Calculus 1: A Large and In Charge Review Solutions

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1 Calculus : A Large and n Charge Review Solutions use the symbol which is shorthand for the phrase there eists.. We use the formula that Average Rate of Change is given by f(b) f(a) b a (a) (b) 5 = 3 = ( ) 3 5 = 3 = 5 7 (c) / / 3 ( ) = / =. (a) 3 = 6 = 6 (b) 5 7 = (c) = ( )( + 3) = 7 + (d) 3 + ( + )( ) ( )( ) = 3 = 3 3 (e) ( )( + + ) ( )( + ) ( 3/ ) (f) (g) (h) (i) + + ( + ) ( 3) = 8 3. (a) + = 3 ( )( + + ) = ( )( + ) + and g() is a continuous etension of f(). (b) This is impossible as = 3 does not eist. (One sided its do not agree). = and ( 7) = = 6 So set g() = does not eist. ndeed since =

2 (c) ( + )( + 6) + g() is a continuous etensions of f().. (a) By the Sandwich Theorem, f() = (b) By the Sandwich Theorem, f() = 3 = 8 So set g() = + 6 and (c) Observe cos ( ( ) cos ( ) for > and cos ) for ( < ). So by the Sandwich Theorem, each directional it is and so cos = (d) Note that + sin ( π for so by the Sandwich Theorem + ) ( ( )) + sin π ( ( )) as > π + sin = 5. (a) Consider + = f(). f() is continuous on (, ) as it is the difference of continuous functions and so f() = < and f() = 3 >. By the ntermediate Value Theorem, c such that f(c) = with < c <, i.e. (c) = c + (b) Again, f() = cos is continuous on (, π/) and f() = > and f(π/) < so by the VT, a solution eists. 6. Use the definition f () h f( + h) f() h 3( + h) + (3 + ) (a) h h + h (b) h h h h 3 + 3h + 3 h + h h( + h + ) h (c) h ( + h) 3 + ( + h) + ( ) h h + 3h + h h + 3 h h = h(3 + 3h + h + = 3 + h h +h + h h h h( + h) h (d) h h h( + h)( + + h) = ( ) = 3/ 3h h h = 3 = + h + h h( + h)( + + h)

3 7. f() being differentiable at a point means that f () eists and is continuous at that point. (a) f () = 6 so both f and f () are continuous and differentiable everywhere. (b) f() is continuous for and f () = + so f() is differentiable for > (c) f() is continuous for any and f () = any ( ) so f() is differentiable for (d) f() is continuous for and f () is continuous on (, ) and (, ). 8. (a) f () = ()( ) ()(+) ( ) = ( ) = ( ) (b) dy du = u(+u ) u( u ) (+u ) = u u3 u+u 3 (+u ) = u (+u ) (c) y = 3/ / y = / (d) y = + y = ( +) ( ) ( = +) ( +) (e) y = a + b (f) g() = + /5 g () = + 5 3/5 (g) u = t /3 + t /3 du dt = 3 t /3 + 3 t /3 (h) s = t 7/ t + t ds dt = 7 t5/ + t 9. Use that the tangent line at a point = c is given by y y = f ()( c) (a) y = (+) () m = =. (+) So the tangent line is y = ( ). (b) y = (+) (+) m = (5) 5 = 3/ 5 = 3 So the tangent line is y 5 = 3 ( ). (c) y = + m = 3. So the tangent line is y = 3 ( ).. Using the rules of derivatives, it is quite simple. (a) (f + g) () = f () + g () = (b) (f g) () = f () g () = (c) (3fg) () = f ()g() + 3f()g () = 5 ( ) f (d) g () = f ()g() f()g () = 7 (g()). 3

