4 due Wednesday, March 8 2017 1 Consider the curve ~r(t) = hcos(4t) cos(t), cos(4t) sin(t)i in the plane, and the surface above 2 y 2 dz it given by the now-familiar equation z = f (x, y) = xx2 +y 2 . We are going to find dt at the point where t = /4 in two different ways. First method: chain rule (a) Find the tangent vector ~r 0 (t) in the plane, and find its value at t = /4. (b) Find the location of ~r(t) when t = /4. (c) Compute the gradient vector f , and evaluate it at the location from (b). (d) Now, use the chain rule to evaluate df dt when t = /4. Second method: brute force (e) Plug the x and y coordinates of ~r into f (x, y), to get f directly as a function of t. (f ) Differentiate f (t) directly, as you would in calc 1, and plug in t = /4. Did you get the same value as part (d)? 2 Another consequence of the chain rule is a way of doing implicit differentiation. In calc 1, we dy ran into implicit differentiation as a method of finding dx when we didn't have an explicit formula for y in terms of x, just an equation relating the two. For example, consider the curve given by x2 y y 2 x + x3 = 3. Method 1: brute force (a) Calculate dy dx by implicit differentiation, the way you learned it in Calc 1. Method 2: using partial derivatives. Let f (x, y) = x2 y y 2 x + x3 , so that our curve is realized as the level curve f (x, y) = 3. Suppose that this curve can also be parametrized as ~r(t) = ht, g(t)i for some function g; note that finding g 0 (t) will be dy . equivalent to finding dx (b) If ~r(t) = ht, g(t)i, find ~r 0 (t). [Just symbolically, you don't have to know what g is.] Since f (x, y) = 3 whenever (x, y) is on the curve parametrized by ~r, we have f (~r(t)) = 3. Differentiating with respect to t and applying the chain rule tells us that f ~r 0 (t) = 0. (c) Show that in this case, we get g 0 (t) = ffxy . (d) Find the partial derivatives of f , and calculate ffxy . Do you get the same function you did in part (a)? 1 due Wednesday, March 8 2017 Page 2 of 2 3 In our section on max / min problems, it has been necessary sometimes to find critical points along the boundary of a region. We have two main methods, which are (i) parametrizing the boundary curve, getting our function in terms of one variable (t), and finding critical points by calc 1 methods, and (ii) Lagrange multipliers. We'll be doing both in this question, and comparing the results. Our aim in both parts will be to find the critical points of the function f (x, y) = 3x2 xy + 2y 2 that lie along the ellipse x2 4 + y2 9 = 1. Method 1: parametrizing the boundary (a) Let ~r(t) = h2 cos(t), 3 sin(t)i. Verify that this curve satisfies the equation x2 4 2 + y9 = 1. (b) Plug in the coordinates of ~r(t) into f (x, y) and get a formula for f as a function of t. (c) Compute f 0 (t) by the usual Calc 1 methods. (d) Setting f 0 (t) = 0, find angles between 0 and 2 where our critical points will lie [Hint: I ran into a place where knowing the formulas for sin(2t) and cos(2t) was convenient.] (e) Use your answer to (d) to find the (four) critical points of f along this curve. Give exact answers, not decimals. Method 2: Lagrange multipliers (f ) Compute f . (g) Let g(x, y) = x2 4 + y2 . 9 Compute g. (h) Since our boundary curve is a level curve of g, our method says that for the critical points of f along this curve, f = g for some number . Set f = g and get two equations in x, y, and . (i) Solve each equation above for ; set your two expressions equal to find a relationship between y and x at our critical points. 2 2 (j) Now use the relationship in (i) and the constraint equation x4 + y9 = 1 to find the critical points of f along the curve. Did you get the same points here and in part (e)? Last compiled at 5:25 P.M. on March 1, 2017