MIDTERM 2 REVIEW: ADDITIONAL PROBLEMS. 1 2 x + 1. y = + 1 = x 1/ = 1. y = 1 2 x 3/2 = 1. into this equation would have then given. y 1.

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1 MIDTERM 2 REVIEW: ADDITIONAL PROBLEMS ) If x + y =, find y. IMPLICIT DIFFERENTIATION Solution. Taking the derivative (with respect to x) of both sides of the given equation, we find that 2 x + 2 y y = 0, which gives y y =. x Notice that the original equation implies y = x, so we can actually write y entirely in terms of x, as y = x x = x + = x /2 +. Taking derivatives again, we then have y = 2 x /2 = 2x x. If you hadn t noticed that y could be written entirely in terms of x, then you would have computed x y 2 y = y y 2 x ( x) 2 Plugging y = y x into this equation would have then given y = x 2 y y x y 2 x x = 2 y 2 x x = x + y 2x x = 2x x. 2)* Find all points on the curve x 2 y 2 + xy = 2 where the slope of the tangent line is. Solution. First we use implicit differentiation to find y at any point. Taking derivatives (with respect to x) on both sides of the given equation, we find which can be rearranged to give Problems marked with a * are extra challenging. 2xy 2 + x 2 2y y + y + x y = 0, y = 2xy2 y 2x 2 y x.

2 We now want to find those points (x, y) that for which y = and which lie on the given curve (i.e., satisfy the equation x 2 y 2 + xy = 2). If we first set y = in the above equation, we find that 2xy 2 y 2x 2 y x =. At this point the algebra gets a bit tricky. First, notice that the above equation can be simplified to which happens to factor as 2x 2 y 2xy 2 + x y = 0, (2xy + )(x y) = 0. So, we must have either xy = or x = y. However, if xy =, then notice that 2 2 the equation for our curve becomes ( /2) 2 + ( /2) = 2, which is not true! So, we must actually have x = y. If we plug this into the equation for our curve, we find that x + x 2 = 2, or equivalently x + x 2 2 = 0. This equation factors as (x 2 )(x 2 + 2) = 0, and so we must either have x 2 = or x 2 = 2. The second case is impossible, so it must be that x 2 =, and hence x = ±. Since x = y, this means that the two solutions are (, ) and (, ). Those are the only two points on our curve where the tangent line has slope. RELATED RATES ) Gold dust is being poured into a conical pile at a rate of 0 cubic feet per minute. The diameter of the base of the pile is always equal in length to the height of the pile. How quickly is the height of the pile increasing when the height is 0 feet? (Hint: Recall that the volume of a cone with base radius r and height h is V = πr2 h.) Solution. The base of the conical pile of gold dust has radius r = h/2, where h is the height of the pile. It follows that the volume of gold dust is V = πr2 h = π ( h 2 ) 2 h = 2 πh. Differentiating this equation with respect to time t gives We are given that dv and so dh = 0 25π dv = dh πh2 = 0 cubic feet per minute. So, when h = 0 feet we have 0 = dh π(0)2, 0.8 feet per minute. 2

3 )* Suppose the minute hand on a clock is 8 inches long, and the hour hand is long. Find the rate at which the distance between the tips of the hands is increasing when the clock reads :00. (Hint: Recall the Law of Cosines, which says that if θ is an angle in a triangle with adjacent edges a, b and opposite edge c, then c 2 = a 2 + b 2 2ab cos(θ).) Solution. Let c be the distance between the tips of the hands of the clock, and let θ be the angle between those hands (measured clockwise from the minute hand to the hour hand). By the Law of Cosines, we have c 2 = cos(θ) = 80 6 cos(θ). Differentiating with respect to time t, we then have 2c dc = 6 sin(θ)dθ. What is dθ? The minute hand travels once around the circular clock face every hour, and so rotates at 2π radians per hour, or 2π/60 = π/0 radians per minute. The hour hand travels once around the circular clock face twelve hours, and so rotates at 2π/2 = π/6 radians per hour, or π/60 radians per minute. The difference between their rates of rotation is therefore π/60 π/0 = π/ radians per minute. So, dθ radians per minute. Now we just need to determine θ and c when the clock reads :00. At that time, the minute hand is pointing straight up (at 2 on the clock face), and the hour hand is pointing at on the clock face. The angle θ between the two is 2π/2 = π/6. So, θ = π/6. If we substitute this into our original formula relating c and θ, we find that c 2 = 80 6 cos(π/6) 2.57, and so c.96 inches. Plugging these values of c, θ, and dθ into our above equation gives 2(.96) dc = 6 sin(π/6)( 0.096), which gives dc 0.0 inches per minute. LINEAR APPROXIMATIONS AND DIFFERENTIALS 5) Estimate (.0) using a linear approximation. Solution. There are many possible solutions to this problem, but the following is probably the most straightforward. If we let f(x) = x, then the quantity we are interested in estimating is f(.0). We can easily compute f() = =, so it seems reasonable to use a linearization of f at a = to make this estimate, i.e., to use the approximation f(x) f ()(x ) + f() for x near, to estimate f(.0). We compute f (x) = x 0, so f () =. We therefore have f(x) (x ) + for x near. In particular, (.0) = f(.0) (.0 ) + = (0.0) + =..