4 (e) (f g) () = f (g())g () = f ()g () = 3 (f) (f g) () = f()f ()g() + (f()) g () = 8 (Chain/Product Rule) (g) ( fg) () = f()g() (fg) () = f()g() (f ()g() + f()g ()) =. Use that d (s(t)) = v(t), d (v(t)) = a(t) and speed is the absolute value of velocity. dt dt The answers here are given so at the first number is velocity, speed then acceleration for t =, given as ( v(), v(), a() ) and the same for time t =. (a) v(t) = 3 and a(t) = so (3, 3, ) and (3, 3, ) (b) v(t) = 9t and a(t) = 8t so (7, 7, 8) and (, 88, ) (c) v(t) = 6t + 6 and a(t) = 6 so (,, 6) and ( 8, 8, ) (d) v(t) = t (+t ). (a) f () = sin + cos (b) dy d = sin sec (c) g (t) = sec t tan t + sec t (d) h (θ) = θ cot θ θ csc θ (e) dy d and a(t) = (+t ) +t((+t )(t)) (+t ) = cos (+cos ) sin ( sin ) (+cos ) (f) y = sec (cos )( sin ) (g) y = sin cos + cos sin (h) y = csc cot csc 3 = cos +cos +sin (+cos ) = 3. This is the same as evaluating the derivative at =. (a) y = 3( + ) ( ) m = 3 (b) y = ( 5) 5 ( ) m = 8 35 (c) y = 3 ( + tan t) /3 m = 3 (d) y = 3 cos ( sin ) m = (e) y = ( + ) 3 ( + ) /3 () m = (f) y = sec (cos )( sin ) m = (g) y = cos(sin(sin )) cos(sin()) cos m = (h) y = cos(sin ( ) sin(sin )) sin cos m =. Use the method of implicit differentiation. so (,, ) and (,, 9 ) cos + = (+cos ) +cos

5 (a) dy d = dy d = y (b) y dy dy + 3y = dy = y d d d 3y ) = dy (c) +y ( ) + dy d + y ( y + dy d 5. This is a combination of problem and 5. = y d +y y + +y = y +y+(+y) y y +y+(+y) y (a) ( +y )(+y dy dy 9 ) = 5( y ) ()(6+m) = 5(6 m) m = d d 3 So the tangent line is y = 9 ( 3) 3 (b) y + y dy = (y + )( d y ) dy + (y + d ) ( y) dy m = d So the tangent line is y = (c) y dy = d 3 m = 8 m = 9 So the tangent line is y = 9 ( ) 6. Just use the rules of taking derivatives. (a) f () = g( ) + g ( )() = g( ) + g ( ) f () = g ( ) + g ( ) g ( ) = 6g ( ) g ( ) (b) f () = g () g() = g () g() f () = g () (c) f () = g ( ) f () = g () g () 3/ 7. V = 3 so dv dt 8. dy dt = = 3 d dt d = d d = + dt 5 5 dt dt g () + g() 3 9. Let s be the distance between the ships so s = + y where is the distance from Ship A to where Ship B was at noon and Y is the same for Ship B. The at the time in question, s = +, =. So: s ds dt = d dt + y dy dt ds dt = ()( 35) + (5) ds dt = 5. Let r be the radius of the water level inside the cone and h its height. By similar triangles, we deduce that h = 3r. Note we must use the same units for all measurement so converted everything into centimeters. Thus the final answer is in cubic centimeters per minute. Thus: V = 3 πr h = 7 πh3 dv dt = dh πh 9 dt dv dt = 8, π 9 π() () = 9 Since this is the overall change in volume, the rate at which water is being pumped into the tank is 8,π +, since the rate of change is the rate in minus the rate 9 out. 5

