Math 148 Exam III Practice Problems
This review should not be used as your sole source for preparation for the exam. You
should also re-work all examples given in lecture, all homework problems, all lab assignment
problems, and all quiz problems.
1. Find the largest possible domain and corresponding range of each function.
2
x+y
p
(b) f (x, y) = 36 − 9x2 − 4y 2
(a) f (x, y) =
2. Draw a contour map showing several level curves of each function.
(a) f (x, y) = xy
(b) f (x, y) = x2 + 9y 2
3. Consider the function f defined by

 3xy
, if (x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0,
if (x, y) = (0, 0)
Determine whether f is continuous at (0, 0). Explain your answer.
4. Find all first-order partial derivatives of each function.
(a) f (x, y) =
x2 y 2
−
y
x
(b) f (x, y) = exy
2
(c) f (x, y) = y ln(xy)
5. Find all first-order partial derivatives of each function.
x
yz
(b) f (x, y, z) = ey cos z sin x
(a) f (x, y, z) = x3 y 2 z +
6. Find all second-order partial derivatives of each function.
√
(a) f (x, y) = x2 y + x y
(b) f (x, y) = sin(x + y) + cos(x − y)
7. Find an equation of the tangent plane to f (x, y) =
p
4 − x2 − 2y 2 at (1, −1, 1).
8. Find the linearization of f (x, y) = y cos(x − y) at (2, 2) and use it to approximate
f (1.9, 2.1). Check the accuracy of your approximation using a calculator.
1
9. Let f (x, y) = ex sin y, where x(t) = t2 and y(t) = 4t. Find
df
when t = 0.
dt
p
10. Compute the directional derivative of f (x, y) = xy − 2x2 at the point P = (1, 6) in
the direction of the point Q = (3, 1).
p
11. In what direction does f (x, y) = x2 − y 2 increase most rapidly at (5, 3)? What is
the largest increase?
12. Find a unit vector that is normal to the level curve of the function
f (x, y) = x2 +
y2
4
at (1, 2).
13. Find all local extrema and saddle points of f (x, y) = −x2 − y 2 + 6x + 8y − 21.
14. Find all local extrema and saddle points of f (x, y) = 2x4 + y 2 − 12xy.
15. Find the absolute extrema of f (x, y) = x2 − y 2 + 4x + y on the rectangular region
R = {(x, y)| − 4 ≤ x ≤ 0, 0 ≤ y ≤ 1}.
16. Find the absolute extrema of f (x, y) = x2 + y 2 + x − y on the disk
D = {(x, y)|x2 + y 2 ≤ 1}.
17. Consider the linear system
x1 (t + 1) = −0.4x1 (t) + 0.2x2 (t)
x2 (t + 1) = −0.3x1 (t) + 0.1x2 (t)
~ = ~0.
Determine the stability of X
18. Find all nonnegative equilibria of
x1 (t + 1) = 2x1 (t)[1 − x1 (t)]
x2 (t + 1) = x1 (t)[1 − x2 (t)]
and determine their stability.
2
Solutions
1. Find the largest possible domain and corresponding range of each function.
(a) f (x, y) =
2
x+y
The expression for f is defined if the denominator is not zero. So the domain of
f is
D = {(x, y)|x + y 6= 0}.
Note that x + y 6= 0 means that point on the line y = −x must be excluded from
the domain. The range of f is
2
z z =
, (x, y) ∈ D .
x+y
Every point (x, y) in the domain D satisfies either (i) 0 < x + y < ∞ in which
case
2
<∞
0<
x+y
or (ii) −∞ < x + y < 0 in which case
−∞ <
2
< 0.
x+y
So the range of f is
{z| − ∞ < z < ∞} = (−∞, ∞) = R.
(b) f (x, y) =
p
36 − 9x2 − 4y 2
The expression for f is defined if the quantity under the square root sign is
nonnegative. So the domain of f is
2
2
x
y
2
2
D = {(x, y)|36 − 9x − 4y ≥ 0} = (x, y) +
≤1
4
9
which is a closed elliptical disk. The range of f is
p
{z| 36 − 9x2 − 4y 2 , (x, y) ∈ D}.
