Math 00 Problem Set VI 1 Find all second partial derivatives of f, y = + y. Find all second order derivatives of gs, t = fs + t, s t. The wave equation u 1 c u = 0 arises in many models involving wave-like phenomena. Let u, t and vξ, η be related by the change of variables u, t = v ξ, t = ct η, t = + ct Show that u 1 u c = 0 is equivalent to v = 0. Show that u 1 u c = 0 if and only if u, t = F ct + G + ct for some functions F and G. Interpret F ct + G + ct in terms of travelling waves. 4 If t 0 is a local minimum or maimum of the smooth function ft of one variable t runs over all real numbers then f t 0 = 0. Derive an analogous necessary condition for 0 to be a local minimum or maimium of the smooth function g restricted to points on the line = a + t d. The test should involve the gradient of g. 5 In a certain community, there are two breweries in competition, so that sales of each negatively affect the profits of the other. If brewery A produces litres of beer per month and brewery B produces y litres per month, then the profits of the two breweries are given by P = + y Q = y 4y + 10 6 10 6 respectively. Find the sum of the two profits if each brewery independently sets its own production level to maimize its own profit and assumes that its competitor does likewise. Find the sum of the two profits if the two breweries cooperate so as to maimize that sum. 6 Equal angle bends are made at equal distances from the two ends of a 100 metre long fence so the resulting three segment fence can be placed along an eisting wall to make an enclosure of trapezoidal shape. What is the largest possible area for such an enclosure? 7 Find the most economical shape of a rectangular bo with no top. 8 An eperiment yields data points i, y i, i = 1,,, n. We wish to find the straight line y = m + b which best fits the data. The definition of best is minimizes the root mean square error, i.e. minimizes n m i + b y i. Find m and b.
MATH 00 PROBLEM SET VI SOLUTIONS 1 Find all second partial derivatives of f, y = + y. Solution. Let f, y = + y. Then f = f y = f = 1 +y +y y f yy = 1 1 +y +y 6 +y / f y = 1 yy +y / f y = 1 y +y / y6 +y / Simplifying, y f = f +y / y = f y = f +y / yy = y +y / Find all second order derivatives of gs, t = fs + t, s t. Solution. gs, t = fs + t, s t g s s, t = f 1 s + t, s t + f s + t, s t g t s, t = f 1 s + t, s t f s + t, s t g ss s, t = 4f 11 + 6f 1 + 6f 1 + 9f = 4f 11 + 1f 1 + 9f g st s, t = 6f 11 4f 1 + 9f 1 6f = 6f 11 + 5f 1 6f g tt s, t = 9f 11 6f 1 6f 1 + 4f = 9f 11 1f 1 + 4f Here f 1 denotes the partial derivative of f with respect to its first argument, f 1 is the result of first taking one partial derivative of f with respect to its first argument and then taking a partial derivative with respect to its second argument, and so on. The wave equation u 1 u c = 0 arises in many models involving wave-like phenomena. Let u, t and vξ, η be related by the change of variables u, t = v ξ, t = ct η, t = + ct Show that u 1 u c = 0 is equivalent to v = 0. Show that u 1 u c = 0 if and only if u, t = F ct + G + ct for some functions F and G. Interpret F ct + G + ct in terms of travelling waves. Solution. By the chain rule u u, t =, t = + + = = c + + c 1
Again by the chain rule ] u, t = + and = v + v = v + v u, t = c = c v c v = c v c v + v + v ] + c + c v + c v + v + c v so that Hence u, t 1 u c, t = 4 v u, t 1 c u, t = 0 for all, t 4 v = 0 for all, t Now v ξ, η = ] = 0. Temporarily rename v ξ, η = 0 for all ξ, η w = w. The equation ξ, η = 0 says that, for ξ, η = each fied η, wξ, η is a constant. The value of the constant may depend on η. That is, wξ, η = Hη, for some function H. As a check, observe that Hη = 0. So the derivative of v viewing ξ as a constant is Hη. Let Gη be any function whose derivative is Hη i.e. an indefinite integral of Hη. Then vξ, η Gη] = Hη Hη = 0. This is the case if and only if, for each fied ξ, vξ, η Gξ, η is a constant, independent of η. That is, if and only if vξ, η Gη = F ξ, for some function F. Hence u, t 1 u c, t = 0 v ξ, η = 0 for all ξ, η vξ, η = F ξ + Gη for some functions F and G u, t = v = F ξ, t + G η, t = F ct + G + ct I ll give the interpretation of F ct. The case G + ct is