Calculus IV
Math 241 Winter 2006
Professor Ben Richert
Exam 1
Solutions
Problem 1. (10 pts) Where is the function
6x3
f (x, y) = 2x3 + y 3

1


continuous?
for (x, y) 6= (0, 0)
for (x, y) = (0, 0)
Solution. The function f (x, y) is continuous at (a, b) if
6a3
6x3
=
2a3 + b3
(x,y)→(a,b) 2x3 + y 3
lim
for (a, b) 6= (0, 0), and if
6x3
= f (0, 0) = 1
(x,y)→(0,0) 2x3 + y 3
lim
for (a, b) = (0, 0).
6x3
Now since
is a rational function, we know that
2x3 + y 3
6a3
6x3
=
2a3 + b3
(x,y)→(a,b) 2x3 + y 3
lim
whenever 2a3 + b3 6= 0. In the case that (a, b) = (0, 0), we can easily see that
6x3
6= 1.
+ y3
lim
(x,y)→(0,0) 2x3
Suppose that we take, for instance, the path to the origin along the x-axis. Then y = 0, and our limit becomes
6x3
= lim 3 = 3 6= 1.
x→0 2x3
x→0
Thus if the limit exists, it equals 3, not 1 as we require.
We conclude that f (x, y) is continuous everywhere except (x, y) such that 2x3 + y 3 = 0. (We don’t have the mention that
the function is not continuous at (0, 0), because this point lies on the curve given).
lim
x
Problem 2. (20 pts) Suppose that the following table gives values for the function f (x, y):
2
3
4
5
y
2
1
1
4
2
3
2
2
2
3
4
3
3
1
1
5
2
6
3
2
(a – 5pts) Give an estimate of fx (3, 4).
Solution. We can estimate the instantaneous change of f (x, y) in the x-direction at (3, 4) by computing the average
rate of change of f (x, y) over a small interval containing x = 3. We take the interval from x = 2 to x = 4, holding
∆z
f (4, 4) − f (2, 4)
1−3
y constant at 4, and obtain that
=
=
= −1. So we say fx (3, 4) ≈ −1.
∆x
4−2
2
(b – 5pts) Give an estimate of fy (3, 4).
Solution. The same procedure works for estimating fy (3, 4). We compute the average rate of change of f (x, y)
f (3, 5) − f (3, 3)
6−2
∆z
=
=
= 2. We
over the interval from y = 3 to y = 5 holding x constant at 3 and obtain
∆y
5−3
2
say that fy (3, 4) ≈ 2.
1
(c – 5pts) Give a linear approximation to f (x, y) at the point (3, 4).
Solution. The equation for the linearization of f (x, y) at the point (a, b) is
L(a,b) (x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b).
We read from the table that f (3, 4) = 3, and then using our estimates from above, get that L (3,4) (x, y) =
3 − 1(x − 3) + 2(y − 4).
(d – 5pts) Use part (c) to give an estimate of f (3.2, 3.9).
Solution. We know that for (a, b) near (3, 4), L(3,4) (a, b) ≈ f (a, b), that is, that the linearization provides a good
estimate for the function. So in this instance, we say that f (3.2, 3.9) ≈ L(3,4) (3.2, 3.9) = 3−1(3.2−3)+2(3.9−4) =
2
2
26
30
−
−
=
.
3 − (.2) + 2(−.1) =
10 10 10
10
Problem 3. (10 pts) During midterm week you always drink too much caffeine and as a result do not sleep soundly. In
r
fact, the night before the calculus midterm, you dreamed that a perfectly cylindrical Coke can on your desk was inexplicably
growing, and threatening to crowd you out of the room. The radius of the can was growing at a rate of 2 in/sec, while the
height was growing at a rate of 3 in/sec. At what rate was the volume of the can increasing when the radius was 2 in and the
height was 4 in?
Solution. Let r, h, and V = πr 2 h denote the radius, height, and volume of the can respectively. We are asked to compute
dV /dt, the rate at which the volume of the can is increasing, when the radius is 2 in, the height is 4 in, the radius is increasing
at a rate of 2 in/sec (so that dr/dt = 2), and the height is increasing at a rate of 3 in/sec (so that dh/dt = 3). The chain
rule tells us that
dV
∂V dr ∂V dh
=
+
,
dt
∂r dt
∂h dt
and we can compute that ∂V /∂r = 2πrh and ∂V /∂h = πr 2 , so that when r = 2 and h = 4, we have ∂V /∂r = 16π and
∂V /∂h = 4π. Thus when r = 2, h = 4, dr/dt = 2, and dh/dt = 3 is dV /dt = 16π(2) + 4π(3) = 44π in/sec.