6 . Draw a triangle with vertices at (, ), (5, ) and the remaining side in quadrant with length. Thus the height of the triangle is h = sin θ and base b = 5. Note θ is the angle between the two fied sides and the final units are in square meters per second. Thus: A = bh = da (5)( sin θ) = sin θ dt = cos θ da dt = cos(π 3 ) = 3. Draw a circle with radius with center at (, ). Then the friend is at the origin and the runner is at the point (, y) on the curve ( ) + y =. Let s be the distance between the runner and his friend. Thus: s = +y s = +( ( ) ) s = + s ds dt = d dt At the time in question, this implies that ds = d d. To find, we need to use some dt dt dt trigometry. Form a triangle between the origin, runner and the center of the circle and let θ be the angle between the -ais and the edge connecting the runner and the center of the circle. Thus by the law of cosines, = + ()() cos θ cos θ = sin θ = 5 By the definition of θ and using opposite interior angles, we obtain d dt 7 5 where v is the runner s velocity. 3. Use the fact that f() f(a) + f (a)( a). = v sin θ d dt = (a) Use the function f() = and a = 36. Then f () = f (36) =. Thus (36. 36) (b) Use the function f() = and a =. Then f () = f () =. Thus. (. ). 99 (c) Use the function f() = 6 and a =. Then f () = 6 5 f () = 9. Thus (.97) (.97 ) (.97) We first find all numbers such that f () = or f () is undefined. These are the critical numbers of the function. n the problem, this is what was meant when it ask to find critical values. (a) f () =. Thus the critical number is { 5 (b) f (t) = 6t + 6t 6. Thus the critical numbers are } { + 5, } 5 6

7 { (c) s = t 3 +t +t = t(t 3+ +3t+). Thus the critical numbers are, 5 3, } 5 (d) f (r) = r + r(r) never zero. (r +) = r (+r ). Thus the critical numbers are {, } since + is (e) g () = 3 / /3 = 3 5/3 ( + ). Thus the critical numbers are {, } (f) g () = 3 ( ) /3 ( ). Thus the critical numbers are { },, 5. The absolute ma and min of a function will always occur at the endpoints of the interval = [a, b] or critical numbers in. (a) f () = 6. The critical number is and so f() = 7, f() = 5 and f(3) =. So ma {f()} = 5 and min {f()} = 7. (b) f () = The critical numbers are {, }, f() =, f() = 9, f( ) = 5 and f( ) = 8. So ma {f()} = 9 and min {f()} =. (c) f () = = ( 3 ). The critical numbers are {, } and so f() = 3, f ( ) = 7 and f() = 5. So ma {f()} = 5 and min {f()} = 3. (d) f () =. The critical numbers are {, } and so f( ) =, f() =, ( +) f() = and f() = 5. So ma {f()} = 5 and min {f()} =. (e) f () = cos sin. The critical number is { } ( π and so f() = and f π ) =. So ma {f()} = and min {f()} =. (f) f () = + sin. The critical numbers are { 5π ( 3, f π ) 6 = π, π 6 6 } ( and so f 5π ) 6 = 5π 6 3, f( π) = π + and f( π) = π +. So ma π + and min {f()} = π {f()} = 6. One must check that f() is continuous on [a, b] and differentiable (a, b), which is easily done. (a) f(b) f(a) b a = 6 ( ) =. f () = 6 + so 6c + = c = (b) f(b) f(a) b a c = = 9 ( ) = 5. f () = 3 + so 3c + = 5 c = ± 3. But only 3 is in the interval [a, b]. (c) f(b) f(a) = b a =. f () = so = c = ±. But only c = 3 /3 3c /3 3 3/ in the interval. 3 3 is 7

