Calculus with Analytic Geometry I Exam 5-Take Home Part Due: Monday, October 3, 2011; 12PM

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1 NAME: Calculus with Analytic Geometry I Exam 5-Take Home Part Due: Monday, October 3, 2011; 12PM INSTRUCTIONS. As usual, show work where appropriate. As usual, use equal signs properly, write in full sentences, and be clear. Try to work on these pages, but if you need more paper, USE MORE PAPER. These instructions should not be there.they make no sense. They came from the in class exam that I used as a template. 1. (Very similar, but not quite the same as Exercise 65, 3.4) If f and g are the functions whose graphs are shown, let u(x) = f(g(x), v(x) = g(f(x)), and w(x) = g(g(x)). Find each derivative if it exists. If it doesn t exist, explain why. I need to see some work here for credit! Solution. The fact that what was required where the values of the derivatives at x = 2. Without that information the exercise becomes considerably more difficult. Unfortunately, with the information I added by , the exercise becomes absurdly easy. The answer is: None of the derivatives exist at x = 2. If you selected x = 1 (as in the textbook), then u (1) = f (g(1))g (1) = f (4)g (1) = ( 1 4 )( 4) = 1 v (1) = g (f(1))f (1) = g (2)f (1) does not exist w (1) = g (g(1))g (1) = g (4)g (1) = 2 3 ( 4) = (Similar, but not the same as Exercise 67 of 3.4) If g(x) = f(x), where the graph of f is shown, evaluate g (2).

2 2 Solution. By the chain rule, dg dx = 1 2 f(x) 1/2 f (x). Setting x = 2, g (2) = 1 2 f(2) 1/2 f (2). Looking at the picture one sees that f(2) = 2. To estimate f (2), there is a part of the picture where we can see that the tangent line goes down four units as the variable x moves three units to the right. This means that it has slope 4/3, thus f (2) = 4/3. All in all, 3. ( 3.5, # 34) g (2) = 1 ( 2 2 1/2 4 ) 3 = (a) Find an equation of the tangent line to the curve with equation y 2 = x 3 + 3x 2 at the point (1, 2). Solution. Step 1. Finding the slope. By implicit differentiation, 2yy = 3x 2 + 6x. Setting x = 1, y = 2, we get 4y = 9, thus y = 9/4. Because solving for y is not terribly hard, one can also do this directly; usually this is much harder or impossible. One has y = ±(x 3 + 3x 2 ) 1/2 and one has to decide which sign to choose. Setting x = 2, we get y = ±2, so we have to choose the negative sign. Now y = (x 3 + 3x 2 ) 1/2 dy, dx = 1 x=1 2 (x3 + 3x 2 ) 1/2 (3x 2 + 6x) = 9 x=1 4. noindentstep 2. Finishing the problem. The point slope form of the tangent is y + 2 = 9 (x 1). 4 Equivalently, y = 9 4 x (b) At what points does this curve have horizontal tangents? (A horizontal tangent is a tangent with 0 slope). Solution. From the work done in part a (the implicit differentiation) we have y = 3x2 + 6x. The 2y tangent line is horizontal when y = 0; a fraction is 0 if and only if its numerator is 0 (and the denominator is not!), so we see we ll have a horizontal tangent line exactly when 3x 2 + 6x = 0. This equation can also be written in the form 3x(x + 2) = 0; the solutions are x = 0 and x = 2. To find the corresponding y values, we go back to the original equation y 2 = x 3 + 3x 2. For x = 0 we get y = 0. However, we should be wary of this point; looking at the equation we have for y we also have a 0 in the denominator. The

3 3 graph will give us more information. Setting x = 2, we get y 2 = 4, thus y = ±2. This gives us two points for sure with horizontal tangent lines, namely ( 2, 2) and ( 2, 2). (c) (Optional) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen. And printing out the result and attaching it to this exam. 4. ( 3.5, # 40) If x 2 + xy + y 3 = 1, find the value of y at the point where x = 1. It should go without saying that this value has to be a number. Solution. By implicit differentiation, 2x + y + xy + 3y 2 y = 0. We could solve this for y, but because we are required to find a specific value, it will be easier to continue differentiating implicitly. Next d dx (2x + y + xy + 3y 2 y ) = 0; thatis,2 + y + y + xy + 6yy y + 3y 2 y = 0. We can rewrite this as 2 + 2y + 6y(y ) 2 + xy + 3y 2 y = 0. Applying once more d/dx, we get 2y + 6(y ) yy y + y + xy + 6yy y + 3y 2 y = 0. We write out all our equations, grouping together some terms. x 2 + xy + y 3 = 1 2x + y + (x + 3y 2 )y = 0 2y + 6y(y )2 + (x + 3y 2 )y = 2 2y + 6(y ) 3 + (18yy + 1)y + (x + 3y 2 )y = 0 Now set x = 1. The first equation becomes 1 + y + y 3 = 1, thus y(1 + y 2 ) = 0, hence y = 0. In the second equation set x = 1, y = 0. It becomes 2 + y = 0, hence y = 2. Now set x = 1, y = 0, y = 2 in the third equation, to get 4 + y = 2, hence y = 2. Finally, setting x = 1, y = 0, y = 2, y = 2 in the last equation gives y = 0, whence y = 42.