Since z is a positive square root, z ≥ 0. Also
36 − 9x2 − 4y 2 ≤ 36
=⇒
p
36 − 9x2 − y 2 ≤ 6.
So the range of f is
{z|0 ≤ z ≤ 6} = [0, 6].
3
2. Draw a contour map showing several level curves of each function.
(a) f (x, y) = xy
The level curves are defined by
xy = k.
If k 6= 0, then x 6= 0 and the level curves are defined by
y=
k
.
x
This is a family of hyperbolas centered at (0, 0). If k = 0, then x = 0 or y = 0.
Thus, the level curve for k = 0 consists of the coordinate axes x = 0 and y = 0.
A contour plot showing the level curves for k = ±1, ±4, ±8 is shown below.
(b) f (x, y) = x2 + 9y 2
The level curves are defined by
x2 + 9y 2 = k.
This is a family of concentric ellipses with center (0, 0). A contour plot showing
the level curves for k = 1, 9, 18 is shown below.
4
3. Consider the function f defined by

 3xy
, if (x, y) 6= (0, 0)
f (x, y) =
x2 + y 2

0,
if (x, y) = (0, 0)
Determine whether f is continuous at (0, 0). Explain your answer.
The function f is defined at (0, 0) as f (0, 0) = 0. However, we will show that the limit
of the function as (x, y) → (0, 0) does not exist.
Along the x-axis (y = 0),
0
3xy
= lim 2 = 0.
2
x→0 x
+y
lim
(x,y)→(0,0) x2
Along the line y = x,
3x2
3
3xy
=
lim
= 6= 0.
2
2
2
x→0 2x
(x,y)→(0,0) x + y
2
lim
Therefore, the limit does not exist. Thus, f is discontinuous at (0, 0).
4. Find the first-order partial derivatives of each function.
(a) f (x, y) =
x2 y 2
−
y
x
For simplicity, we rewrite the function as
f (x, y) =
x3 − y 3
.
xy
Using the Quotient Rule, we have
3x2 (xy) − y(x3 − y 3 )
2x3 y + y 4
2x3 + y 3
=
=
x2 y 2
x2 y 2
x2 y
−3y 2 (xy) − x(x3 − y 3 )
−2xy 3 − x4
−2y 3 − x3
fy (x, y) =
=
=
.
x2 y 2
x2 y 2
xy 2
fx (x, y) =
(b) f (x, y) = exy
2
Using the Chain Rule, we have
fx (x, y) = y 2 exy
2
2
fy (x, y) = 2xyexy .
5
(c) f (x, y) = y ln(xy)
Using the Chain Rule, we have
fx (x, y) = y
y
xy
=
y
.
x
Using the Product and Chain Rules, we have
x
fy (x, y) = ln(xy) + y
= ln(xy) + 1.
xy
5. Find all first-order partial derivatives of each function.
(a) f (x, y, z) = x3 y 2 z +
x
yz
For simplicity, we rewrite the function as
f (x, y, z) = x3 y 2 z + xy −1 z −1 .
Therefore, we have
1
yz
x
3
fy (x, y, z) = 2x yz − 2
y z
x
3 2
fz (x, y, z) = x y − 2 .
yz
fx (x, y, z) = 3x2 y 2 z +
(b) f (x, y, z) = ey cos z sin x
Treating y and z as constants, we have
fx (x, y, z) = ey cos z cos x.
Using the Chain Rule, we have
fy (x, y, z) = cos zey cos z sin x
fz (x, y, z) = −y sin zey cos z sin x.
6
6. Find all second-order partial derivatives of each function.
√
(a) f (x, y) = x2 y + x y
The first-order partial derivatives of f are
fx (x, y) = 2xy +
√
y
x
fy (x, y) = x2 + √ .
