MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q1. Let
 3
2
3
x + x − y
f (x, y) =
x2 + y 2

c
if (x, y) ̸= (0, 0)
if (x, y) = (0, 0)
where c ∈ R is a fixed constant.
(a) Find the partial derivative fx (1, 1).
(b) Find, with justification, a value for c that guarantees that the partial derivative fx (0, 0)
exists.
Solution:
(a) In a neighborhoud of (1, 1), f is given by
f (x, y) =
x3 + x2 − y 3
.
x2 + y 2
So we can use the single-variable rules of differentiation (in particular, the quotient rule)
to find
(3x2 + 2x)(x2 + y 2 ) − (x3 + x2 − y 3 )(2x)
fx (x, y) =
(x2 + y 2 )2
for all (x, y) near (1, 1). Thus,
fx (1, 1) =
10 − 4
= 2.
2
(b) Using the definition of the partial derivative, we have
f (h, 0) − f (0, 0)
fx (0, 0) = lim
= lim
h→0
h→0
h
h3 +h2
h2
h
−c
h+1−c
.
h→0
h
= lim
We can take c = 1 to guarantee that this limit exists, and in which case fx (0, 0) = 1.
[Note: In fact, the only value of c that makes the above limit exist is c = 1. Can you
prove this?]
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q2. Let f (x, y) be a differentiable function. You are told that f (1, 2) = 1, f (0, 0) = −1, ∇f (1, 2) =
(1, 4) and ∇f (0, 0) = (1, 0). Using this information, determine
d
(f (1 + 3t, 2 − t)) at t = 0.
dt
Solution:
Let’s use the chain rule to find
d
(f (x(t), y(t))) where x(t) = 1 + 3t and y(t) = 2 − t:
dt
d
dx
dy
(f (x(t), y(t))) = fx (x(t), y(t))
+ fy (x(t), y(t))
= 3fx (x(t), y(t)) − fy (x(t), y(t)).
dt
dt
dt
When t = 0, we have
fx (x(0), y(0)) = fx (1, 2)
and fy (x(0), y(0)) = fy (1, 2).
Since ∇f (1, 2) = (1, 4), we have
fx (1, 2) = 1
and
fy (1, 2) = 4.
Consequently,
d
(f (x(t), y(t)))
dt
t=0
= 3(1) − 4 = −1.
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q3. Let f (x, y) = ex cos(xy).
(a) Find the rate of most rapid increase in f at the point (1, π). Give a unit vector in the
direction that produces this increase.
(b) Find an equation for the tangent plane to f at the point (0, 0).
Solution:
(a) The rate of most rapid increase is ∥∇f (1, π)∥. Here,
∇f (x, y) = (fx (x, y), fy (x, y)) = (ex cos(xy) − yex sin(xy), −xex sin(xy)).
Thus,
∥∇f (1, π)∥ = ∥(−e, 0)∥ = |e|∥(1, 0)∥ = e.
This increase occurs in the direction of the unit vector
u
b=
1
∇f (1, π)
= (−e, 0) = (−1, 0).
∥∇f (1, π)∥
e
(b) The equation of the tangent plane to f at (0, 0) is given by
z = f (0, 0) + fx (0, 0)(x − 0) + fy (0, 0)(y − 0)
= 1 + (1)(x − 0) + (0)(y − 0)
= 1 + x.
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q4. Let f (x, y) = 1 + 2xy − x2 − y 2 .
(a) Find the critical points of f and determine whether they are local maxima, minima or
saddle points.
(b) Find the global maximum and minimum values of f if they exist, or prove that they
don’t exist.
[Hint: Complete the square.]
Solution:
(a) We want the points where
∇f (x, y) = (0, 0) ⇐⇒ (2y − 2x, 2x − 2y) = (0, 0) ⇐⇒ y = x.
That is, the critical points of f are all the points of the form (x, y) = (a, a) where a ∈ R.
Let’s apply the Second Derivative Test to these points. The Hessian matrix of f is
−2 2
fxx fxy
.
=
Hf (x, y) =
2 −2
fyx fyy
The determinant of this matrix is 0. Thus all the critical points are degenerate! So the
Second Derivative Test is of no help.
