Solution for Final Review Problems 1


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1 Solution for Final Review Problems 1 (1) Compute the following its. (a) ( ) ( ) ( )( ) = (b) (c) ( 1)( 2 + ) 1 ( 1)( ) = (1 )(2 3) (1 )(2 3) (d) (e) t 2 t t / (1/ 1)(2 3/) = 3 ( )( ) ( ) 2 ( ) = 1 t t 2 t 2 2 t 2t t 2 t 2 t 2 1 2t = ichen/math1143f/fp1sol.pdf 1
2 2 (f) sin(4) tan(3) sin(4) tan(3) ( ( ) sin(4) cos(3) sin(3) ) sin(4) sin(3) cos(3) ( ) ( ) sin(4)/(4) 4 = 4 sin(3)/(3) 3 3 (2) Find the derivative of each of the following functions. (a) f() = tan() + cos( 2 ) f () = ( tan()) + (cos( 2 )) = tan() + sec 2 () 2 sin( 2 ) (b) f() = f () = (3 + 1) ( 3 1) ( 3 + 1)( 3 1) ( 3 1) 2 (c) f(t) = 6 3 t 5 = 32 ( 3 1) 3 2 ( 3 + 1) ( 3 1) 2 = 62 ( 3 1) 2 (d) f() = cos(sin()) f (t) = (6t 5/3 ) = 1t 8/3 f sin(sin()) cos() () = 2 cos(sin()) (3) Find local and absolute maima and minima of the function f() = 3 3 on the interval [ 2, 2]. Take the derivative of f(): f () = Solve f () = and we obtain two critical points = 1 and = 1. Compare f( 2), f( 1), f(1) and f(2) and we see that f() takes the absolute maimum 2 when = 2 or 1 and f() takes the absolute minimum 2 when = 2 or 1. And f() has a local maimum at = 1 and a local minimum at = 1. (4) A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is his shadow on the building decreasing when he is 4 m from the building?
3 Let be the distance between the spotlight and the man and y be the length of his shadow on the building (both in meters). Then 2/y = /12 and hence y = 24. Take derivative on both sides of y = 24 with respect to t: ( dy dt ) + ( d dt ) y = dy dt = y ( ) d dt Plug in d/dt = 1.6, = 8 and y = 3 and we obtain dy/dt =.6. So his shadown is decreasing at a rate of.6 m/s. (5) Use Intermediate Value Theorem and Mean Value Theorem to show that the equation = has eactly one solution. Let f() = Since f() = 1 and f( 1) = 2, f() = has a solution in (, 1) by Intermdediate Value Theorem. Net, we will show this solution is unique. Suppose that f() = has two solutions = a and = b. So f(a) = f(b) =. By Mean Value Theorem (or Rolle s Theorem), there eists c in (a, b) such that f (c) = f(b) f(a) b a =. On the other hand, f () = > for all. This is a contradiction. Hence f() = has only one solution. (6) The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. If the area of printed material on the poster is fied at 384 cm 2, find the dimension of the poster with the smallest area. Let and y be the length and width of the poster (measured in cm). Then the printed area is ( 8)(y 12) = 384. So y = The area of the poster is S = y = Solve S () = : ( ) = ( 8) 2 = 256 = 24 ( 8) 2 Since 8 + S() S() =, S() takes the minimum at = 24 on (8, ). So the smallest poster has dimension
4 4 (7) A cylindrical can is made out of materials costing 6 dollar/m 2 for the top and bottom and 1 dollar/m 2 for the sides. Assume that its volume is fied at 1 m 3. Find the dimension that minimizes the cost. Let r be the radius of the base of the can and h be its height (in meters). Then the total cost is 6(2πr 2 ) + 1(2πrh) = 12πr 2 + 2πrh with volume πr 2 h = 1. Substitute h = 1/(πr 2 ) and the cost function becomes f(r) = 12πr r Solve f (r) = : 24πr 2 r 2 = r = π Since r + f(r) r =, f(r) takes the minimum at r = So the cheapest can has base radius 3 25 m and 3π 3π height 3 72 m. 5π (8) Two cars start moving from the same point. One travels south at 6 mph and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later? Let f(t) be the distance between two cars after t hours. Then f(t) = (6t) 2 + (25t) 2 = 65t. So f (t) = 65 and the distance between the cars is increasing at a