2 y
Therefore, we find that
fxx (x, y) = 2y
1
fxy (x, y) = 2x + √
2 y
1
fyx (x, y) = 2x + √
2 y
x
fyy (x, y) = − 3/2
4y
(b) f (x, y) = sin(x + y) + cos(x − y)
The first-order partial derivatives of f are
fx (x, y) = cos(x + y) − sin(x − y)
fy (x, y) = cos(x + y) + sin(x − y).
Therefore, we find that
fxx (x, y)
fxy (x, y)
fyx (x, y)
fyy (x, y)
=
=
=
=
− sin(x + y) − cos(x − y)
− sin(x + y) + cos(x − y)
− sin(x + y) + cos(x − y)
− sin(x + y) − cos(x − y)
7
7. Find an equation of the tangent plane to f (x, y) =
p
4 − x2 − 2y 2 at (1, −1, 1).
Using the Chain Rule, we have
−x
fx (x, y) = p
4 − x2 − 2y 2
−2y
.
fy (x, y) = p
4 − x2 − 2y 2
At the point (1, −1), the partial derivatives are fx (1, −1) = −1 and fy (1, −1) = 2.
Therefore, an equation of the tangent plane is
z − z0
z−1
z−1
x − 2y + z
=
=
=
=
fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
−1(x − 1) + 2(y + 1)
−x + 2y + 3
4.
8. Find the linearization of f (x, y) = y cos(x − y) at (2, 2) and use it to approximate
f (1.9, 2.1). Check the accuracy of your approximation using a calculator.
Using the Product and Chain Rules, we have
fx (x, y) = −y sin(x − y)
fy (x, y) = cos(x − y) + y sin(x − y).
At the point (2, 2), the partial derivatives are fx (2, 2) = 0 and fy (2, 2) = 1. Moreover,
z = f (2, 2) = 2. Therefore, the linearization of f at (2, 2) is
L(x, y) = z0 + fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 )
L(x, y) = 2 + 0(x − 2) + 1(y − 2)
L(x, y) = y.
To approximate f (1.9, 2.1), we use the linearization
f (1.9, 2.1) ≈ L(1.9, 2.1) = 2.1.
Using a calculator, we find that
f (1.9, 2.1) = 2.1 cos(−0.2) ≈ 2.06.
8
9. Let f (x, y) = ex sin y, where x(t) = t2 and y(t) = 4t. Find
df
when t = 0.
dt
Using the Chain Rule for functions of one parameter,
df
∂f dx ∂f dy
=
+
dt
∂x dt
∂y dt
x
= e sin y(2t) + ex cos y(4).
If t = 0, then x = y = 0 and
df
= 0 + 4 = 4.
dt
10. Compute the directional derivative of f (x, y) =
the direction of the point Q = (3, 1).
p
xy − 2x2 at the point P = (1, 6) in
The vector which defines the direction is
−→
~v = P Q = h2, −5i.
A unit vector in the direction of ~v is
~v
=
~u =
||~v ||
2
5
√ , −√
29
29
.
The gradient vector of f is
*
∇f (x, y) =
y − 4x
x
p
, p
2 xy − 2x2 2 xy − 2x2
+
.
At the point P = (1, 6),
∇f (1, 6) =
1 1
,
2 4
.
Therefore, the directional derivative of f at P in the direction of Q is
1
5
1
D~u f (1, 6) = ∇f (1, 6) · ~u = √ − √ = − √ .
29 4 29
4 29
9
11. In what direction does f (x, y) =
the largest increase?
p
x2 − y 2 increase most rapidly at (5, 3)? What is
The function increases most rapidly in the direction of the gradient vector:
*
+
x
−y
,p
∇f (x, y) = p
.
x2 − y 2
x2 − y 2
At the point (5, 3), we have
∇f (5, 3) =
5 3
,−
4 4
.
The largest increase is given by
r
||∇f (5, 3)|| =
9
25
+
=
16 16
r
34
=
16
√
34
.
4
12. Find a unit vector that is normal to the level curve of the function
f (x, y) = x2 +
y2
4
at (1, 2).