At (x, y) = (a, a), the value of f is f (a, a) = 1 + 2a2 − a2 − a2 = 1. Notice that
f (x, y) = 1 − (x2 − 2xy + y 2 ) = 1 − (x − y)2 ≤ 1
for all (x, y) ∈ R2 . This shows that the value 1 is actually the maximum value of f . So
the critical points (a, a) are all local (in fact, global) maxima.
(b) In part (a) we proved that 1 is the global maximum value of f . We can also conclude
from our work in part (a) that f has no global minimum, since if it did have one, it
would occur at a critical point—but we saw that all the critical points are actually
maxima. [Alternatively: We can see that f has no global minimum value because
f (x, 0) = 1 − x2 can be made arbitrarily small.]
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q5. The plane x + y + 2z = 2 intersects the paraboloid z = x2 + y 2 in an ellipse E.
(a) Determine an equation for E.
(b) Use the Lagrange Multipliers Algorithm to find the point(s) on E that are closest to the
origin.
Solution:
(a) x + y + 2(x2 + y 2 ) = 2. [Note that this is actually a circle! We can re-write the equation
3
.]
as (x + 41 )2 + (y + 14 )2 = 89 . This is a circle centred at (− 41 , − 14 ) of radius 2√
2
(b) To find the
ppoints on E closest to the origin, we want to minimize the distance function
d(x, y) = (x − 0)2 + (y − 0)2 for (x, y) on E. This gives us a constrained optimization
problem. It will be easier to work with the distance-squared function f (x, y) = x2 + y 2 .
Thus our problem is
minimize
subject to
f (x, y) = x2 + y 2
x + y + 2(x2 + y 2 ) = 2.
|
{z
}
g(x,y)
Let’s use the Lagrange Multipliers Algorithm, as instructed.
• Step 1: We must solve the system of equations
∇f (x, y) = λ∇g(x, y)
g(x, y) = 2.
or equivalently
2
2x = λ(1 + 4x)
(1)
2y = λ(1 + 4y)
(2)
2
(3)
x + y + 2(x + y ) = 2.
We can re-write equations (1) and (2) as
(2 − 4λ)x = λ
(2 − 4λ)y = λ
Notice that 2 − 4λ ̸= 0, since otherwise we get 0 = 12 . Thus we can divide through
λ
by 2 − 4λ to find that x = y = 2−4λ
. If we plug this into equation (3), we get
2x + 4x2 = 2 ⇐⇒ 2x2 + 2x − 1 = 0 ⇐⇒ (2x − 1)(x + 1) = 0.
Thus x = 12 or x = −1 (and then consequently y = x and we can solve for λ using
λ
x = 2−4λ
).
This gives us the solutions (x, y) = ( 12 , 12 ) and (−1, −1).
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
• Step 2: We must solve the system of equations
∇g(x, y) = (0, 0)
g(x, y) = 2.
or equivalently
1 + 4x = 0
(4)
1 + 4y = 0
(5)
2
2
x + y + 2(x + y ) = 2.
(6)
From (4) and (5) we find that x = y = − 41 which is not a solution to (6). (In Step
1 we’d determined the solutions to (6) where x = y and this is not one of them.)
So there are no solutions.
• Step 3: We must look for end points on the constraint curve g(x, y) = 2. This is
our ellipse (circle) E, which has not end points.
• Step 4: Let’s evaluate f at the solutions found in Step 1: f ( 12 , 12 ) = 12 and
f (−1, −1) = 2. Consequently, the point on E closest to the origin is the point
( 21 , 21 ).
[Note: Since E is a circle, we can actually verify this result using elementary
geometry. The point on E closest to (0, 0) will lie on the radial line through (0, 0).
Since the centre of E is at (− 14 , − 41 ), the radial line through (0, 0) has equation y = x.
The points of intersection of this line and E give us the closest and furthest points
from (0, 0). This is exactly what the Lagrange Multipliers Algorithm determined!]
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q6. (7 marks) Determine whether each statement is true or false. Circle the correct answers. No
justification is needed.
