Mathematics 1210 PRACTICE EXAM II Fall 2018 ANSWER KEY

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1 Mathematics 1210 PRACTICE EXAM II Fall 2018 ANSWER KEY 1. Calculate the following: a. 2 x, x(t) = A sin(ωt φ) t2 Solution: Using the chain rule, we have x (t) = A cos(ωt φ)ω = ωa cos(ωt φ) x (t) = ω 2 A sin(ωt φ) = ω 2 x(t) b. f x, f(x) = ( ) x 2 3 x π Solution: Using the chain rule an the quotient rule, we habe ( ) ( ) x 2 2 (x π) f x (x 2) (x 2) x (x π) (x) = 3 x π (x π) 2 ( ) x 2 2 (2 π) = 3 x π (x π) 2 y c. x, cos xy = y2 + 2x Solution: The implicit erivative will be use in this problem. Implicit ifferentiation treats y as a function of x even though it is not known explicitly. First ifferentiate the equation efining the relationship between x an y with respect to x an apply the usual ifferentiation rules (chain rule an prouct rule in the first term) to obtain sin(xy)(y + xy ) = 2yy + 2. Then, collect terms which inclue a factor of y from the chain rule an factor it out. y sin(xy) 2 = (2y + x sin(xy))y Finally, ivie to solve for y : y t = (2 + y sin xy) y = (2y + x sin xy). 1 cos x Solution: Since 1 cos x = 0 0,

2 which tells us that we can apply l Hopital s Rule. Hence, we have 1 cos x = x x () (1 cos x) = (sin x sec 2 x) + tan x cos x (sin x) (tan x/ cos x) + tan x cos x = = 0 (sin x) 0, which once again tells us that we can apply l Hopital s Rule. So, we have 1 cos x = = = x x t Hence, we can conclue that () (1 cos x) [(tan x/ cos x) + tan x cos x] t (sin x) cos x sec x tan x( sin x) + tan x( sin x) + cos x sec 2 x] cos 2 x = 2. cos x 1 cos x = 2. e. f tan x, f(x) = sin x 1 + x 2 Solution: This is of the form f(g( u(x) v(x) ), where f(x) = sin x, g(x) = x1/2, u(x) = tan x, an v(x) = 1 + x 2. So, using the chain rule (twice) an quotient rule, we will have ( ) tan x f 1 tan x 1/2 (1 + x 2 ) sec 2 x 2x tan x (x) = cos 1 + x x 2 (1 + x 2 ) 2 f. f x, f(x) = x2 sin 2 (x 3 ) Solution: This is of the form p(x)q(r(s(x))) where p(x) = x 2, q(x) = x 2, r(x) = sin x, s(x) = x 3. So, using the prouct rule an the chain rule (twice) gives f (x) = 2x sin 2 (x 3 ) + 6x 4 sin (x 3 ) cos (x 3 ). g. m v, m(v) = m 0 1 v 2 /c 2, m 0 is rest mass, an c = m/s. Solution: Notice that m 0 is constant, an we can rewrite m(v) as ( ) 1 m(v) = m 0 1 v2 2. c 2

3 Now, we have the function in the form f(g(v)), wo we can use the power rule an chain rule, which gives m v = 1 2 m 0 (1 v2 c 2 ) 3 ( 2 2v ) c 2 = m 0v c 2 (1 v2 c 2 ) 3 2 h. 1 cos x x Solution: Notice that 1 cos x x = 0 0, which implies that we can use l Hopital s Rule. Hence, we have 1 cos x = x x (1 cos x) sin x x (x) = = 0. 1 i. h 0 (x + h) 3 x 3 h Solution: Notice that this is the efinition of erivative, it follows that (x + h) 3 x 3 h 0 h = x x3 = 3x 3 j. sin x x x 3 Solution: Using LHopital rule 3 times, we have sin x x cos x 1 x 3 = 3x 2 sin x = 6x cos x = = A circular oil slick spreas so that its raius increases at the rate of 1.5 feet/secon. How fast is the area of the enclose oil increasing at the en of two hours? Solution: The area an the raius are relate statically by A = πr 2 In this situation they are relate ynamically by A(t) = πr(t) 2. Taking erivatives with respect to time using implicit erivative, we get A (t) = 2πr(t)r (t).

4 Given r (t) = 1.5 feet/secon an t = 2 hours = 7200 secons. Hence, after 2 hours the area increasing at rate We get A (t) = 2π(1.5)(1.5)(7200) = 32400π. Alternatively, we can irectly substitute r(t) = 1.5t into the area formula, A(t) = 2.25πt 2 an A (t) = 4.5πt so at t = 2 hours or 7200 secons, A = A balloon initially on the groun 150 feet away from an observer is release an rises at a rate of 8 f/s. How fast is the istance between the observer an the balloon changing when the balloon is 50 feet above the groun? Solution: Let r be the istance between observer an balloon an h is the istance of balloon above the groun. Consier the following figure, The Pythagorean Theorem tells us that r 2 = h Calculating the rate of change of istance between the observer an balloon using implicit erivative an power rule, we get 2r r t = 2hh t + 0 r t = h h r t. Hence, at h = 50, we have r = = It follows that r t = 50(8) =