The gradient vector is normal to the level curve:
D
yE
∇f (x, y) = 2x,
.
2
At the point (1, 2), we have
∇f (1, 2) = h2, 1i .
A unit vector in the direction of ∇f (1, 2) is
∇f (1, 2)
~u =
=
||∇f (1, 2)||
10
2 1
√ ,√
5 5
.
13. Find all local extrema and saddle points of f (x, y) = −x2 − y 2 + 6x + 8y − 21.
We first locate the critical points by setting the partial derivatives equal to zero:
fx = −2x + 6 = 0
fy = −2y + 8 = 0.
and
The only critical point is (3, 4).
Next we compute the second partial derivatives:
fxx = −2
fyy = −2.
fxy = 0
The discriminant is D(x, y) = 4. Since D(3, 4) = 4 > 0 and fxx (3, 4) = −2 < 0, there
is a local maximum at (3, 4). The local maximum value is f (3, 4) = 4. There are no
local minima or saddle points.
14. Find all local extrema and saddle points of f (x, y) = 2x4 + y 2 − 12xy.
We first locate the critical points by setting the partial derivatives equal to zero:
fx = 8x3 − 12y = 0
fy = 2y − 12x = 0.
and
It follows from the second equation that y = 6x. Substituting this into the first
equation, we have
8x3 − 12(6x)
8x3 − 72x
8x(x2 − 9)
x
=
=
=
=
0
0
0
0, ±3.
The three critical points are (0, 0), (3, 18), and (−3, −18).
Next we compute the second partial derivatives
fxx = 24x2
fxy = −12
fyy = 2.
The discriminant is
D(x, y) = 48x2 − 144.
Since D(0, 0) = −144 < 0, (0, 0, 0) is a saddle point. Since D(±3, ±18) = 288 > 0 and
fxx (±3, ±18) = 216 > 0, there are local minima at (3, 18) and (−3, −18). The local
minimum value is f (±3, ±18) = −162. There are no local maxima.
11
15. Find the absolute extrema of f (x, y) = x2 − y 2 + 4x + y on the rectangular region
R = {(x, y)| − 4 ≤ x ≤ 0, 0 ≤ y ≤ 1}.
Since f is a polynomial, it is continuous on the closed, bounded rectangular region R,
there is both an absolute maximum and an absolute minimum.
We first find the critical points by setting the partial derivatives equal to zero:
fx = 2x + 4 = 0
and
fy = −2y + 1 = 0.
The only critical point is −2, 21 and the value of f there is f −2, 12 = − 15
.
4
Next we look at the values of f on the boundary of R which consists of four edges.
On E1 , we have y = 0 and
f (x, 0) = x2 + 4x
− 4 ≤ x ≤ 0.
To find the critical values of f on E1 , set
f 0 (x) = 2x + 4 = 0.
So (−2, 0) is a critical point and f (−2, 0) = −4.
On E2 , we have x = 0 and
f (0, y) = −y 2 + y
0 ≤ y ≤ 1.
To find the critical values of f on E2 , set
f 0 (y) = −2y + 1 = 0.
So 0, 21 is a critical point and f 0, 21 = 14 .
On E3 , we have y = 1 and
f (x, 1) = x2 + 4x
− 4 ≤ x ≤ 0.
To find the critical values of f on E3 , set
f 0 (x) = 2x + 4 = 0.
So (−2, 1) is a critical point and f (−2, 1) = −4.
On E4 , we have x = −4 and
f (−4, y) = −y 2 + y
0 ≤ y ≤ 1.
To find the critical values of f on E4 , set
f 0 (y) = −2y + 1 = 0.
So −4, 21 is a critical point and f −4, 21 = 14 .
Evaluating the function at each of the vertices on the boundary, we have f (0, 0) = 0,
f (−4, 0) = 0, f (0, 1) = 0, and f (−4, 1) = 0. Therefore, the absolute maximum value
is 1/4 and the absolute minimum value is −4.