4 6)* Use differentials to estimate the volume of a thin cylindrical shell with height h, inner radius r, and thickness r. What is the error in this estimate? Solution. The exact volume of the shell would be the difference between the volume of the outer cylinder (with radius r+ r) and the volume of the inner cylinder (with radius r), namely volume of shell = π(r + r) 2 h πr 2 h. If V (r) denotes the volume of a cylinder with radius r and (fixed) height h, then we can rewrite this as volume of shell = V (r + r) V (r) = V. In particular, we can estimate this volume using the approximation dv V. A quick computation gives dv = 2πrh dr (recall that the height h is some fixed constant). So, our estimate is volume of shell 2πrh dr = 2πrh r. What is the error in this estimate? Well, the exact volume of the shell is V = π(r + r) 2 h πr 2 h, so the error in our estimate is error = V dv = ( π(r + r) 2 h πr 2 h ) 2πrh r = ( π(r 2 + 2r r + ( r) 2 )h πr 2 h ) 2πrh r = ( 2πrh r + π( r) 2 h ) 2πrh r = π( r) 2 h. ABSOLUTE MAXIMA AND MINIMA 7) Find the absolute maximum and minimum of the function f(x) = x 5 x + 2 on the interval [, ]. Solution. We first find the critical points of f on the given interval. We compute f (x) = 5x x 2 = x 2 (5x 2 ).

5 So, f (x) = 0 exactly when either x = 0 or x = ± /5 ±0.77. We now compare the values of the function f at these critical points and the end points: f( ) = ( ) 5 ( ) + 2 = ( ) + 2 = 2 f( /5) = ( /5) 5 ( /5) + 2 = = f(0) = = 2 f( /5) = ( /5) 5 ( /5) + 2 = 9 25 f() = = 2. So, the maximum value of f on the interval is 6 value is = , and the minimum )* Find the absolute maximum and minimum of the function f(x) = x x +x 2 x on the interval [0, 2]. Solution. We first find the critical points of f on the given interval. We compute f (x) = x 9x 2 + 6x. We need to determine when f (x) = 0, which means we need to find the roots of the above polynomial. For a general cubic polynomial, this can be quite difficult. In this case, however, we are going to be a bit luck. First, notice that f () = = 0, so we know x is one of the factors of the f (x). If we perform polynomial long division, we find that f (x) = (x )(x 2 5x + ) = (x )(x )(x ) = (x ) 2 (x ). So, we can now see that f (x) = 0 exactly when x =,. We now compare the values of f at the endpoints and these critical points: f(0) = 0 f(/) = (/) (/) + (/) 2 (/) = 27/ f() = + = 0 f(2) = 2 (2) + (2) 2 2 = 2. So, the maximum value of f on the interval is 2, and the minimum value is 27/ MEAN VALUE THEOREM 9) Two runners start a race at the same time and end in a tie. Prove that at some point during the race they were running at exactly the same speed. Solution. Let f(t) and g(t) be the positions of the first and second runners at time t, respectively. Let h(t) = f(t) g(t), and suppose the runners finish the race at time T. We are given that h(0) = 0 (since the runners start the race at the same position) and h(t ) = 0 (since the runners end in a tie, and so are at the same position at 5

6 the end time T ). Assuming the positions of the runners define continuous and differentiable functions (which seems reasonable), it follows from Rolle s Theorem that h (c) = 0 at some time c between time 0 and T. This implies f (c) = g (c), i.e., the two runners were running at exactly the same speed at time c. GRAPHING 0) Follow our usual checklist of properties to sketch y = ( x 2 ) 5. Solution. We first note that the function f(x) = ( x 2 ) 5 is defined for all real numbers. When x = 0, we have y = 5 = 02, so the y intercept is (0, 02). When y = 0, we have 0 = ( x 2 ) 5, whose only solutions are x = ±2. So, the x-intercepts are ( 2, 0) and (2, 0). Next note that f( x) = ( ( x) 2 ) 5 = ( x 2 ) 5 = f(x), so the function f is even. This means that the graph will be symmetric about the y-axis. As for asymptotes, there can be no vertical asymptotes (since f is defined everywhere). There are also no horizontal asymptotes, since lim x ± f(x) =. We now move on to checking where f is increasing and decreasing. We first compute f (x) = 5( x 2 ) ( 2x) = 0x( x 2 ). It follows that f (x) = 0 exactly when x = 0, ±2. These points divide up our domain as (, 2) ( 2, 0) (0, 2) (2, ). Checking test points in each interval, we find f ( ) > 0, f ( ) > 0, f () < 0, and f () < 0. It follows that f is increasing on (, 2) and ( 2, 0), and decreasing on (0, 2) and (2, ). There is a local maximum at x = 0 (where f(0) = 5 = 02). Lastly, we check the concavity of the graph. Computing the second derivative, we find f (x) = 0( x 2 ) 0x( x 2 ) ( 2x) = 0( x 2 ) + 80x 2 ( x 2 ) = 0( x 2 ) ( ( x 2 ) + 8x 2 ) = 0( x 2 ) (9x 2 ). In particular, we can see that f (x) = 0 exactly when x = ±2, ± 2. These points divide up our domain as (, 2) ( 2, 2) ( 2, 2) ( 2, 2) (2, ). Checking test points, we find that f ( ) < 0, f ( ) > 0, f (0) < 0, f () > 0, and f () < 0. So, f is concave down on (, 2), ( 2, 2 ), and (2, ), and concave up on ( 2, 2) and ( 2, 2). There are inflection points at x = ±2, ± 2. Putting all of the information we ve discovered together, we should have a graph that looks approximately like this: 6