5 4. Approximate 66 an sin π 100 using linear approximation (i.e., the ifferential). Solution: Using the approximation of erivative, or Consiering function, We have y y(x + x) y(x) x x y(x + x) = y(x) + y x x. y(x) = x, y, (x) = 1 2 x So The approximation of 66 is y(66) = y(64 + 2) In the secon one, consier the function We have Hence, the approximation of sin π 100 = y(64) + y, (64)(2) = y(x) = sin x, y, (x) = cos x. sin(π/100) = sin 0 + π 100 = sin 0 + cos 0( π 100 ) = 0 + π 100 = π 100. is given by π Use the ifferential to approximate the increase in volume of a spherical bubble as its raius increases from 3 to inches. Solution: Similar to the previous problems, using the approximation of erivative y(x + x) = y(x) + y, (x) x over function v = y(r) = 4 3 πr3 with y(r) = 4/3πr 3, y (r) = 4πr 2 an y(3) = 36π, an coincientally, y (3) = 36π. So, we get y(0.025) = y(3.025) y(3) = y( ) y(3) = y(3) + y, (3)(0.025) y(3) = 36π(0.025) Hence, the volume of spherical bubble in crease approximately 0.9π cubic inches.

6 6. Consier the following functions. In each cases, fin all local maxima an minima of f, where f is increasing an ecreasing, where f is concave up an concave own, an all inflection points. Does f have a global maximum or a global minimum? Sketch the graph of f(x). (a) f(x) = x 3 12x + 1. Solution: For f(x) = x 3 12x + 1, f (x) = 3x 2 12 an f (x) = 6x. Then f = 0 when x 2 = 4 or x = ±2. We have f is increasing for x < 2, f is ecreasing for 2 < x < 2 an f is increasing for x > 2. Moreover, f( 2) = = 17 f(2) = = 15 So f has a local maximum at x = 2 an a local minimum at x = 2. There is no global maximum or minimum. f is concave up where f > 0, i.e., for x > 0, an concave own for x < 0. There is one inflection point where f changes concavity, at x = 0 since f (0) = 0. From the information, which we obtaine we can sketch the graph as following, (b) f(x) = x 2 Solution: For f(x) = x 2, f (x) = 2x (1+x 2 ) 2. So f is increasing where f > 0, i.e., for x < 0 (the enominator of f is always positive) an f is ecreasing where f < 0, i.e., for x > 0. There is one local maximum at x = 0 where f changes from increasing to ecreasing. This is also a global maximum. There are no local or global minima as f approaches zero but remains positive as x tens to infinity in both irections. By the quotient rule, f (x) = 6x2 2 (1+x 2 ) 3, so 3 3, f < 0 an f is concave own f = 0 an f has inflection points when x = ± between these points, an f > 0 an f is concave up outsie these points. From the information, which we obtaine we can sketch the graph as following,

7 7. Consier f(x) = 2 sin(x π 4 ) on the interval [ π 2, 5π 4 ] Fin where f is increasing an ecreasing. Fin maximum an minimum value of f in the given interval. Solution: Notice that f (x) = 2 cos(x π 4 ), which is positive for π 2 < x < 3π 4 (where f is increasing) an negative for 3π 4 < x < 5π 4 (where f is ecreasing.) In the interval [pi, 5pi], we know that f, (x) = 0 at x = 3π 4. There is a local maximum at x = 3π 4 where f = 2, f = 0, an f < 0, an no other critical points in the interval. Comparing with the enpoints f( π 2 ) = 2 an f( 5π 4 ) = 0, the global maximum is 2 an the global minimum is 0 on this interval. 8. A rectangle has two corners on the x-axis an the other two on the parabola y = 12 x 2, with y 0. What are the imensions of the rectangle of this type with maximum area? Solution: From the problem, we can raew the figure as following, Since the corners are (x, 0), ( x, 0)(x, 12 x 2 ), ( x, 12 x 2 ) the area of rectangle will be given by A = 2xy = 2x(12 x 2 ) = 2x x.

8 A x = 2(12 3x2 ), which is zero when x = ±2. We take x to be the vertex on the right, so x = 2, y = 8, the imensions are 4 by 8 (an the maximum area is 32.) 9. An object is propelle upwar from the groun with initial velocity v 0 = 32 f/s. After time t, the height x(t) of the object is given by x(t) = 16t 2 + v 0 t + x 0, where x 0 is the initial position, which is assume to be x 0 = 0. (a) What is the velocity of the object when it reaches its maximum height? Solution: The object will reach its maximum height when the rate of change of height or its velocity is 0 (v(t) = 0 f/s. (b) At what time oes the object reach its maximum height? Solution: Notice that v(t) = 32t + v 0. The object reach its maximum at v(t) = 32t + v 0 = 0 or t = v 0 32 = = 1 sec. (c) What is the maximum height reache by the object? Solution: We know that the object will reach its maximum when t = 1 or at height x(1) = = 16 f.