Multiple Choice. at the point where x = 0 and y = 1 on the curve

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1 Multiple Choice 1.(6 pts.) A particle is moving in a straight line along a horizontal axis with a position function given by s(t) = t 3 12t, where distance is measured in feet and time is measured in seconds. What is the distance travelled by the particle in the time period 0 t 3 seconds? Solution: Since the particle is moving in a straight line, we first compute the intervals when the particle moving forward (positive derivative) and when the particle moving backwards (negative derivative). Note s (t) = 3t 2 12, hence s (t) = 0 when t = ±2. Only t = 2 is in the inverval we are considering. The total distance is the distance traveled from t = 0 to t = 2 plus the distance traveled from t = 2 to t = 3, that is total distance = f(2) f(0) + f(3) f(2) = = 23 feet. 2.(6 pts.) Find the value of dy dx at the point where x = 0 and y = 1 on the curve (x + 1)y 2 = cos 2 (πy). Solution: Differentiating implicitly, we use the product rule and the chain rule: d((x + 1))y 2 + (x + 1)d(y 2 ) = 2 cos(πy)d(cos(πy)) dx(y 2 ) + (x + 1)2ydy = 2 cos(πy)( sin(πy)πdy) We then collect the dy and dx on each separate side Now, solving for dy dx Plugging in the point (0,1) gives us dy(2y(x + 1) + 2π cos(πy) sin(πy)) = dx(y 2 ) dy dx = y 2 2y(x + 1) + 2π cos(πy) sin(πy) dy dx (0, 1) = 1 2 2(1)(0 + 1) + 2π cos(π) sin(π) =

2 3.(6 pts.) Let f(x) = x 3 6x 2 + 9x. Find the absolute maximum and absolute minimum of f on the interval [ 1, 3]. (That is find the maximum and minimum value of f(x) on the given interval). Solution: We see that f (x) = 3x 2 12x + 9 = 3(x 2 4x + 3) = 3(x 1)(x 3), so f has critical points in x = 1 and x = 3. To find absolute minimums and maximums it is enough to check critical points and the end points of the closed interval. We see that f( 1) = ( 1) 3 6( 1) 2 + 9( 1) = = 16 f(1) = 1 3 6(1) 2 + 9(1) = 4 f(3) = 3 3 6(3) 2 + 9(3) = = 0. We can now conclude that the absolute maximum is 4 and the absolute minimum is (6 pts.) The cost function for the production of jphones is given by C(x) where x is the number of jphones produced and cost is given in dollars. We know that C(x) is continuous on the interval 0 x < and differentiable on the interval (0, ). We have C(0) = 500, ( these are some overhead costs that must be paid even if we produce no jphones). If we know that 0 C (x) 100 for x > 0, which of the following must be true? ( only one must be true, the remaining ones might be false) Solutions: The minimum value of C(99) occurs if C (x) 0 (is constantly 0). Thus C(99) C(0) + (0)(99 0) = 500. Likewise, the maximum value of C(99) occurs if C (x) 100 (is constantly 100), yielding C(99) (99) = 10, 400. Putting these together: $500 C(99) $10, (6 pts.) Find the linearization of the function f(x) = 1 at a = x10 3

3 Solutions: The value of f(x) at x = 1 is 1 2. Note that f (x) = 10x9 (1 + x 10 ), thus 2 f (1) = Therefore the linear approximation is L(x) = (x 1) or equivalently 4 L(x) = 10x (6 pts.) Which of the following gives a complete list of the critical numbers/points of the function f(x) = 1 3 x1/3 x. Solution: Note that f (x) = 1 9 x 2/3 1. Since 0 is in the domain of f, but not in the domain of f, then x = 0 is a critical point. Setting f (x) = 0 gives x 2/3 = 9 or x 2 = This gives two more critical points x = ± (6 pts.) Find all of the the local maxima and minima of f(x) = (x + 1) 4 (x 2) 2. Solution: When looking for local maxima and minima it is enough to check the functions critical values, we therefore start by taking its derivative. Using the product rule and the chain rule we find that f (x) = 4(x + 1) 3 (x 2) 2 + (x + 1) 4 2(x 2) = 2(x + 1) 3 (x 2) ( 2(x 2) + (x + 1) ) = 2(x + 1) 3 (x 2)(3x 3) = 6(x + 1) 3 (x 2)(x 1), from here we easily read off the critical points to be x = 1, x = 1, and x = 2. We now want to check the sign of f (x) in the intervals between the critical points. interval 6(x + 1) 3 (x 1) (x 2) f (x) (, 1) ( 1, 1) + + (1, 2) + + (2, )

