Final practice, Math 31A - Lec 1, Fall 2013 Name and student ID: Question Points Score Total: 90

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1 Final practice, Math 31A - Lec 1, Fall 13 Name and student ID: Question Points Score Total: 9

2 1. a) 4 points) Find all points x at which the function fx) x 4x x 1/3 is not differentiable. First x 4x + 3 x 3)x 1) isn t differentiable at x 1 and x 3 because x 4x + 3 crosses the x-axis with nonzero slope there. Second x 1/3 isn t differentiable at x because it has a vertical slope there. Thus f isn t differentiable at x, x 1 and x 3. b) points) Evaluate the indefinite integral 1 sin x) cos x) cos x) dx. 1 sin x) cos x) cos x) c) 4 points) Show that the function is concave up on [, 1]. fx) ) dx 1 sin x + sin x + cos x + cos x dx ) sin x + cos x + dx x + cos x + sin x + C cos t dt, Let F θ) θ cos t dt, then F θ) cos θ by the fundamental theorem of calculus and fx) F x ). Thus f x) xf x ) x cos x, and f x) cos x 8x cos x sin x which is positive for x. Page

3 . a) 5 points) Compute the limit lim x sin x 1 + x 1. lim x sin x sin x 1 + x + 1) lim 1 + x 1 x x sin x lim x x lim x 1 + x + 1) b) 5 points) Compute the limit lim x x ) 1 x. lim x x ) 1 x lim x 1 lim x 1 lim x 1 ) 1 x 1 x ) x 1 1 x 1 + x + ) 1 x 1 + x 1 + x + Page 3

4 3. Consider fx) a) 4 points) Find all the critical points in [, π]. sin x cos x + 1. f x) cos x sin xcos x + 1) + sin x cos xsin x) cos x + 1) cos x sin xcos x + sin x + 1) cos x + 1) 4 cos x sin x cos x + 1) The critical point are where f x), that is at x, π, π, 3π, π. b) 3 points) Find the maximum value of fx) in [, π]. Thus the maximum value of f is 1. c) 3 points) Show that fx) 4x in [, π]. f) f π ) 1 fπ) f 3π ) 1 We have that fx) f) x )f c) for some c by the Mean Value Theorem. Note that f) and f c) 4 since cos c sin c 1 and cos c + 1) 1. Thus fx) 4x for x in [, π]. Page 4

5 4. 1 points) Sketch the graph of the function fx) x + 1 x + ). You need to do all the steps in the analysis of the shape of the function. Answer: We ll find vertical and horizontal asymptotes of f, and then we ll determine the areas of positive and negative slope and concavity by looking at f and f. By inspecting the denominator, we see that the function has a vertical asymptote at x. Also, we note that x + 1 lim x ± x + ) lim 1/x + 1/x x ± 1 + 4/x + 4/x, so y is a horizontal asymptote. Note that f has a zero at x 1. Next, we ll compute f and f, and we ll use them to determine the behavior of f on certain intervals. Using the quotient rule: And, again by the quotient rule: f x) x + ) x + 1)x + ) x + ) 4 x ) x + 1) x + ) 3 x x + ) 3. f x) x + )3 + 3xx + ) x + ) 6 x + ) + 3x x + ) 4 x x + ) 4 x 1) x + ) 4. By testing f x) at x 3, 1, 1, we see that f has positive slope on, ) and negative slope on, ) and, ). Also, by testing f x) at x 3,,, we see that f is concave up on 1, ) and down on, ) and, 1). Combining these properties, we obtain the graph from WolframAlpha): Page 5

6 5. 1 points) a) Find the point on the graph of fx) x x + which is the closest to the point 1, ).What is the distance of this point to 1, )? b) Answer a) with fx) x x 7/. Answer: a) At a point x, the distance between 1, ) and x, fx)) is given by: dx) x 1) + fx) ) x 1) + x x + ). We would like to minimize this quantity. Note that the minimum of dx) will occur at the same x as the minimum of gx) x 1) + x x + ) since dx) gx)), so it s enough to minimize gx) by finding its critical points. We take the derivative of gx): g x) x 1) + x x + )x 1)) x 1)1 + x x + )) 4x 1) x x + 5 ). By testing the discriminant b 4ac) of x x + 5, we see that 4 1 <, so x 1 is the only root of g x). So the only critical point of g is x 1. We note that as x goes to ±, the distance between 1, ) and x, fx)) approaches infinity. Thus, the minimal value of g and of dx)) must occur at x 1. The minimum occurs at 1,1), and the distance from 1,) at this point is given by d1) 1 1) + f1)) f1) 1. b) We employ the same strategy as before, where dx) x 1) + fx) ) x 1) + x x 7 ) and gx) x 1) + x x 7 ). We look for critical points of gx): g x) x 1) + x x + 7 ) x ) x 1)1 + x 4x 7) 4x 1)x x 3) 4x 1)x 3)x + 1) So the critical points of g are x 1, 3, 1. By the same reasoning as before, atleast one of these points is the point at which dx) is minimal. So we test these points: d3) d 1) + 1 ) 17 and d1) f1) 9. So the minima occur at x 3, 1, with corresponding points 3, 1 ) and 1, 1 ), and the distance from 1, ) at each of these points is 17. Picture from WolframAlpha) Page 6

