Solutions to Homework 5
1. Let z = f (x, y) be a twice continously differentiable function of x and
y. Let x = r cos θ and y = r sin θ be the equations which transform
polar coordinates into rectangular coordinates. Show that
∂2z
∂2z
∂2z
1 ∂ 2 z 1 ∂z
+
=
+
+
.
∂x2 ∂y 2
∂r2
r2 ∂θ2
r ∂r
Solution:
fr = fx xr + fy yr = fx cos θ + fy sin θ
frr = (fx cos θ)r + (fy sin θ)r = (fx )r cos θ + (fy )r sin θ
= (fxx xr + fxy yr ) cos θ) + (fyx xr + fyy yr ) sin θ
= (fxx cos θ + fxy sin θ) cos θ) + (fyx cos θ + fyy sin θ) sin θ
= fxx cos2 θ + 2fxy cos θ sin θ + fyy sin2 θ
fθ = fx xθ + fy yθ = fx (−r sin θ) + fy (r cos θ)
fθθ = (−fx r sin θ)θ + (fy r cos θ)θ
= −(fx )θ r sin θ + −fx (r sin θ)θ + (fy )θ r cos θ + fy (r cos θ)θ
= (−fxx xθ − fxy yθ )r sin θ − fx r cos θ + (fyx xθ + fyy yθ )r cos θ − fy r sin θ
= (−fxx (−r sin θ)−fxy r cos θ)r sin θ−fx r cos θ+(fyx (−r sin θ)+fyy r cos θ)r cos θ−fy r sin θ
= fxx r2 sin2 θ − 2fxy r2 cos θ sin θ + fyy r2 cos2 θ − fx r cos θ − fy r sin θ
Therefore, frr + (1/r2 )fθθ upon applying the using the identity
cos2 θ + sin2 θ is equal to −(1/r)(fx cos θ + fy sin θ), which is equal to
−(1/r)fr . The expression for the Laplace equation in polar
coordinates follows immediately.
2. A function f is harmonic in a domain Ω if f is defined on Ω and f is
a solution to the Laplace equation: fxx + fyy = 0. Use the conclusion
of the previous exercise to show that the function below given in
polar coordinates is harmonic on the domain Ω = {(r, θ) : r > 0}:
f (r, θ) = ln r
Solution: fr = 1/r. frr = −1/r2 . fθ = fθθ = 0. Now plug into the
Laplace equation expressed in polar coordinates:
frr + (1/r2 )fθθ + (1/r)fr = −1/r2 + 0 + (1/r)(1/r) = 0.
A nice application of the Laplace equation in polar coordinates is a
complete answer to the following question: if f is harmonic and is
radially symmetric (i.e. constant on circles), what can be said of f ?
Well, the symmetry implies that fθ = 0 and so fθθ = 0. Moreover, f
is essentailly a function only of r. Thus, the Laplace equation
becomes the following ordinary differential equation:
f 00 (r) + (1/r)f 0 (r) = 0.
Let y = f 0 (r), so that the equation becomes y 0 + (1/r)y = 0. This is
a separable ODE: write y 0 = dy/dr and separate the variables as
shown:
dy
1
= − dr.
y
r
Integration yields ln y = − ln r + C. From this it follows that
y = A/r for some positive constant A. Since y = f 0 (r), integration
yields f (r) = A ln r + B. Thus, we have completely characterized all
radially symmetric solutions to the Laplace equation.
3. Suppose that one side of a triangle is increasing at a rate of 3 cm/s
and that a second side is decreasing at a rate of 2 cm/s. If the area of
triangle remains constant, at what rate does the angle between these
two sides change when the first side is 20 cm long, the second side is
30 cm long, and the angle between the two sides is π/6? Is it possible
for this to be a right triangle? (Hint: the area can be computed
using the formula Area = (1/2)ab sin C, where a and b are the
lengths of the side incident at the angle C.)
Solution: Let the first side be a, the second b, and the angle
between them be C. We are given that (1/2)ab sin C = A, where A is
a constant. Let f (a, b, C) = (1/2)ab sin C. The goal is to determine
the value of dC/dt when a = 20, b = 30, da/dt = 3, db/dt = −2, and
C = π/6. So, we compute d/dt of the equation f = A:
fa
da
db
dC
+ fb + fC
= 0 =⇒
dt
dt
dt
1
da 1
db 1
dC
b sin C
+ a sin C + ab cos C
= 0.
2
dt
2
dt 2
dt
Plugging
in the given values and solving for dC/dt yields a value of
√
− 3/36 (radians per second).
Note that these are not the sides of a right triangle: a and b are not
both legs because C is not a right angle and, so, b, being greater
than a must be the hypotenuse. But then in a 30-60-90 triangle,
b = 2a, which is not true.