7 ) Follow our usual checklist of properties to sketch y = x x 2 9. Solution. We first note that the domain of the function f(x) = x is all real numbers except ±, i.e., the domain is (, ) (, ) (, ). When x = 0, we x 2 9 have y = 0, and so the y-intercept is (0, 0). When y = 0, we have x = 0, so the point (0, 0) is also the only x-intercept. This function is neither even nor odd, so the graph doesn t have any special symmetry. As for asymptotes, there are vertical asymptotes at x = ±, where one can check lim f(x) = lim x lim f(x) = x + f(x) = x lim f(x) =. x + We also have a horizontal asymptote at y = 0, since lim f(x) = 0. x ± We now move on to checking where f is increasing and decreasing. We first compute f (x) = (x2 9)() (x)(2x) (x 2 9) 2 = (2x2 + 9) (x 2 9) 2. Since the numerator is never zero, there are no points where f (x) = 0. In fact, we always have f (x) < 0. So, the function f is always decreasing. 7

8 Lastly, we check the concavity of f. We first compute f (x) = (x2 9) 2 ( x) ( 2x 2 9)(2(x 2 9)(2x)) (x 2 9) = x(x2 9) 2 + x(2x 2 + 9)(x 2 9) (x 2 9) = x(x2 9)( (x 2 9) + (2x 2 + 9)) (x 2 9) = x(x2 + 8) (x 2 9). It follows that f (x) = 0 exactly when x = 0. This divides up our domain into (, ) (, 0) (0, ) (, ). Cecking test points in each interval, we find f ( ) < 0, f ( ) > 0, f () < 0, and f () > 0, so f is concave down on (, ) and (0, ), and concave up on (, 0) and (, ). There is an inflection point at x = 0 (where f(0) = 0). Putting all of this information together, we should sketch a graph that looks approximately like the following: ) Follow our usual checklist of properties to sketch y = + x + x 2. Solution. We first note that the domain of the function f(x) = + + = x2 +x+ x x 2 x 2 is all real numbers except 0, i.e., the domain is (, 0) (0, ). There is no y- intercept, since the function is not defined for x = 0. When y = 0, we have x 2 + x + = 0 (using the second form of f), which has no roots. So, there is also no x-intercept. This function is neither even nor odd, so the graph doesn t have any special symmetry. As for asymptotes, the graph has a vertical asymptote at x = 0, where one can check lim f(x) =. The graph also has a horizontal asymptote at x 0 y =, since lim f(x) =. x ± 8

9 We now move on to finding where f is increasing and decreasing. We first compute f (x) = x2 (2x + ) (x 2 + x + )(2x) (x 2 ) 2 = x2 2x x = x 2 x. So, there is a single critical point at x = 2. Recalling that our domain is (, 0) (0, ), this divides up our domain into (, 2) ( 2, 0) (0, ). Checking some test points, observe that f ( ) = 2 = < 0, f ( ) = 2 = > 0, and f () = 2 = < 0. So, f is decreasing on (, 2) < 0, increasing on ( 2, 0), and decreasing on (0, ). Notice that f has a local minimum at x = 2 (where the value is f( 2) = ). Lastly, we check the concavity of f. We compute f (x) = x ( ) ( x 2)(x 2 ) (x ) 2 = 2x + 6x 2 x 6 = 2(x + ) x. So, the only time f (x) = 0 is when x =. This divides up our domain as (, ) (, 0) (0, ). Checking some test points, observe that f ( ) = 2( +) < 0, f ( ) = 2( +) > 0, and f () = 2(+) > 0. So, f is concave down on ( ) ( ) () (, ), concave up on (, 0), and concave up on (0, ). Notice that f has an inflection point at x = (where the value is f( ) = 7). 9 Putting all of this information together, we should sketch a graph that looks approximately like the following: It is a bit difficult to see in this graph, but there is an inflection point at x = and a local minimum at x = 2. 9

MATH 10550, EXAM 2 SOLUTIONS. 1. Find an equation for the tangent line to. f(x) = sin x cos x. 2 which is the slope of the tangent line at

MATH 10550, EXAM 2 SOLUTIONS. 1. Find an equation for the tangent line to. f(x) = sin x cos x. 2 which is the slope of the tangent line at MATH 100, EXAM SOLUTIONS 1. Find an equation for the tangent line to at the point ( π 4, 0). f(x) = sin x cos x f (x) = cos(x) + sin(x) Thus, f ( π 4 ) = which is the slope of the tangent line at ( π 4,