Solutions to Homework 5

1. Let z = f(x, y) be a twice continously differentiable function of x andy. Let x = r cos θ and y = r sin θ be the equations which transformpolar coordinates into rectangular coordinates. Show that

∂2z

∂x2+∂2z

∂y2=∂2z

∂r2+

1

r2∂2z

∂θ2+

1

r

∂z

∂r.

Solution:fr = fxxr + fyyr = fx cos θ + fy sin θ

frr = (fx cos θ)r + (fy sin θ)r = (fx)r cos θ + (fy)r sin θ

= (fxxxr + fxyyr) cos θ) + (fyxxr + fyyyr) 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 = fxxθ + fyyθ = fx(−r sin θ) + fy(r cos θ)

fθθ = (−fxr sin θ)θ + (fyr cos θ)θ

= −(fx)θr sin θ +−fx(r sin θ)θ + (fy)θr cos θ + fy(r cos θ)θ

= (−fxxxθ−fxyyθ)r sin θ−fxr cos θ+ (fyxxθ +fyyyθ)r cos θ−fyr sin θ

= (−fxx(−r sin θ)−fxyr cos θ)r sin θ−fxr cos θ+(fyx(−r sin θ)+fyyr cos θ)r cos θ−fyr sin θ

= fxxr2 sin2 θ − 2fxyr

2 cos θ sin θ + fyyr2 cos2 θ − fxr cos θ − fyr sin θ

Therefore, frr + (1/r2)fθθ upon applying the using the identitycos2 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 polarcoordinates follows immediately.

2. A function f is harmonic in a domain Ω if f is defined on Ω and f isa solution to the Laplace equation: fxx + fyy = 0. Use the conclusionof the previous exercise to show that the function below given inpolar 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 theLaplace 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 acomplete answer to the following question: if f is harmonic and isradially symmetric (i.e. constant on circles), what can be said of f?Well, the symmetry implies that fθ = 0 and so fθθ = 0. Moreover, fis essentailly a function only of r. Thus, the Laplace equationbecomes the following ordinary differential equation:

f ′′(r) + (1/r)f ′(r) = 0.

Let y = f ′(r), so that the equation becomes y′ + (1/r)y = 0. This isa separable ODE: write y′ = dy/dr and separate the variables asshown:

dy

y= −1

rdr.

Integration yields ln y = − ln r + C. From this it follows thaty = A/r for some positive constant A. Since y = f ′(r), integrationyields f(r) = A ln r +B. Thus, we have completely characterized allradially symmetric solutions to the Laplace equation.

3. Suppose that one side of a triangle is increasing at a rate of 3 cm/sand that a second side is decreasing at a rate of 2 cm/s. If the area oftriangle remains constant, at what rate does the angle between thesetwo sides change when the first side is 20 cm long, the second side is30 cm long, and the angle between the two sides is π/6? Is it possiblefor this to be a right triangle? (Hint: the area can be computedusing the formula Area = (1/2)ab sinC, where a and b are thelengths of the side incident at the angle C.)

Solution: Let the first side be a, the second b, and the anglebetween them be C. We are given that (1/2)ab sinC = A, where A isa constant. Let f(a, b, C) = (1/2)ab sinC. The goal is to determinethe value of dC/dt when a = 20, b = 30, da/dt = 3, db/dt = −2, andC = π/6. So, we compute d/dt of the equation f = A:

fada

dt+ fb

db

dt+ fC

dC

dt= 0 =⇒

1

2b sinC

da

dt+

1

2a sinC

db

dt+

1

2ab cosC

dC

dt= 0.

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 notboth legs because C is not a right angle and, so, b, being greaterthan 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 thefunction

T (x, y, z) = 200 exp (−x2 − 3y2 − 9z2),

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 ismeasured in C and x, y, and z in meters.

(a) Determine the rate of change of the temperature at the pointP (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−−→PQ– this needs to be scaled so as to become a

unit vector:1

‖−−→PQ‖

−−→PQ =

1√6〈1,−2, 1〉.

Let u equal this unit vector. The directional derivativeDuT (2,−1, 2) is computed by first computing the gradient:

∇T = 200 exp(−x2 − 3y2 − 9z2)〈−2x,−6y,−18z〉.

∇T (2,−1, 2) = 200 exp(−43)〈−4, 6,−36〉.

Then dot with u:

DuT (2,−1, 2) = 200 exp(−43)〈−4, 6,−36〉 · 1√6〈1,−2, 1〉.

= −52(200)√6

e−43 ≈ −8.98× 10−16.

(b) In which direction does the temperature increase at the fastestrate at P? What is magnitude of this maximal rate of increase?

Solution: The gradient direction is the direction of fastestincrease. Take the gradient vector and divide by its length.Answer: (1/

√337)〈−2, 3,−18〉.

The magnitude of the rate of increase is the length of thegradient vector. Answer: 200

√1348e−43.

(c) What are the units of the directional derivative of T? Explainor give an example which illustrates your answer.

Solution: The units are degrees Celsius per unit length (orwhatever units represent each of x, y, and z individually): thedirectional derivative is computed by taking a limit of the ratioof a difference of temperatures (so the numerator is in degreesCelsius) divided by a 1-dimensional variation of the inputvariables.

Perhaps an easier way to see this is to appeal to the observationthat the directional derivative in the i direction is the partialderivative with respect to x. This partial derivative has units ofchange of T per change in x, or degrees Celsius per units of x(presumably length since the phrasing of the problem is suchthat x represents the position of a point).

5. In this problem, I would like you to use the fact that the gradientvector ∇f(a, b, c) is normal to the the level surface f(x, y, z) = 0provided 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 surfacex4 + y4 + z4 = 3x2y2z2 at the point (1, , 1, 1).

Solution: Let g(x, y, z) = x4 + y4 + z4 − 3x2y2z2. Thus, the surfaceis the level surface g = 0. As such, the gradient vector ∇g(1, 1, 1) isnormal to the surface. So, we compute this vector:

∇g = 〈4x3 − 6xy2z2, 4y3 − 6x2yz2, 4z3 − 6x2y2z〈

∇g(1, 1, 1) = 〈−2,−2,−2〉.

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 − 0.01y2,

where x, y, and z are measured in meters. Currently you arestanding at the point P (60, 40, 966). The positive x-axis representseast and the positive y-axis represents north.

(a) If you walk due south, will you start to ascend or descend? Atwhat rate?

Solution: The question is asking for the value of the directionalderivative in the direction of −j. This is the negative of partialy. Since zy = −0.02y, the value is equal to 0.02(40) = 0.8. Beingpositive, this means that the hill slopes upward in this direction.

(b) If you walk northwest, will you start to ascend or descent? Atwhat rate?

Solution: As above, except u = 1√2〈−1, 1〉 is the direction:

Duz(60, 40) = 〈−0.01(60),−0.02(40)〉 · u = −1/5√

2 = −√

2/10.

Since this is negative, again the hill slopes downward in thisdirection.

(c) In which direction is the slope largest? What is the rate ofascent in that direction? At what angle above the horizontaldoes the path in that direction begin?

Solution: The gradient direction is the steepest:

∇z(60, 40) = 〈−0.6,−0.8〉

Fortunately, this is already a unit vector! However, a moreuseful answer would be a compass direction: S37W, i.e. walk37 degrees to the West of due South.

The slope in this direction is the magnitude of the gradientvector, which happens to be 1. Therefore, the angle of ascent is45 degrees.