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Solutions to midterm 2 dy by implicitly di erentiating the following: x2 + y = 1: 1. (a) Find dx 4 9 x + 2y dy = 0. Solve Solution: By implicitly di erentiating we get 2 9 dx dy 9 x this equation to get dx = , 4y . 2 (b) Find the equation of the tangent line to the ellipse x4 + y9 = 1: at the p 3 point (,1; 2 3 ). Solution: The slope at the point is obtained by substituting x = p ,1; y = 3 2 3 in to your formula for dy=dx, hence slope m = 3=2. Now use the point-slope formula to get 2 2 p p 2. T T T F F p p p p y , 3 2 3 = 23 (x + 1) =) y = 23 x + ( 23 + 3 2 3 ) p p which can be simpli ed to y = ( 3=2) x + 2 3. \A cubic function always has one point of in ection." \The linear approximation to the function f (x) = x2 at x = 0 is the line y = 0." \A point (c; f (c)) on a graph y = f (x) can be both a maximum value and a minimum value." f (x) = 6x3 + 12 is never zero. If f (x) is a quadratic function then f (x) ! 1 as x ! 1. 3. Consider the function y = f (x) = x2 x+ 1 on the interval ,2 x 2. (a) Find f (x). 0 Solution: By the quotient rule we have 2 1 , x (2x) = 1 , x2 : f (x) = (x + (1)x2 + 1)2 (x2 + 1)2 0 (b) Find all of the critical points (we use only real numbers!). Solution: Critical points are (i) endpoints, so x = 2 are endpoints; (ii) stationary points (i.e., points where f (x) = 0) so x = 1 are stationary points; and nally (iii) singular points (i.e., where f (x) is unde ned), but the denomenator of f (x) is never zero, so there are no singular points. Thus, the critical points occur at x = 1; 2. (c) Compute the maximum and minimum values that f (x) attains on the interval ,2 x 2. Solution: Substitute all four critical points into f (x) are choose the margest, namely f (1) = 1=2, and the smallest, namely f (,1) = ,1=2. 0 0 0 1 4. A light in a lighthouse 1 mile o shore from a straight shoreline is rotating at 2 revolutions per minute, i.e., at 4 radians per minute! How fast is the beam moving along the shoreline when it passes the point half a mile from the point opposite the lighthouse? Lighthouse 1 h shore Solution: We know d=dt = 4 , we want dh=dt, so we must relate and h. The picture shows that tan = h. Implicitly di erentiate to give dh = sec2 () d : dt dt We want the value when h = 1=2, i.e., when tan = 1=2. There are several ways to compute sec2 (). One is to compute cos() using Pythagorus, another is to use the identity sec2 () = 1 + tan2 () = 1 + 1=4 = 5=4. Sub in to the 5 above to give dh dt = 4 4 = 5 . 5. Let f be a continuous function and let its derivative f have the graph shown: y 0 (0; 1) bc b b (,2; 0) (2; 0) (0; ,1) x c b (a) Where is f increasing/decreasing ? Solution: Increasing when f (x) > 0, i.e., when x < ,2, 0 < x < 2 and x > 2. It's decreasing when f (x) < 0, i.e., when ,2 < x < 0. (b) Where is f concave up/down ? Solution: Concave up when f (x) is increasing, i.e., when x > 2. Concave down f (x) is decreasing, i.e., when x < ,2, ,2 < x < 0 and when 0 < x < 2. (c) Assume that f (0) = 0. Draw the graph of y = f (x). Solution: I'll do it in class. The point to watch out for is that the slope of the graph moving to the right from the origin is +1, while the slope of the graph moving to the left from the origin is ,1 0 0 0 0 2