(a) If f (x, y) is differentiable at (a, b) then f (x, y) is continuous at (a, b).
Circle the correct answer:
True
False
(b) If f (x, y) is continuous at (a, b), then
Circle the correct answer:
p
|f (x, y)| is continuous at (a, b).
False
True
(c) The degree 2 Taylor polynomial of f (x, y) = xy at (1, 1) is
P2,(1,1) (x, y) = 1 + (x − 1) + (y − 1) +
Circle the correct answer:
True
1
2(x − 1)(y − 1) .
2
False
(d) If f (x, y) doesn’t have a global minimum in D ⊆ R2 then either f is not continuous or
D is either not closed or not bounded.
Circle the correct answer:
True
(e) If x = 2u − v and y = 3u + 3v, then
Circle the correct answer:
False
∂(x, y)
1
= .
∂(u, v)
9
True
False
(f) In spherical coordinates, the equation ρ cos ϕ = 2 describes a sphere.
Circle the correct answer:
True
False
Z
2π
Z
2Z 1
(g) In cylindrical coordinates, the integral
r dz dr dθ represents the volume of a
0
1
0
solid enclosed between a cylinder and two planes.
Circle the correct answer:
True
False
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Justifications:
(a) Theorem 1 in §5.2.
(b) Continuity Theorems. Since | · | and
continuous functions.
√
· are continuous,
p
|f (x, y)| is a composition of
(c) The easy way is to multiply out the given P2 and show that it actually is equal to xy!
(Think back to the WA2 problem about Taylor polynomials of polynomials to see why
this makes sense.)
(d) This is just the (contrapositive of the) Extreme Value Theorem.
∂(x, y)
∂(u, v)
(e)
= 9. (So what was calculated here is actually
.)
∂(u, v)
∂(x, y)
(f) Since z = ρ cos ϕ, the given equation is simply z = 2, which is a plane.
(g) The given statement is almost correct. What the integral represents is the volume of the
solid enclosed between the cylinders r = 1 and r = 2 and the two planes z = 0 and z = 1.
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
ZZ
xy dA, where D is the triangle with vertices (0, 2), (1, 1), and (3, 2).
Q7. (a) Evaluate
D
(b) Let D = {(x, y) : x2 + y 2 ≤ 2, x ≤ y}. Use polar coordinates to evaluate
ZZ p
x2 + y 2 + 1 dA.
D
Solution:
(a) Here is a sketch of the region D.
y
2
y-slice
x = 2y − 1
1
x=2−y
x
1
2
3
From this we see that we should use the order of integration dx dy. Thus,
ZZ
ymax
Z
Z
xmax (y)
xy dx dy
xy dA =
ymin
xmin (y)
Z 2 Z 2y−1
D
=
xy dx dy
1
2−y
2
1
y[(2y − 1)2 − (2 − y)2 ] dy
2
Z
2
Z
=
1
3
=
2
y 3 − y dy
1
3 1 3 1 2 y=2
=
y − y
2 4
2
y=1
27
= .
8
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
(b) Here is a sketch of the region D.
5π
4
π
4
√
− 2
From this we see that
ZZ p
Z
2
2
x + y + 1 dA =
D
θmax
Z
θmin
5π
4
Z
=
π
4
√
2
rmax (θ) p
r2 + 1 r dr dθ
rmin (θ)
√
2 p
Z
r
r2 + 1 dr dθ
0
√
Z
=π
2
r
p
r2 + 1 dr
0
Z
=π
1
3
1√
u du
2
π
= (33/2 − 1).
3
(u = r2 + 1)
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
3 bounded by the planes x + y + z = 3, x = 2 and
Q8. Let E be the region in the first octant of R
ZZZ
y = 1. (See the picture below.) Evaluate
x dV .