7 6. 1 points) A boat is moving at the velocity 1 meters per second directly towards a wall. The movement of the boat is perpendicular to the wall. There is a rotating reflector on the boat which rotates at the constant rate of 8 revolutions per second. It turns out that when the boat is exactly 5 meters away from the wall, the light beam forms the angle of 45 o with the direction of the boat, and the light beam is pointing to the left when looking from the boat at the wall. Determine the velocity with which the light spot is moving along the wall at that moment. Answer: We relate the distance from the boat to the wall x to the angle θ of the light with the direction of the boat and the distance y of the light spot from the center of the wall i.e. the spot directly in front of the boat on the wall). We have y x tan θ. We re told in the problem statement that, at the given moment, x 5 m, dx dt 1 m/s, dθ dt 8 revs/sec π radians/rev)8 revs/sec) 16π rads/sec. We re interested in knowing the speed of the light spot along the wall - that is, dy Using implicit differentiation, we have: dt. dy dt dx dt tan θ + x sec θ dθ dt. Plugging in the given information and the fact that sec 45, we see that dy dt 116π + 1) )16π) Page 7

8 7. a) 5 points) Compute the first derivative and the second derivative of F x) : b) 5 points) Find the point x x where F x) : obtains its minimum in [, 1]. Solution to a). Note that F x) x x t) sin tdt. 4 x sin t dt tdt t sin t dt. By the Fundamental Theorem of Calculus and the product rule we have F x) sin t dt + x sin x x sin x cos x + 1. Hence F x) sin x. Solution to b). First compute F x). Suppose Gx) is an antiderivative of t. Then F x) Gx 4 ) Gx ) by the Fundamental Theorem of Calculus. Differentiating using the chain rule yields F x) 4x 3 G x 4 ) xg x ) 4x 3 x 4 x x 4x 5 x. We now find the critical points by setting this equal to zero. If 4x 5 x then x x 3 1) so x or x 3 1. Thus the minimum of G must occur at either x, x 1 or x 3 1. We note that if x 1 then x 4 x so 4 t dt. x Since the only critical point occurs at x 3 1, we thus know that F 3 1 ) <. On the other hand F ) 1 t dt and F 1) 1 t dt. Hence F obtains it s minimum at x 3 1. Page 8

9 8. 1 points) At the point c, c ) on the graph of y x, draw a tangent line to y x. Compute the area between this tangent line, the curve y x and the y-axis in terms of c). Solution. First find the equation of the tangent line at the point c, c ). The slope of the tangent line at that point is y c) c. Using point slope we see the equation of the tangent line is y c ) cx c), which we rewrite as A picture is shown below y cx c. Since y x is concave up, the tangent line lies below the curve, so the area of the region bounded by y x, it s tangent line at c, c ) and the y-axis is c x cx c )) dx c x cx + c ) dx [ x 3 3 cx + c x ] c c3 3 c3 + c 3 +. c3 3. Page 9

10 9. 1 points) Compute the volume of the solid formed by rotating the region bounded by y x and y x around the line x 1. Solution. First find the points where the curve x intersects y x. Note that x x so we simply need to solve x x. This factors as x + ) x 1), which has two solutions, namely x ±1. We sketch a figure of the region in question. We will use cylindrical shells to evaluate the volume. The volume is then. 1 1 πrx)hx) dt where rx) is the radius and hx) is the height. The height is given by hx) x x. The radius is rx) 1 x. Hence the integral becomes 1 π 1 1 π1 x) x x ) dx π1 x) x + x) dx x 3 x x + ) dx + π π π 3 4. π1 x) x x) dx 1 x 3 3x + ) dx Page 1