8 (d) f(b) f(a) = 3 3 b a = 9. f () = + (+) = (+) (c + ) = 9 8 = (c + ) c = ±3 But only c = 3 is in the interval. 7. f() = 3 and f( ) = 8 and since f() is continuous everywhere, the ntermediate Value Theorem gives the eistence of a zero. Since f () = 5 + > for all, the zero is unique. 8. Suppose there is such a function. Then f() f() = 5 so by the Mean Value Theorem, c such that f (c) = 5. But this contradicts the fact that f () for all. Thus no such function can eist. 9. A function is increasing when f () >, decreasing when f () <, is concave up when f () > and concave down when f () <. (a) f () = 6 6 = 6( )(+). So f() is increasing on (, ) (, ) and decreasing on (, ). t has a local ma at (, 7) and a local min at (, 3). f () = 6. So f() is concave up on (, /) and concave down on (/, ). t has an inflection point at (, ) 3 (b) f () = 3 = ( 3). So f() is increasing on ( 3, ) ( 3, ) and decreasing on (, 3) (, 3). t has a local ma at (, ) and local mins at ( 3, 8) and ( 3, 8). f () =. So f() is concave up on (, ) (, ) and concave down on (, ). t has an inflection point at (, 7) and (, 7). (c) h () = 6( ). So f() is increasing on (, ) (, ) and decreasing on (, ) (, ) and has no local ma and a local min at (, ). h () = 6( ) + ( ) = 6( )( + ). So f() is concave up on (, 5) (, + 5) (, ) and concave down on ( 5, ) ( + 5, ). t has inflection points at = ± and = ± 5 (d) P () = + + = + ( +) / ( +). So f() is increasing everywhere and has no local ma or min = ( +3) + + P () =. So P () is concave up on (, ) ( +) 3/ and concave down on (, ). t has an inflection point at =. (e) Q () = + + = + (+) / (+). So f() is increasing on ( /, ) and decreasing on (, /). t has a local min at =. 8 so

9 Q () = + + (+) / = 3+. So Q() is concave up everywhere + (+) (+) 3/ and no inflection points. (f) f () = /3 = /3 ( /3 ). So f() is increasing on (, ) (, ) and decreasing on (, ). t has a local ma at = and local min at =. f () = 3 5/3. So f() is concave up on (, ) and concave down on (, ). t has an inflection point at =. (g) f (t) = sin t. So f(t) is increasing everywhere and has no local ma or min. f (t) = cos t. So f(t) is concave up on ( 3π, ) ( π π, ) 3π and concave down on ( ) ( π, 3π π, ) ( π 3π, π). f(t) has inflection points at t = ± 3π and t = ± π. 3. l Hôpital s rule is useful for these sort s of its. The tricks in b., c. and e. are particually useful. (a) (b) ( )( 3) + + Thus the it is. = L = 3 + ( + ) = L L = (c) By using l Hôpital s Rule, the it is same as Now = L the it is 3. (d) + + (e) ( ) = L Thus L =. Hence = by l Hôpital s Rule. = L = L L = 3. y = and S = + y. Subsituting for y yields S = + S =. The critical number with = is a minimum and so = y =. 3. A = y = 3. The perimeter of the fence P is given by P = y + 3 P = Minimizing this function yields P = and the critical number = yields a minimum. Hence = and y = 3. The units are in feet. 9

10 33. V = y and the cost function is given by C = ( ) + 6()(y) + 6()(y) = + 36y since there is one side of area at $ per square foot, sides of area y and sides of area y at $6 per quare foot. Thus C = + 8 C = 8. The critical number = 6/3 corresponds to the minimum which makes the minimal cost, C min $ Don t forget +C! (a) F () = C (b) F () = C (c) F () = 5/ 7/ + C (d) F () = C (e) F () = C 35. For the solutions, C α, β and γ are arbitrary constants. (a) f () = α f() = α + β (b) f () = /5 + α f() = /5 + α + β (c) f (t) = t 3 + α f (t) = 5t + αt + β f(t) = t 5 + αt + βt + γ (d) f (t) = t 3 t3/ + α f (t) = 6 t3 5 t5/ + αt + β f(t) = t 8 5 t7/ + αt + βt + γ (e) f() = + C. Since f() =, we have that c = hence f() = +. (f) f() = 3 sin 5 cos + C. f() = 3 sin 5 cos + 9 Since f() =, we have that c = 9 and so (g) f () = α f() = 3 = 6 3 +α+β. Since f() =, we have that β = and since f() = we have α = 3. Thus f() = Use the formula that the Riemann Sum is the sum of the area of the rectangles. n each case, assume you use n, an aribtrary number of rectangles. Note that have used Right Handed Sums. (a) We have that the width of each rectangle is ( 8 and the height is given by the n n ) 8 3 k function value. Thus the area is given by. n n n (b) The width of each rectangle is π and so the area is given by n ( n π [ (π kπ ) ( kπ ) ]) + + sin π + n n n n k= k=

