Solutions to Homework 5

Transcription

1 Solutions to Homework 5 1. Let z = f(x, y) be a twice continuously 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 2 z x z y 2 = 2 z r z r 2 θ z r r. Solution: f r = f x x r + f y y r = f x cos θ + f y sin θ f rr = (f x cos θ) r + (f y sin θ) r = (f x ) r cos θ + (f y ) r sin θ = (f xx x r + f xy y r ) cos θ) + (f yx x r + f yy y r ) sin θ = (f xx cos θ + f xy sin θ) cos θ) + (f yx cos θ + f yy sin θ) sin θ = f xx cos 2 θ + 2f xy cos θ sin θ + f yy sin 2 θ f θ = f x x θ + f y y θ = f x ( r sin θ) + f y (r cos θ) f θθ = ( f x r sin θ) θ + (f y r cos θ) θ = (f x ) θ r sin θ + f x (r sin θ) θ + (f y ) θ r cos θ + f y (r cos θ) θ = ( f xx x θ f xy y θ )r sin θ f x r cos θ + (f yx x θ + f yy y θ )r cos θ f y r sin θ = ( f xx ( r sin θ) f xy r cos θ)r sin θ f x r cos θ+(f yx ( r sin θ)+f yy r cos θ)r cos θ f y r sin θ = f xx r 2 sin 2 θ 2f xy r 2 cos θ sin θ + f yy r 2 cos 2 θ f x r cos θ f y r sin θ Therefore, f rr + (1/r 2 )f θθ upon applying the using the identity cos 2 θ + sin 2 θ is equal to (1/r)(f x cos θ + f y sin θ), which is equal to (1/r)f r. The expression for the Laplace equation in polar coordinates follows immediately.

2 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: f xx + f yy = 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: f r = 1/r. f rr = 1/r 2. f θ = f θθ = 0. Now plug into the Laplace equation expressed in polar coordinates: f rr + (1/r 2 )f θθ + (1/r)f r = 1/r (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 f θ = 0 and so f θθ = 0. Moreover, f is essentially a function only of r. Thus, the Laplace equation becomes 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 is a separable ODE: write y = dy/dr and separate the variables as shown: dy y = 1 r dr. Integration yields ln y = ln r + C. From this it follows that y = A/r for some positive constant A. Since y = f (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.)

3 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: da f a dt + f db b dt + f C dc dt = 0 = 1 da b sin C 2 dt + 1 db a sin C 2 dt + 1 dc ab cos C 2 dt = 0. Plugging in the given values and solving for dc/dt yields a value of 3/36 (radians per second). This triangle could not be a right-triangle for all time t because C is not a right angle and, so, b, being greater than a must be the hypotenuse. But then in a triangle, b = 2a, which is not true. The question of whether such a triangle could at some time t be a right triangle is much more delicate. Please refer to the notes at the end of these solutions for an answer to this question. 4. Suppose that the temperature at a point (x, y, z) is given by the function T (x, y, z) = 200 exp ( x 2 3y 2 9z 2 ), where exp (x) is just a way or writing e x 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 = 6 1, 2, 1. P Q Let u equal this unit vector. The directional derivative D u T (2, 1, 2) is computed by first computing the gradient: T = 200 exp( x 2 3y 2 9z 2 ) 2x, 6y, 18z.

4 Then dot with u: T (2, 1, 2) = 200 exp( 43) 4, 6, 36. D u T (2, 1, 2) = 200 exp( 43) 4, 6, 36 = 52(200) 6 e , 2, 1. (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) 2, 3, 18. The magnitude of the rate of increase is the length of the gradient vector. Answer: e 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 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 x 4 + y 4 + z 4 = 3x 2 y 2 z 2 at the point (1,, 1, 1).

5 Solution: Let g(x, y, z) = x 4 + y 4 + z 4 3x 2 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 = 4x 3 6xy 2 z 2, 4y 3 6x 2 yz 2, 4z 3 6x 2 y 2 z g(1, 1, 1) = 2, 2, 2. So, an equation of the tangent plane is 2(x 1) 2(y 1) 2(z 1) = Suppose you are climbing a hill whose shape is given by the equation z = x y 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 z y = 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 = 1 2 1, 1 is the direction: D u z(60, 40) = 0.01(60), 0.02(40) 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) = 0.6, 0.8

6 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. On whether the triangle can be a right triangle at some instance of time. I interpret a (t) = 3 and b (t) = 2 to meant that these hold for all time t. We can then solve the differential equations to obtain a(t) = 3t + a(0) and b(t) = 2t + b(0). We are given that a(0) = 20 and b(0) = 30. However, so that we can study this problem in slightly more generality, I will keep the values of a(0) and b(0) as unspecified constants. By scaling the triangle by 1/a(0), we may assume that a(0) = 1 and b(0) = λ, where λ > 0. In our specific problem, we have a(0) = 20 and b(0) = 30 so that λ = 3/2 and the lengths of the sides of the triangle have been scaled by a factor of 1/20. Rather than call the angle between a(t) and b(t) by the letter C, I will use θ(t). Let A be the area of the scaled triangle. This is a constant. Since A = (1/2)a(t)b(t) sin θ(t) and θ(0) = π/6 (given), we have that A = (1/2)(1)(λ)(1/2) = λ/4. Note that the true area has been scaled by 1/a(0) 2 = 1/400. Since λ = 3/2, we have that the scaled area is A = (3/2)/4 = 3/8. This means that the true area is a(0) 2 (3/8) = 400 (3/8) = 150. Let f(t) = sin θ(t) = 2A/(a(t)b(t)). Thus, f(t) = λ 2(3t + 1)( 2t + λ). We consider the component of the domain of f(t) which contains t = 0. Therefore, we restrict our attention to t ( 1/3, λ/2).

7 Since 1 sin θ(t) 1, there are further restrictions. Since θ(t) represents an angle of a triangle, θ(t) (0, π); therefore, 0 < sin θ(t) 1. The condition f(t) > 0 holds for all t ( 1/3, λ/2). The condition f(t) 1 implies (after solving f(t) 1 above) that 12t 2 + (4 6λ)t + λ 0. The quadratic formula can be used to determine additional restrictions on t in terms of λ. In the specific case where λ = 3/2, this implies that t is approximately between 0.2 and Since 1/3 < 0.2 and 0.62 < λ/2 = 3/4, there will, in fact, be two times when θ(t) = π/2, i.e. the triangle is a right triangle. Solving f(t) = 1 with λ = 3/2, we have that are times when θ(t) = π/2. t = The unscaled version for our specific problem is perhaps a little easier to understand. Let g(t) = sin θ(t) = 300/[(3t + 20)( 2t + 30), where t ( 20/3, 15). A graph of g(t) is shown below. As we can see, there are two values of t for which g(t) = 1, meaning that at these times θ(t) = π/2.

8 We can also consider the problem of whether other angles of the triangle could become right angles. The angle opposite side b(t) can be a right angle. From some simulations that I have done with a computer, this occurs for some value of t (0, 1).