , 3 i × h 1 , 0 , 1 i = h 2 , 2 , − 2 i_._

Making this a unit vector and considering both possible directions, we

obtain

4. Use the method of Lagrange multipliers to find the minimum

value of z = x

2

• y

2 subject to x

2 y = 1. At which point or points does

the minimum occur?

6. Let R = { x

2

• y

2 ≤ x }. Evaluate

Solution: We switch the order of integration. The region of integration

R is given as

R = { 0 ≤ y ≤ 1 ,

y ≤ x ≤ 1 }.

Bounding x by constants gives 0 ≤ x ≤ 1. Next, we bound y ( x ) in terms

of x. From

y ≤^ x , we obtain y ≤^ x

2

. Therefore

R = { 0 ≤ x ≤ 1 , 0 ≤ y ≤ x

2 }.

The double integral becomes

Z

1

0

Z

x

2

0

sin( πx

2 )

x

dydx

This evaluates to

Z

1

0

sin( πx

2 )

x

[ x

2 − (^) 0] dx =

Z

1

0

x sin( πx

2 ) dx

1

0

cos( πx

2 )

2 π

π

7. Let E be region bounded by planes y = 0, y = 2, y + z = 3 and the

surface z = x

2

. Consider

ZZZ

I = dV.

E

Find the limits of integration for the following orders of integration:

(a)

RRR

E dzdxdy (b)

RRR

E dxdydz

Solution: We use polar coordinates: x = r cos θ , y = r sin θ. The in-

equality defining R becomes

r

2 ≤ r cos θ

and so 0 ≤^ r ≤^ cos θ. Since cos θ ≥^0 , we must have −^ π/ 2 ≤^ θ ≤^ π/ 2.

The integral is then

Z

π/ 2

− π/ 2

Z

cos θ

0

r cos θ

r

2

rdrdθ =

Z

π/ 2

− π/ 2

Z

cos θ

0

cos θdrdθ

Z

π/ 2

− π/ 2

(cos θ )

2 dθ

Z

π/ 2

− π/ 2

(1+ cos 2 θ ) dθ

π/ 2

− π/ 2

θ

sin2 θ

π

• (^) x ( y,z ) bounds: here we have x

2 ≤ (^) z. Thus

z ≤^ x ≤^

z.

Case 1: 2 ≤^3 −^ z. In this case, the upper bound for y to keep is y ≤^ 2.

This case also implies the restriction z ≤^1. Thus we have the contribution

Z

1

0

Z

2

0

Z

√ z

−

√ z

dxdydz.

Case 2: 2 ≥^3 −^ z. In this case, the upper bound for y to keep is y ≤^3 −^ z.

This case implies the restriction z ≥ 1. Thus we have the contribution

Z

3

1

Z

3 −^ z

0

Z

√ z

−

√ z

dxdydz.

Altogether, the integral is

Vol( E )=

Z

1

0

Z

2

0

Z

√ z

−

√ z

dxdydz +

Z

3

1

Z

3 −^ z

0

Z

√ z

−

√ z

dxdydz.

8. 10 marks Let E be the solid bounded below by the paraboloid z = x

2

• y

2

and above by the cone z =

p x

2

• y

2

. Let

(a) Write I in cylindrical coordinates. Do not evaluate.

(b) Write I in spherical coordinates. Do not evaluate.

(c) Evaluate I.