11 37. will use a right hand sum to estimate the areas. Note that for n =, this is given by the definite integral. (a) (b) (c) i. The width of each rectangle is and the heights are and so the area is 8. ii. The width of each rectangle is and the heights are,, 7 and so the area is. 3 iii. ( + 3) d = = ( (9) + 3 ) = 6 i. The width of each rectangle is and the heights are and so the area is. ii. The width of each rectangle is and the heights are 3, and so the area is iii.. ( ) d = 3 3 = 8 3 = 3 i. The width of each rectangle is and the heights are 7 and 9 so the area is 9. ii. The width of each rectangle is and the heights are 3, 7, 55 and 9 so the area is. iii. ( + 3 ) d = + = + 8 = Drawing a picture will help. (a) The area of the square is 9 and the area of the triangle is so the integral is 3. (b) The area of the rectangle is 3 and the area of the semicircle is 9π is 3 + 9π. (c) The area of the outside rectangles are both rectangle is so the integral is. 39. This is the Fundamental Theorem of Calculus. (a) g () = + (b) g (y) = y sin y (c) f() = cos( ) since F () = (d) tan (e) + cos(t ) dt so the integral and the area of the inside

12 (f) 3 sin( 3 ). (a) = = 6 3 (b) = 3 y6 + 3 y (c) = 3 t3 + t (d) = (e) = (f) = csc = 6 = 9 5 ( ) d = = d = = π/ π/3 = 3 3. The total area bound by a curve y = f() is given by (a) (b) /3 3 d = 3 d + = 3 /3 + 3 = 85 3 /3 3 3 d = = /3 d d b a f() d d = d

13 . These are done using a u substitution. Don t forget the +C! (a) Let u = = u 5 du = u6 + C = ( ) 6 + C (b) Let u = = u 6 du = 7 u7 = 7 ( )7 + C (c) Let u = + = u 3/ du = 5 u5/ = 5 ( + ) 5/ + C (d) Let u = 3 = 3 u du = 9 u 3 + C = 9 ( 3) 3 + C (e) Let u = 3 5 = u /5 du = 5 6 u6/5 + C = 6 (3 5)6/5 + C (f) Let u = 3 = sec u du = 3 3 tan u + C = 3 tan(3) + C (g) Let u = + = u 9 du = 5 u + C = 5 ( + ) + C (h) Let u = π = cos u du = π π sin u + C = ( π ) π sin + C (i) Let u = sec = u du = 3 u3 + C = 3 sec3 + C 3. These are also done using a u substitution. Don t forget to change the its of integration! (a) Let u = + 3 = 3 (b) Let u = = (c) Let u = t = π π 5 (d) Let u = + = 5/ (e) Let u = πt = π π (f) Let u = a u = 5 u du = 9 u3/ = 9 cos u du = sin u π = sin u du = cos u π = 5/ u du = 3 u3/ = 3 cos u du = π sin u π = a u du = a ( 5 5 ) 8 u du = a 3 u3/ = a3 3 3

14 π/. Observe that f() = sin sin is an odd function since f( ) = f() so + 6 π/ + d = 6 5. Let u =. Then d = u du. The last integral is the area of a quater-circle with radius. Thus the integral is π Let u = then 7. Observe that where f() = 3 +h f() d = + t3 dt = 9 +h f(u) du = + t3 dt + t3 dt = f( + h) f() + t3 dt. Thus by the definition of the derivative: h h +h + t3 dt = f () By the Fundamental Theorem of Calculus, f () = + 3 and so h h +h + t3 dt = + 8 = 3