+ 5 x x x (30 pts) tan -1 x 5 + y = x sin x. = tan-1 x 5. (I) Calculate the derivatives of the following functions. NO NEED TO SIMLIFY!

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1 MIDTERM 2 - AU 4.nb 3 MIDTERM 2 - SOLUTIONS. (30 pts) (I) Calculate the derivatives of the following functions. NO NEED TO SIMLIFY! (a) (8 pts) y x tan - x 5 d x tan - x 5 + x d tan - x 5 2 x tan - x 5 + x + x x 4 tan- x 5 2 x + 5 x x 4 + x 0 (b) (8 pts) y x sin x

2 2 SOLUTIONS - MIDTERM 2.nb y x sin x ln y ln x sin x Differentiate both sides : d Hln yl d J ln xsin x N y d Hsin x ln xl y cos x ln x + Hsin xl x y cos x ln x + Hsin xl x xsin x cos x Hln xl + Hsin xl x HIIL H4 ptsl The curve with equation x 2 +xy +y 2 4 is a "rotated ellipse". HaL Use implicit differentiation to find the derivative. Differentiate both sides of the equation x 2 +xy +y x +y + x + 2 y 0 Hx + 2 y L -2 x - y -2 x - y x + 2 y

3 SOLUTIONS - MIDTERM 2.nb 3 HbL Consider the two points H3, -L and H2, -2L. Show that one of these points lies on the ellipse defined above, and one does not. Check whether the point H3, -L satisfies the equation : H3L 2 +H3L H-L +H-L ! 4 It does not. Therefore, the point H3, -L does not lie on the ellipse. Check whetherthe point H2, -2L satisfies the equation : H2L 2 +H2L H-2L +H-2L ! It does. Therefore, the point H2, -2L does lie on the ellipse. HcL For the point in part HbL which lies on the ellipse, find the equation of the line tangent to the ellipse at this point. y - H-2L x2, y-2 Hx - 2L We have to evaluate x2, y-2. x2, y-2-2 x - y x + 2 y x2, y-2-2 H2L - H-2L H2L + 2 H-2L H2L Therefore, the equation of the line tangent to the ellipse at this point is y + 2 Hx - 2L or y + 2 x H6 ptsl EXPLANATION IS NOT REQUIRED!

4 4 SOLUTIONS - MIDTERM 2.nb The graph of a function g' HtL, the derivative of g is given below. Assume that the function g is continuous on its domain H-2, 4L. y g'htl 2 y g' ( t ) t - -2 HIL Find all critical points of g. x 0, x, and x 3. Because g' H0L 0, g' H3L 0, and g' is not defined at x. HIIL On what interval Hor intervalsl is the function g increasing? The function g is increasing on H-2, L and H3, 4L. 2. HCONTINUEDL HIIIL At what point Hor pointsl does the function g have a LOCAL MAXIMUM?

5 SOLUTIONS - MIDTERM 2.nb 5 At x. Because x is a critical point, g is continuous on its domain, and g' changes the sign from positive to negative as x increases through. HIVL On what interval Hor intervalsl is the function g CONCAVE UP? The function g is concave up on H0, L and H, 4L. Because g' is increasing on those intervals. HVL Find all inflection points of g. At x 0. Because g' changes from decreasing to increasing at x 0. Therefore, g changes from cocave down to concave up at x H20 ptsl An observer stands 200 meters from the launch site of a hot air balloon. The balloon rises vertically at a constant rate of 5 m ê s. How fast is the distance between the observer and the balloon changing 20 seconds after the launch? Make sure to label the figure!

6 6 SOLUTIONS - MIDTERM 2.nb Given dt 5 m ê s, we have to find dt. t20 s We have to find an equation relating x and y. From the figure above and the Pythagorean Theorem, it follows that y x 2 By differentiating both sides of the equation with respect to t, we get 2 y dt 2 x dt, and, therefore y dt x dt. At the time t 20 s, x 20 H5L 00 m, and y So, y m at the time t 20 s. By substituting these values into the equation, we get dt t20 00 H5L Solving for dt t20, we get dt t20 00 H5L H5L m ê s. 00 H5L 00 2 I M

7 SOLUTIONS - MIDTERM 2.nb 7 4. H8 ptsl Show your work. Suppose at time t 0. s HtL t is the position of an object moving along a line HIL Find the instantaneous velocity, v inst HtL, at t 2. v inst H2L s ' H2L H2 tl t 2 4 HIIL Alternatively, we can find v inst H2L by following the steps below : STEP : Find the average velocity, v av HhL, over the time + h, 2D, if - < h < 2 + hd, if 0 < h <. Let - < h < 0 v av HhL s H2L - s H2 + hl 2 - H2 + hl 9 - H2 + hl2-5 -h 9 - H2 + hl2-5 -h h - h2-5 -h 4 h + h2 h 4 + h Let 0 < h < s H2 + hl - s H2L v av HhL H2 + hl - 2 H2 + hl h 4 +4 h +h h So, v av HhL 4 + h 4 h + h2 h 4 + h STEP 2 : Using the result in step, compute the limit. Show your work! lim hø0 v av HhL lim hø0 H4 + hl 4

8 8 SOLUTIONS - MIDTERM 2.nb STEP 3 : Using the result in step 2, find v inst H2L. v inst H2L lim hø0 v av HhL 4 5. H6 ptsl DO NOT USE GRAPHS TO JUSTIFY YOUR ANSWER! NO CREDIT WITHOUT WORK! Let f HxL 5 + x ln x HaL H8 ptsl Find the critical points of f. Since f' HxL ln x + x x ln x +, the only critical points are those where f' HxL 0. f' HxL 0 ï ln x + 0 ln x - x e - is the only critical point for f. HbL H8 ptsl Find the absolute minimum and the absolute maximum values of f on the ed. Justify your answer! Since there are no critical points on the ed, the max and min values are obtained at the end points. f HL 5 + ln 5 f HeL 5 + e ln e 5 + e > 5 Therefore, absolute minimum value f HL 5, and absolute maximum value f HeL 5 + e.