E
z
1
y
2
x
Solution:
Let’s use the order of integration dz dy dx. Thus,
ZZZ
xmax
Z
Z
ymax (x) Z zmax (x,y)
x dz dy dx.
x dV =
xmin
E
ymin (x)
zmin (x,y)
Clearly xmin = 0 and xmax = 2. Next, consider an x-cross section:
z
z =3−x−y
y
3−x
1
y-slice
In this cross-section, we see that ymin (x) = 0 and ymax (x) = 1. Then, taking a y-cross section,
we find that zmin (x, y) = 0 and zmax (x, y) = 3 − x − y. So the given integral is
Z
2 Z 1 Z 3−x−y
Z
2Z 1
0
0
Z
3x − x − xy dy dx =
x dz dy dx =
0
2
0
0
0
2
7
5
x − x2 dx = .
2
3
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
Q9. A solid ring is formed using the region E lying inside the sphere x2 + y 2 + z 2 = 4 and outside
the the cylinder x2 + y 2 = 3. (See the picture below.) The ring is made out of a material
whose mass-density function is given by f (x, y, z) = z 2 .
(a) Set up (but do not evaluate) iterated integrals that compute the mass of the ring...
(i) ...in cylindrical coordinates, with order of integration dz dr dθ.
(ii) ...in spherical coordinates, with order of integration dρ dϕ dθ.
(b) Find the mass of E by evaluating one of your integrals from part (a).
Solution:
(a)
(i) We have
ZZZ
Z
2
mass(E) =
θmax
Z
rmax (θ) Z zmax (r,θ)
z dV =
θmin
E
rmin (θ)
z 2 r dz dr dθ.
zmin (r,θ)
Clearly, θmin = 0 and θmax = 2π. Here is a sketch of a θ-cross section of E.
z
r=
√
r-slice
3
r
In cylindrical coordinates, the sphere x2 +y 2 +z 2 = 4 becomes r2 +z 2 = 4, and this
gives the circle of radius 2 in the θ-cross section above.
The cylinder x2 + y 2 = 3
√
2
becomes r = 3 and this gives
√ the vertical line r = 3 in the θ-cross section above.
Consequently, rmin (θ)
√ = 3 and rmax (θ) = 2.
√ Next, if we take an r-slice, we find
that zmin (r, θ) = − 4 − r2 and zmax (r, θ) = 4 − r2 occur on the circle. Thus,
Z
mass(E) =
0
2π
Z
2
√
3
√
Z
−
4−r2
√
z 2 r dz dr dθ.
4−r2
MATH237: Calculus 3
Practice Final Exam Solutions
Spring 2022
(ii) We have
ZZZ
Z
z 2 dV =
mass(E) =
θmax
θmin
E
ϕmax (θ) Z ρmax (ϕ,θ)
Z
ϕmin (θ)
(ρ cos ϕ)2 ρ2 sin ϕ dρ dϕ dθ.
ρmin (ϕ,θ)
Clearly, θmin = 0 and θmax = 2π. Here is a sketch of a θ-cross section of E.
z
ϕ-slice
ϕmin
r
ϕmax
In spherical coordinates, the sphere x2 + y 2 + z 2 = 4 becomes ρ = 2, and this gives
the circle√of radius 2 in the θ-cross
√ section above.
√ Since r = ρ sin ϕ, the vertical
line r = 3 is given by ρ sin ϕ = 3 ⇐⇒ ρ = 3 csc ϕ. This line intersects the
circle ρ = 2 when
√
√
3
2 sin ϕ = 3 =⇒ sin ϕ =
.
2
Consequently, ϕmin (θ) =
π
3
and ϕmax (θ) =
2π
3 .
Next, if we take a ϕ-slice, we find that ρmin (ϕ, θ) =
and ρmin (ϕ, θ) = 2 is on the circle. Thus,
Z
2π
2π
3
Z
mass(E) =
Z
3 csc ϕ is on the vertical line
2
√
π
3
0
√
ρ4 cos2 ϕ sin ϕ dρ dϕ dθ.
3 csc ϕ
(b) Let’s use part (a)(i):
Z
mass(E) =
=
=
=
2π
Z
2
√
Z
4−r2
u
du
z 2 r dz dr dθ
√
2
0
3 − 4−r
Z 2 Z √4−r2
2π √
z 2 r dz dr
√
3 − 4−r2
Z
4π 2
r(4 − r2 )3/2 dr
3 √3
Z
4π 1 1 3/2
3
4π
=
.
15
√
0
2
(u = 4 − r2 )