similar. Suppose that u, t = F ct. Think of u, t as the height of water at position and time t. Pick any number z. All points, t in space time for which ct = z have the same value of u, namely F z. So if you move so that your position is = z + ct i.e. move the right with speed c you always see the same wave height. Thus F ct represents a wave moving to the right with speed c. Similarly, G + ct represents a wave moving to the left with speed c. 4 If t 0 is a local minimum or maimum of the smooth function ft of one variable t runs over all real numbers then f t 0 = 0. Derive an analogous necessary condition for 0 to be a local minimum or maimium of the smooth function g restricted to points on the line = a + td. The test should involve the gradient of g. Solution. Define ft = g a + t d and determine t 0 by 0 = a + t 0 d. Then f t = g a + t d d. Then 0 is a local ma or min of the restriction of g to the specified line if and only if t 0 is a local ma or min
of ft. If so, f t 0 necessarily vanishes. So if 0 is a local ma or min of the restriction of g to the specified line, then g 0 d and 0 = a + t 0 d for some t0. The second condition is to ensure that 0 lies on the line. 5 In a certain community, there are two breweries in competition, so that sales of each negatively affect the profits of the other. If brewery A produces litres of beer per month and brewery B produces y litres per month, then the profits of the two breweries are given by P = + y 10 6 Q = y 4y + 10 6 respectively. Find the sum of the two profits if each brewery independently sets its own production level to maimize its own profit and assumes that its competitor does likewise. Find the sum of the two profits if the two breweries cooperate so as to maimize that sum. Solution. If A adjusts to maimize P for y held fied and B adjusts y to maimize Q for held fied then and y are determined by P = 4 10 = 0 = = 1 6 106 Q y = 8y 10 = 0 = y = 1 6 106 = P + Q = + y 1 10 6 5 + y = = 10 6 1 + 1 5 8 4 = 5 8 106 On the other hand if A, B adjust, y to maimize P + Q = + y 1 10 6 5 + y, then and y are determined by P + Q = 5 10 6 = 0 = = 5 106 P + Q y = 6y 10 6 = 0 = y = 1 106 = P + Q = + y 1 5 10 6 + y = = 10 6 4 5 + 5 1 = 11 15 106 6 Equal angle bends are made at equal distances from the two ends of a 100 metre long fence so the resulting three segment fence can be placed along an eisting wall to make an enclosure of trapezoidal shape. What is the largest possible area for such an enclosure? Solution. Here is a figure of the fence. θ 100 θ sin θ It has area A, θ = 100 sin θ + 1 sin θ cos θ = 100 sin θ + 1 sinθ
The maimize the area, we need to solve 0 = A = 100 4 sin θ + sinθ = 100 4 + cos θ = 0 0 = A θ = 100 cos θ + cosθ = 100 cos θ + cosθ = 0 Here we have used that the fence of maimum area cannot have sin θ = 0 or = 0. The first equation forces cos θ = 100 4 and hence cosθ = cos θ 1 = 100 4 1. Subbing these into the second equation gives 100 100 4 + 100 4 1 = 0 = 100 100 4 + 100 4 = 0 = 6 00 = 0 = = 100 cos θ = 100/ 00/ = 1 θ = 60 A = 100 100 100 + 1 100 = 500 7 Find the most economical shape of a rectangular bo with no top. Solution. We wish to minimize A = y + yz + z for any given fied volume V = yz. Subbing in z = V y. A = y + V + V y A = y V z y A y = V y To minimize, we want A = A y = 0, which is the case when y = V, y = V. This forces y = y. Since V = yz is nonzero, neither nor y may be zero. So = y = V 1/, z = / V 1/. 8 An eperiment yields data points i, y i, i = 1,,, n. We wish to find the straight line y = m + b which best fits the data. The definition of best is minimizes the root mean square error, i.e. minimizes n m i + b y i. Find m and b. Solution. We wish to choose m and b so as to minimize the rms error Em, b = n m i + b y i. 0 = E m = n n ] n ] m i + b y i i = m i + b i n i y i ] 0 = E b = n n ] n ] n ] m i + b y i = m i + b y i These are two linear equations on the unknowns m and b. They may be solved in the usual way. Define S = n i, S y = n y i, S = n i and S y = n iy i. The two equations are after dividing by two So which gives n1 S : S 1 + S : S m + S b = S y 1 S m + nb = S y ns S m = ns y S S y ns S b = S S y + S y S m = nsy SSy ns S, b = SSy SyS ns S 4