Calculus 3
Quiz # 4
September 27, 2012
Solutions
Instructions.
• SHOW WORK! and write clearly.
• Try to write in the form of full sentences. Eventually I may penalize you for the failure to use full sentences.
A sentence must contain a verb. For example “5” or “x2 + y 2 are NOT sentences. “x2 + y 2 = 5” is a sentence;
the symbol “=” would be read “is equal,” and “is” is a verb.
• Do not use an equal symbol to connect quantities that are not equal, except when setting up equations.
• Use extra paper as necessary.
The Test.
1. Let f (x, y, z) =
√
x+
√
y+
√
z + ln(4 − x2 − y 2 − z 2 )
(a) (5 points) Evaluate f (1, 1, 1).
√
√
√
√
Solution. f (1, 1, 1) = 1 + 1 + 1 + ln(4 − 1 − 1 − 1) = 3 1 − ln 1 = 3.
(b) (15 points) Find and describe the domain of f . Your description should be precise; anybody hearing
it should be able to know exactly what is in the domain, and what isn’t.
Solution. Preferred description: The domain is the intersection of the interior of the ball of radius
2 centered at the origin with the first octant.
Acceptable description: The set {x2 + y 2 + z 2 < 4, x ≥ 0, y ≥ 0, z ≥ 0}.
(c) (5 bonus points) Describe the range of f .
Solution. This was really tough and we have not developed yet the means needed to solve it. One
can show that the maximum value of the function happens at the point x = y = z = µ where µ is the
first (and only) positive root of the equation 3µ4 + 4µ3 − 4 = 0. The range works out then to (−∞, b],
where b = f (µ, µ, µ).
2. The following table gives values of the height h of waves (in feet) in the open see as functions of the wind
speed v and the length of time t that the wind has been blowing at that speed. For example, if the wind
speed is 15 km/h and it has been blowing at that speed for 20 hours, then the waves are 5 ft. high.
(a) (5 points) What are the meanings of the partial derivatives
∂h
∂v
and
∂h
∂t ?
∂h
Solution.
∂v is the rate of change in h with respect to v; roughly how much h changes when v changes
by a unit, while t remains fixed.
∂h
∂t is the same, with the roles of v and t reversed.
2
(b) (10 points) Estimate the values of fv (40, 15) and ft (40, 15).
Solution.
fv (40, 15). Coming from the left,
f (30, 15) − f (40, 15)
16 − 25
−9
=
=
= 0.9;
30 − 40
30 − 40
−10
from the right
f (50, 15) − f (40, 15)
36 − 25
11
=
=
= 1.1.
50 − 40
50 − 40
10
The best possible approximation from the information given is
fv (40, 15) ≈
1
(0.9 + 1.1) = 1.0.
2
ft (40, 15). Coming from the left,
f (40, 10) − f (40, 15)
21 − 25
−4
=
=
= 0.8;
10 − 15
10 − 15
−5
from the right
f (40, 20) − f (40, 15)
28 − 25
3
=
= = 0.6.
20 − 15
20 − 15
5
The best possible approximation from the information given is
ft (40, 15) ≈
1
(0.8 + 0.6) = 0.7.
2
∂h
? JUSTIFY!
t→∞ ∂t
Solution. As t gets larger, h seems to stabilize. In fact at low values of v it becomes constant quite
soon, but even at the highest value of v appearing in the table, the rate at which h increases steadily
diminishes. In otherwords, the table indicates that for a fixed wind speed, h tends to become constant
as time increases, thus
∂h
lim
(t) = 0.
t→∞ ∂t
(c) (5 points) What appears to be the value of the following limit: lim
3. Find the first derivatives of the following functions
(a) (15 points) f (x, y) = x4 y 3 + 8x2 y.
Solution.
∂f
(x, y) = 4x3 y 3 + 16xy,
∂x
∂f
(x, y) = 3x4 y 2 + 8x2 .
∂y
(b) (10 points) f (x, y, z) = xz − 5x2 y 3 z 4 .
Solution.
∂f
∂f
(x, y, z) = z − 10xy 3 z 4 ,
(x, y, z) = −15x2 y 2 z 4 ,
∂x
∂y
Z x
(c) (10 points) F (x, y) =
cos(et ) dt.
∂f
(x, y, z) = x − 20x2 y 3 z 3 .
∂z
y
Solution.
(d) (10 points) f (x, y) = xy .
Solution.
4. (15 points) If u = erθ sin θ, find
Solution.
∂F
(x, y) = cos(ex ),
∂x
∂f
(x, y) = yxy−1 ,
∂x
∂3u
.
∂r2 ∂θ
∂3u
(r, θ) = (2 + r)θ2 erθ sin θ + θ2 erθ cos θ.
∂r2 ∂θ
∂F
(x, y) = − cos(ey ).
∂y
∂f
(x, y) = (ln x)xy .
∂y