rate of 65 mph. (9) Sketch the graphs of each of the following functions. You must follow the steps AH as in Sec. 4.5: (A) Domain (B) Intercepts (C) Symmetry (D) Asymptotes (E) Intervals of Increases and Decreases (F) Local maimum and minimum (G) Concavity and points of inflection (H) Sketch the curve. (a) f() = The domain of f() is (, ). Solve f() = and we obtain the intercepts: ( 2 2, ), (, ) and (2 2, ). And the yintercept is (, f()) = (, ). Since f( ) = 8( ) 2 ( ) 4 = = f(), f() is even. It is not odd and not periodic. The graph y = f() does not have any vertical asymptotes. Since (8 3 ) =
5 and (8 3 ) = it does not have any slant and horizontal asymptotes. Take the first derivative of f(): f () = = 4( 2)( + 2). Therefore, f() is increasing for in (, 2) (, 2) and decreasing for in ( 2, ) (2, ). Since f () changes from positive to negative at 2, f() has a local maimum at 2. Since f () changes from negative to positive at, f() has a local minimum at. By symmetry, f() has a local maimum at 2. Take the second derivative of f(), f () = = 12( 2/ 3)( + 2/ 3). Hence f() is concave upward for in ( 2/ 3, 2/ 3); f() is concave downward for in (, 2/ 3) (2/ 3, ). And 2/ 3 and 2/ 3 are points of inflection. (b) f() = The domain of f() is (, 1) (1, ). The intercept is ( 1, ) and the yintercept is (, 1). It is not even, odd and periodic. The graph of y = f() has a vertical asymptote = 1. Since f() f() = and f() f() = 1 it has a horizontal asymptote y = 1. Take the first derivative of f(): f () = 2/(1 ) 2. So f() is always increasing and it does not have any local maima/minima. Take the second derivative of f(): f () = 4/(1 ) 3. So f() is concave upward for in (, 1) and concave downward for in (1, ). It does not have any points of inflection. (1) Evaluate the following integrals. (a) 1 (2 )d 1 ( 2 )d = ( ) 1 = 1 6 5
6 6 (b) π/2 (sin() cos())d π/2 (sin() cos())d = ( cos() sin()) π/2 = (c) d Substitute t = 1 + 3, i.e., = (t 1)/ d = ( ) t 1 t 1/22 d 3 = t23/22 + C = (1 + 3)23/22 + C 2 (d) ( 3 + 1) d 2 Substitute t = ( 2 d = 1 3 d(3 + 1)) 2 ( 3 + 1) d = ( 3 + 1) 2 d(3 + 1) = t dt 2 = 1 3t + C = 1 3( 3 + 1) + C (11) Epress the following integral as a it of Riemann sums. Do not evaluate the it. 6 (a) d 5 Divide [2, 6] into n intervals of length = 4/n: f()d = d 5 f( i ) 4 ( ) 2n + 4i f n n f(2 + i ) 4n 3 (2n + 4i) n 5 + (2n + 4i) 5 (b) 2π 2 sin()d
7 2π Divide [, 2π] into n intervals of length = 2π/n: f()d = 2π 2 sin()d 2π n f( i ) ( ) 2πi f n f( + i ) ( ) 8π 3 i 2 2πi sin n 3 n (12) The table gives the values of a function obtained from eperiment. Use them to estimate 6 f()d using midpoint rule with n = f() The midpoint approimation of the integral is 2(f(1) + f(3) + f(5)) = 2( ) = 18 (13) Starting with 1 = 1 use Newton s method to find 3, the third appromiation to the root of the equation =. By Newton s method, So and n+1 = n f( n) f ( n ) = n 3 n + n n = = 2( 1)3 3 3( 1) = 23 n n = = 2( 5/4)3 3 3( 5/4) = (14) For what values of does the graph of f() = + sin() have a horizontal tangent? Solve f () = 1 + cos() = and we obtain = 2kπ + π. So the graph of f() has a horizontal tangent when = (2k + 1)π for all integers k. (15) Let F () = 3 f() and G() = f( 3 ). If f(1) = 2 and f (1) = 3, find F (1) and G (1). Since F () = 1 3 [f()] 2/3 f (), F (1) = 2 2/3 = 3 2/2. Since G () = f ( 3 ) 2/3 /3, G (1) = 1. (16) The displacement of a particle is given by s(t) = A cos(bt + C) with A, B, C constants. Find the velocity and acceleration of the particle at time t. 7
8 8 The velocity is f (t) = AB sin(bt+c) and the acceleration is f (t) = AB 2 cos(bt + C).
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