4. Suppose that the temperature at a point (x, y, z) is given by the
function
T (x, y, z) = 200 exp (−x2 − 3y 2 − 9z 2 ),
where exp (x) is just a way or writing ex without using a superscript.
(This is handy when formulas get to be quite complicated!) Here T is
measured in ◦ C and x, y, and z in meters.
(a) Determine the rate of change of the temperature at the point
P (2, −1, 2) in the direction towards the point Q(3, −3, 3).
Solution: The goal is to compute the directional derivative in
−−→
the direction of P Q– this needs to be scaled so as to become a
unit vector:
1 −−→
1
−−→ P Q = √ h1, −2, 1i.
6
kP Qk
Let u equal this unit vector. The directional derivative
Du T (2, −1, 2) is computed by first computing the gradient:
∇T = 200 exp(−x2 − 3y 2 − 9z 2 )h−2x, −6y, −18zi.
∇T (2, −1, 2) = 200 exp(−43)h−4, 6, −36i.
Then dot with u:
1
Du T (2, −1, 2) = 200 exp(−43)h−4, 6, −36i · √ h1, −2, 1i.
6
=−
52(200) −43
√
e
≈ −8.98 × 10−16 .
6
(b) In which direction does the temperature increase at the fastest
rate at P ? What is magnitude of this maximal rate of increase?
Solution: The gradient direction is the direction of fastest
increase. Take
√ the gradient vector and divide by its length.
Answer: (1/ 337)h−2, 3, −18i.
The magnitude of the rate of √
increase is the length of the
gradient vector. Answer: 200 1348e−43 .
(c) What are the units of the directional derivative of T ? Explain
or give an example which illustrates your answer.
Solution: The units are degrees Celsius per unit length (or
whatever units represent each of x, y, and z individually): the
directional derivative is computed by taking a limit of the ratio
of a difference of temperatures (so the numerator is in degrees
Celsius) divided by a 1-dimensional variation of the input
variables.
Perhaps an easier way to see this is to appeal to the observation
that the directional derivative in the i direction is the partial
derivative with respect to x. This partial derivative has units of
change of T per change in x, or degrees Celsius per units of x
(presumably length since the phrasing of the problem is such
that x represents the position of a point).
5. In this problem, I would like you to use the fact that the gradient
vector ∇f (a, b, c) is normal to the the level surface f (x, y, z) = 0
provided f (a, b, c) = 0 (i.e provided (a, b, c) lies on this level surface).
Determine an equation for the tangent plane to the surface
x4 + y 4 + z 4 = 3x2 y 2 z 2 at the point (1, , 1, 1).
Solution: Let g(x, y, z) = x4 + y 4 + z 4 − 3x2 y 2 z 2 . Thus, the surface
is the level surface g = 0. As such, the gradient vector ∇g(1, 1, 1) is
normal to the surface. So, we compute this vector:
∇g = h4x3 − 6xy 2 z 2 , 4y 3 − 6x2 yz 2 , 4z 3 − 6x2 y 2 zh
∇g(1, 1, 1) = h−2, −2, −2i.
So, an equation of the tangent plane is
−2(x − 1) − 2(y − 1) − 2(z − 1) = 0.
6. Suppose you are climbing a hill whose shape is given by the equation
z = 1000 − 0.005x2 − 0.01y 2 ,
where x, y, and z are measured in meters. Currently you are
standing at the point P (60, 40, 966). The positive x-axis represents
east and the positive y-axis represents north.
(a) If you walk due south, will you start to ascend or descend? At
what rate?
Solution: The question is asking for the value of the directional
derivative in the direction of −j. This is the negative of partial
y. Since zy = −0.02y, the value is equal to 0.02(40) = 0.8. Being
positive, this means that the hill slopes upward in this direction.
(b) If you walk northwest, will you start to ascend or descent? At
what rate?
Solution: As above, except u = √12 h−1, 1i is the direction:
√
√
Du z(60, 40) = h−0.01(60), −0.02(40)i · u = −1/5 2 = − 2/10.
Since this is negative, again the hill slopes downward in this
direction.
(c) In which direction is the slope largest? What is the rate of
ascent in that direction? At what angle above the horizontal
does the path in that direction begin?
Solution: The gradient direction is the steepest:
∇z(60, 40) = h−0.6, −0.8i
Fortunately, this is already a unit vector! However, a more
useful answer would be a compass direction: S37◦ W, i.e. walk
37 degrees to the West of due South.
The slope in this direction is the magnitude of the gradient
vector, which happens to be 1. Therefore, the angle of ascent is
45 degrees.