Students' answers to review problems for the second exam in Math 291, fall 2002

Students have so far (11/14) requested no formulas for the second exam. O.k.: I got (11/23) one request for one formula: less work for "the management"! Now (11/25) a draft formula sheet exists. And today (11/26) an updated version has been posted.

 1. Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+y2 subject to the constraint x2+2y2=4. ```f(x,y)=x+y^2 g(x,y)= x^2+2y^2=4 df/dx= 1 df/dy= 2y dg/dx= 2x dg/dy= 4y``` Using Lagrange Multipliers: (Note L is used to represent lambda) L= 2x 2yL = 4y x^2+2y^2= 4 From equation 2, we can solve and find that L = 2 or y=0. If L= 2 then by plugging this value into equation 1, you get x=1. By putting x=1 into the last equation, y= (3/2)^(1/2) or -(3/2)^(1/2). If y=0 then x= +2 or -2. You must then plug these values into the function, x+y^2. The minimum is therefore at (-2,0) and equals -2. The maximum is therefore at both (0,sqrt(3/2)) and (0,-sqrt(3/2)) and equals 5/2. Sent by Jordana Ellway on Thu, 14 Nov 2002 02:09:15

 12. The constraint x4+x2y2+2y4+z4=1 defines a closed and bounded set in R3, and thus the continuous function f(x,y,z)= xyz attains its maximum value on that set. What is the maximum value of xyz subject to this constraint? Be sure to analyze carefully and completely any system of equations you solve. First, we use Lagrange Multipliers, by taking the gradient of the constraint, multiplying it by L (which I will use for lambda here), and setting it equal to the gradient of the function we want to maximize. This yields 3 equations, the constraint equation, and four unknowns. I will label each equation as: (4x^3 + 2xy^2)L = yz (2yx^2 + 8y^3)L = xz (4z^3)L = xy x^4 + (x^2)(y^2) + 2y^4 + z^4 = 1 I will multiply both sides of equation (1) by x, both sides of equation (2) by y, and both sides of equation (3) by z. If L=0, then at least one of the variables is 0 and then the function xyz is zero, but it has positive values, so such a point can't be a max. Then, we divide both sides by L. This will give me: 4x^4 + 2(x^2)(y^2) = xyz/L 2(y^2)(x^2) + 8y^4 = xyz/L (4z^4) = xyz/L We can set equation (5) = equation (6), and we get either both x and y = 0, or x = +/- Wy where W is 2^(1/4) (approximately 1.18921). If the first case is true, we have all three variables equal to zero. This can't happen because of the constraint equation (4). which is the first candidate for a min or a max (but we know its neither, because we know we can have both positive and negative values of the function). If the second case is true, we move on. We then set (5) = (7), and plug in Wy for x. From this, we get z= +/- Vy where V is (8+2sqrt(2))^(1/4) (approximately 1.28270). Now, since we have all the variables in terms of y, we can plug in for x and z into the constraint equation (4). When we plug in, we get 8.12132y^4 = 1, and y= +/- .59237. We plug this back into (eq. 8) to get x = +/- .704451. Then we plug in y into (9) and get z = +/- .759836. We have eight candidates for max and min, but 4 of them will give the same value, and the other four will give the negative of that value. We plug all these candidates into the original f(x,y,z)=xyz equation. Since these 8, and (0,0,0) are the only candidates for min and max, we know that the min is -.317076 where we have (+,+,-), (+,-,+), (-,+,+), or (-,-,-). And the max is .317076 where we have (+,+,+), (+,-,-), (-,+,+), or (-,-,+). Sent by Alexander Pergament on Sun, 17 Nov 2002 21:30:22