12
16. Find the absolute extrema of f (x, y) = x2 + y 2 + x − y on the disk
D = {(x, y)|x2 + y 2 ≤ 1}.
Since f is a polynomial, it is continuous on the closed, bounded disk D, there is both
an absolute maximum and an absolute minimum.
We first find the critical points by setting the partial derivatives equal to zero:
fx = 2x + 1 = 0
and
fy = 2y − 1 = 0.
The only critical point is − 21 , 21 and the value of f there is f − 12 , 21 = − 12 .
Next we look at the values of f on the boundary of D which consists is the circle
x2 + y 2 = 1. The circle can be parameterized using polar coordinates
x = cos θ,
y = sin θ,
0 ≤ θ ≤ 2π.
On the boundary of D, we have
f (cos θ, sin θ) = 1 + cos θ − sin θ,
0 ≤ θ ≤ 2π.
To find the critical values of f on the boundary of D, set
f 0 (θ) = − sin θ − cos θ = 0.
It follows that
− sin θ
sin θ
−
cos θ
− tan θ
tan θ
= cos θ
= 1
= 1
= −1.
Therefore, θ = 3π
and θ = 7π
give critical points and the endpoints of the interval
4
4
θ = 0, 2π give a point of interest.
√ √ √
√ 2
2
2
3π
7π
At θ = 4 , (x, y) = − 2 , 2 . At θ = 4 , (x, y) = 2 , − 22 . At θ = 0, 2π,
(x, y) = (1, 0). Evaluating f at each of these points, we have
√ √ !
√
2 2
f −
,
= 1− 2
2 2
√
√ !
√
2
2
f
,−
= 1+ 2
2
2
f (1, 0) = 2.
√
The absolute maximum value is 1 + 2 and the absolute minimum value is − 12 .
13
17. Consider the linear system
x1 (t + 1) = −0.4x1 (t) + 0.2x2 (t)
x2 (t + 1) = −0.3x1 (t) + 0.1x2 (t)
~ = ~0.
Determine the stability of X
~
Let X(t)
= hx1 (t), x2 (t)i. The system can be written in matrix form as
−0.4
0.2
~ + 1) =
~
X(t
X(t).
−0.3 0.1
To determine stability, we find the eigenvalues of the matrix A. That is,
−0.4 − λ
0.2
= 0
−0.3
0.1 − λ (−0.4 − λ)(0.1 − λ) + (0.2)(0.3) = 0
λ2 + 0.3λ + 0.02 = 0
(λ + 0.1)(λ + 0.2) = 0.
The eigenvalues are λ1 = −0.1 and λ2 = −0.2. Since |λ1 | < 1 and |λ2 | < 1, it follows
that ~0 is locally stable.
14
18. Find all nonnegative equilibria of
x1 (t + 1) = 2x1 (t)[1 − x1 (t)]
x2 (t + 1) = x1 (t)[1 − x2 (t)]
and determine their stability.
To find the equilibria we need to solve
x1 = 2x1 (1 − x1 )
x2 = x1 (1 − x2 ).
Simplifying the first equation, we have
2x21 − x1 = 0
x1 (2x2 − 1) = 0.
Thus, x1 = 0 or x1 = 12 . If x1 = 0, then the second equation gives x2 = 0. If x1 = 12 ,
then the second equation gives x2 = 13 . Therefore, the equilibria are
1 1
(0, 0)
and
,
.
2 3
The Jacobian matrix is
2 − 4x1 0
J(x1 , x2 ) =
.
1 − x2 −x1
At (0, 0),
2 0
J(0, 0) =
.
1 0
The eigenvalues are λ1 = 2 and λ2 = 0. Since |λ1 | > 1, it follows that (0, 0) is unstable.
At 21 , 31 ,
1 1
0 0
,
= 2
J
.
− 12
2 3
3
The eigenvalues are λ1 = 0 and λ2 = − 12 . Since |λ1 | < 1 and |λ2 | < 1, it follows that
1 1
,
is locally stable.
2 3
15