4 So we can conclude that f has a local minima when x = 1 and x = 2, and a local maxima when x = 1. 8.(6 pts.) Let f(θ) = θ2 4 + cos(θ) where 0 θ 2π. For which of the following intervals is the graph of f(θ) concave up on the entire interval. Solution: We calculate the second derivative in order to determine the concavity of the graph. f (θ) = 1 4 (2θ) sin(θ) = θ 2 sin(θ) f (θ) = 1 2 cos(θ) The graph of f(θ) is concave up when f (θ) > 0, which is when cos(θ) < 1. This happens ( 2 π in the interval 3, 5π ). 3 9.(6 pts.) Let f be a function of x. The table below shows whether the functions f (x) and f (x) are positive, negative or have value 0 at each of the given values of x. x f (x) = 0 = 0 = 0 f (x) < 0 = 0 > 0 Which of the graphs shown below is a feasible graph of f(x)? (Note that the label for each graph is given on the lower left of the graph.) Solution: Since f ( 2) < 0 then f(x) is concave up at x = 2, and since f (1) > 0 then f(x) is concave up at x = 1. The only graph satisfying this condition is 5

5 (b) 10.(6 pts.) The graph below shows the velocity of a jogger running back and forth along a straight jogging path over a 3 hour period. Here v(t) denotes the velocity of the jogger after t hours. Consider the following time intervals: which of the following is true: A : [1.5, 1.7], B : [2, 2.2], C : [2.6, 2.8] Solution: Speed is the absolute value of velocity. Based on the graph, we see that, in the interval A = [1.5, 1.7], the graph of the velocity is decreasing towards 0, so his speed is decreasing. On the interval B = [2.0, 2.2], the graph is decreasing but becoming more negative, so the absolute value is increasing (speed). On the interval C = [2.6, 2.8] the graph is increasing towards zero, so the absolute value is decreasing. Thus the jogger is speeding up in the interval B and slowing down in the intervals A and C. 6

6 Partial Credit You must show your work on the partial credit problems to receive credit! 11.(12 pts.) The volume of a cone is given by V = 1 (area of base)(height). 3 Water in poured into the inverted cone shown on the right at the rate of 1 cubic meters per minute. Note the base of this cone has a radius of 4 meters and its height is 20 meters. (a) Find a formula relating the volume of water in the cone (V) to the height of the water in the cone (h). Hint: Similar triangles may help. Solution: We will start by relating the radius of the surface of the water r to the height of the water h. We see by appealing to similar triangles that r 4 = h which implies that 20 r = h 5. We know that the area of the base of the water is A = πr2 and hence A = π h2 25. So we now have that V = 1 πh3 Ah = (b) What is dh dt when the height of the water in the cone is at 2 meters. 7

7 Solution: We know that dv dt = 1m3 /min. So using implicit differentiation with respect to t we see that 1 = dv dt = d ( ) πh 3 dt 75 = π d 75 dt (h3 ) = h2 dh dt = πh2 25 Now setting h = 2 we get that 1 = 4π 25 dh dt dh dt dh so we see that dt = 25. So we can conclude 4π that when the water is at height 2m the height changes with 25 4π m3 /min. 12.(12 pts.) Show that the equation x 7 + x 5 + x 3 + x + 1 = 0 has one and exactly one real solution. Identify the theorem(s) you are using and verify that the required conditions to apply the theorems are true to gain full credit. Solution: Note that f(x) = x 7 + x 5 + x 3 + x + 1 is a polynomial and therefore a continuous function. Since f( 1) = 3 and f(1) = 5, by the intermediate value theorem, there exists a c in ( 1, 1) such that f(c) = 0. So f(x) has at least one real solution. Observe that f (x) = 7x 6 + 5x 4 + 3x 2 + 1, which is always positive, for any choice of x in R. This means that f(x) is a monotone increasing function. Thus we can conclude that f(x) has only one real solution. 13.(14 pts.) A function f(x) is defined for 1 x 5 and its derivative is given by f (x) = x(x 3) 2 for x in ( 1, 5) (a) Find the critical numbers/points for f in the interval ( 1, 5). Solution: The critical points are when f (x) = 0, thus we get x = 0 and x = 3 as critical points of f(x). 8

8 (b) Find the intervals where f is increasing and decreasing. (justify your answer) Solution: We make the following table to determine when is the derivative positive or negative. interval x (x 3) 2 f (x) ( 1, 0) + (0, 3) (3, 5) We conclude that the function is decreasing in ( 1, 0) and increasing in (0, 5). (c) Classify the critical points as local maxima or local minima or neither and say which test you are using to make your conclusions in each case. Solution: From part (b) we have that at x = 0 the derivative changes from negative to positive, thus x = 0 is a local minima. At x = 3 the derivative does not change sign, thus x = 3 is neither a local maxima nor a local minima. (d) Find the intervals where f is concave up and concave down (recall that f(x) is only defined for 1 x 5. ) Solution: We first compute f (x). Note that f (x) = (x 3) 2 + 2x(x 3) = (x 3)(x 3 + 2x) = 3(x 3)(x 1). The possible inflection points are x = 3 and x = 1. To verify the sign of f (x) at the intervals ( 1, 1), (1, 3) and (3, 5) we compute interval 3(x 3) (x 1) f (x) ( 1, 1) + (1, 3) + (3, 5) We can conclude that f(x) is concave up in the intervals ( 1, 1) and (3, 5) and concave down in the interval (1, 3). 14.(2 pts.) You will earn 2 points if your instructor can read your name easily on the front page of the exam and you mark the answer boxes with an X (as opposed to a circle or any other mark). 9