 14. Find the line integral intC F·dr where F=2xyi+x2jand C is an arc of a circle centered at (0,1), of radius 1 joining the points P1(0,0) and P2(0,2). Here F= 2xyi+x^2j. First we notice that the field is conservative. Since P = 2xy and Q = x^2, dP/dy = 2x and dQ/dx = 2x. Since dP/dy and dQ/dx are both 2x, we suspect that the field is conservative. Therefore, we now must get an f so that F = grad(f). Since f_x = 2xy, if we get the antiderivative with respect to x, f(x,y) = x^2  y + C_1(y). Since f_y = x^2, if we get the antiderivative with respect to y, f(x,y) = x^2 y+ C_2(x). Logically, we can now deduce that C_1(y) and C_2(x) are not necessary (since we are computing a line integral and the difference of two values of f will just cause the resulting constants to cancel) and therefore equal 0 and f(x,y) = (x^2)y. Now, because the function is conservative and F = grad(f), we can say that the integral(over the path C) of F dot dr equals the integral(over the path C) of grad(f) dot dr. Then we can apply the fundamental theorem for line integrals and say that the above integral is equal to f(r(b)) - f(r(a)) where r is the position vector and a<=t<=b. But, since the beginning and final points are provided in the problem, we can simply rewrite this to be f(0,2) - f(0,0). Therefore the answer is (0^2)(2) - (0^2)(0) which equals 0. Note from the management: you can also "do" this problem by a direct parameterization using x=cos(t) and y=1+sin(t), where t goes from -Pi/2 to Pi/2. The parameterized integral can be done, and is really not too bad. The answer will again be 0. The computation above is much shorter, though! Sent by Greg Ryslik on Mon, 18 Nov 2002 21:45:04

 15. Suppose R is the trapezoid with vertices (1,0), (2,0), (0,-2), and (0,-1). Use the change of variables u=x+y and v=x-y to compute the double integral over R of e^(x+y)/(x-y) dA. First, solve for x and y in terms of u and v. You get: x=(u+v)/2 y=(u-v)/2. By sketching R, you'll notice that the trapezoid is made up of four lines (x=0, y=0, y=x-1, and y=x-2), which can be written in terms of u and v: x=0 => (u+v)/2=0 => u=-v y=0 => (u-v)/2=0 => u=v y=x-1 => (u-v)/2=(u+v)/2+1 => u-v=u+v-2 => v=1 y=x-2 => (u-v)/2=(u+v)/2+2 => u-v=u+v-4 => v=2 The Jacobian is |x_u*y_v-x_v*y_u|. So the Jacobian in this case is 1/2. Since it is constant, we can just pull it out of the integrand, and multiply through by 1/2 at the end. So, the double integral becomes this iterated integral: (1/2) int(1 to 2) [int(-v to v) e^(u/v) du] dv. First, we can solve the inner integral by using the substitution w=u/v, therefore making dw=1/v du, and the limits of integration change from u=-v to w=-1 and u=v to w=1. So the inner integral is: int(-1 to 1) e^w v dw=v int(-1 to 1) e^w dw = v(e-(1/e)). Now use this to compute the outer integral. We can factor out the constant e-(1/e) and multiply through at the end to make it easier: int(1 to 2) v dv = 4/2-1/2 = 3/2. So, double integral over R of e^(u+v)/(u-v) dA, (by multiplying by the Jacobian and (e-1/e), which were factored out in the above steps) turns out to be: (1/2)(e-(1/e))(3/2) = 3e/4-3/(4e) Sent by Jason Sullivan on Mon, 18 Nov 2002 19:08:02

 18. Use Green's Theorem to evaluate intT(x2y) dx - (xy2 dy, where T is the circle x2+y2=4. Green's Theorem states that intCP dx + Q dy] = int intRdQ/dx - dP/dy) dA where C is the boundary of R. For this problem P=x^2*y and Q=x*y^2. So dQ/dx=-y^2 and dP/dx=x^2. By Green's Theorem the integral can be written as int[intR -1(x^2+y^2) dA where R is the inside of the circle of radius 2 centered at (0,0). We use polar coordinates for this double integral. If we sub in the values x=r cos(w) and y=r sin(w), the integrand becomes -r^2. Now if we plug in the limits, we get inttheta=0theta=2*piintr=02-r^2 r dr d(theta). This is then evaluated as follows: inttheta=0theta=2*pi(-1/4)*r^4 |r=0r=2 d(theta) = inttheta=0theta=2*pi(-4) d(theta)=-4*theta |theta=0theta=2*pi=-8*pi. Sent by Nicholas Vander Valk on Mon, 18 Nov 2002 22:25:37