4 4 Note: So far, these are fairly straightforward exercise. If you do them all right, you get 90points. For 100 points you have to do at least one of the following exercises. Each one adds 10 points to your total, so your grade here could be more than 100 points. 5. ( 3.5, #46) Show that the sum of the x- and y-intercepts of any tangent line to the curve x + y = c is equal to c. Solution. Suppose we have the tangent line at a point (x 0, y 0 ) of the curve. By implicit differentiation 1 2 x 1/ y 1/2 y = 0 so that y = x 1/2 y y =. The slope at (x 1/2 0, y 0 ) is thus y 0 / x 0 and the equation x of the tangent line is y y 0 = (x x 0 ); solving for y, etc. y = x + x 0 + y 0. x0 x0 (We used that x 0 x0 = x 0.) The y-intercept is obtained by setting x = 0 and works out to x 0 +y 0. The x intercept is obtained setting y = 0; we get x0 x = ( x 0 + y 0 ) = x 0 + x 0, so that the sum of the two intercepts works out to x 0 + y x 0. For the final touch, one has to recognize that this is a binomial expansion of the form a 2 + b 2 + 2ab, if a = x 0, b = y 0. That is, x 0 + y x 0 = ( x 0 + y 0 ) 2 = ( c) 2 = c. Since (x 0, y 0 ) was an arbitrary point on the curve, that is all it takes. 6. Exercise #80, Section 3.5 of the textbook: Solution. One thing that makes this problem look more difficult than it really is, is a typical magical trick: Your attention is drawn in one direction, things are happening elsewhere. Here your attention is drawn to the lamp. Forget about the lamp for a moment; it is hardly relevant. The problem could easily have been phrased as follows: Draw a tangent to the ellipse from ( 5, 0). Where does this tangent intersect the line x = 3? Let us call (x 0, y 0 ) the point on the ellipse where this tangent is actually tangent to the ellipse. By implicit differentiation 2x + 8yy = 0, thus y = x/4y; at (x 0, y 0 ) we get a slope m = x 0 /4y 0. Notice m > 0 since we are taking y 0 > 0 and x 0 < 0. The equation of the tangent line is thus y = m(x + 5). But we still don t know what m is. But we have two further conditions. One is that the point (x 0, y 0 ) is on the line, second that it is on the ellipse. Being on the line means that y 0 = m(x 0 + 5) = ( x 0 /4y 0 )(x 0 + 5); multiplying by 4y 0 and rearranging we get x = 5x 0. Being on the ellipse means x = 5 so that we must have 5 = 5x 0 or x 0 = 1. Going back to the equation for the ellipse, we now have = 5, hence y 0 = 2 (being positive). Thus m = 1/4 and now setting x = 3 in y = (1/4)(x + 5), we get the answer y = 2.

5 5 7. Problem 8 from the Problems Plus Section of Chapter 3 (p.270). A car is traveling at night along a highway shaped like a parabola with its vertex at the origin (see the figure). The car starts at a point 100m west and 100 m north of the origin and travels in an easterly direction. There is a statue located 100m east and 50m north of the origin. At what point on the highway will the car s headlights illuminate the statue. Your answer, of course, should be something like at a point x east/west and y meters north/south of the origin. And it should be clear how you arrived at your answer. Solution. A lot of information is sort of hidden. For example, the parabola has the vertex at (0, 0) and has the y-axis as its axis. It stays in the upper half-plane. The only parabolas satisfying all these conditions have equations y = ax 2, with a > 0. We are also told that the point ( 100, 100) is on the parabola; thus 100 = a( 100) 2 = 10, 000a. So a = 1/100. The equation of the parabola is y = x2. Suppose that the car s headlights illuminate the statue at the point of coordinates (x 0, y 0 ). That means that the tangent line of the parabola at that point goes through the point (100, 50). We know: y 0 = x2 0. The slope at that point is 1 50 x 0, so the equation of the tangent line is y y 0 = 1 50 x 0(x x 0 ). Finally, that the tangent line goes through (100, 50) tells us that 50 y 0 = 1 50 x 0(100 x 0 ). This last equation can be rewritten as 50 y 0 = 2x x2 0. Using that y 0 = x2 0, we can simplify this to y 0 = 2x But, from the parabola equation, y 0 = x 2 0/100, so now we have that (multiplying by 100) x x = 0. Solving we have two solutions x = 100 ± The one with the minus sign is the one we are looking for, the other one corresponds to when the tailights of the car, if they were powerful enough, would be illuminating the statue. Squaring and dividing by 100, we get the y 0 value. The car is ( )m east and m north of the origin. Notice The pictures in the posted version are in color.