Problem 4. (20 pts) Consider the function f (x, y) = x3 + 8y 3 − 24xy. Classify each critical point as either a maximum, a
minimum, or a saddle. (You may be interested to know that 6 · 48 · 8 − 24 2 = 1728).
Solution. First we compute several partial derivatives:
fx (x, y)
fy (x, y)
fxx (x, y)
fyy (x, y)
fxy (x, y)
= 3x2 − 24y
= 24y 2 − 24x
= 6x
= 48y
= −24
Now the critical points occur where fx (x, y) = 0 = fy (x, y), so the next step is to solve the system:
3x2 − 24y
= 0
2
24y − 24x = 0
for x and y. The first equation tells us that 3x2 = 24y, or that x2 = 8y. The second equation tells us that y 2 = x. Putting
these two together implies that 8y = x2 = (y 2 )2 = y 4 , that is, that y 4 − 8y = 0, or y(y 3 − 8) = 0. We conclude that y = 0 or
y = 2. Now if y = 0, then x = y 2 = (0)2 so that x = 0 and (0, 0) is a critical point. If y = 2, then x = y 2 implies that x = 4,
and thus (4, 2) is a critical point.
Recall that D(x, y) = fxx (x, y)fyy (x, y) − [fxy (x, y)]2 and that if D(a, b) > 0 and fxx (a, b) > 0, then (a, b) is a min, if
D(a, b) > 0 and fxx (a, b) < 0, then (a, b) is a max, and if D(a, b) < 0, then (a, b) is a saddle point.
We see that D(x, y) = 6x48y −(−24)2, so that D(0, 0) < 0 and (0, 0) is a saddle point. Furthermore, D(4, 2) = (6)(4)(48)(2)−
(−24)2 = 1728 > 0 (using the hint above), and clearly fxx (4, 2) = 6(4) = 24 > 0, so that (4, 2) is a minimum.
√
2
Problem 5. (20 pts) Suppose that f (x, y) = 10 x − y measures the height (in feet above sea level) x feet east and y feet
north of the center of the crops unit.
(a – 10pts) In which direction is the ground rising the fastest if you are standing at the point (4, 2) (4 feet east and 2 feet
north of the center of the crops unit)?
Solution. We know that ∇f (a, b) gives the direction of maximum increase at (a, b). So we compute that
1
5
5
5
∇f (x, y) = h10 x−1/2 , −2yi = h √ , −2yi, and thus ∇f (4, 2) = h √ , −2(2)i = h , −4i is the direction of
2
x
2
4
maximal increase.
(b – 10pts) How fast is the ground rising towards the southwest (from your vantage point at (4, 2))?
Solution. The rate of change towards the southwest from (4, 2) is given by the directional derivative D →
f (4, 2) =
u
√
→
→
∇f (4, 2) · u where u is a unit vector pointing southwest. Now h−1, −1i points southwest, and it has length 2,
−1 −1
→
and thus u = h √ , √ i works for our calculation. We computed ∇f (4, 2) above, so we get that D →
(f (4, 2)) =
u
2
2
−5
4
−5
8
3
5
−1 −1
→
∇f (4, 2) · u = h , −4i · h √ , √ i = √ + √ = √ + √ = √ feet per foot southwest.
2
2
2
2 2
2
2 2 2 2
2 2
Problem 6. (10 pts) Match the following functions and graphs by filling up the table below (this is the only problem for
which you need show no work):
Function
Graph
(a) f (x, y) = (x2 − y 2 )2
II
(b) f (x, y) = x2 − y 2
I
(c) f (x, y) = (x2 + y 2 )2
III
600
20
z
z400
200
0
4
0
2
-20
0
-4
-2
2
-2
(II)
2000
z
1000
y
-2
0
x
-4
4
(I)
0
-4
y
-2
0
x
4
2
2
-4
4
4
2
0
0 y
-4
-2
-2
0
x
(III)
2
-4
4
Solution. An easy way to see this is that x2 − y 2 has upward parabolas if y = 0 and downward parabolas if x = 0, so that it
must be I. The (x2 − y 2 )2 is zero if x = y, so must be II (because III has upward parabola shapes if one looks at the plane
x = y, while I is zero on that plane).