Review Problems for the Final

Size: px

Start display at page:

Download "Review Problems for the Final"

Percival Baldwin
6 years ago
Views:

1 Review Problems for the Final Math 6-3/6 3 7 These problems are intended to help you study for the final However, you shouldn t assume that each problem on this handout corresponds to a problem on the final Nor should you assume that if a topic doesn t appear here, it won t appear on the final Compute the following derivatives: (a) d ( ) d ( + ) (c) d ln + ln ( + ln)] d (d) (e) d d sin ( ) (g) d d ( ln)(sin + 37) (i) dy, where sin y cos y + cosy = d (j) dy d, where y = (e + ) +3 (b) d ( ln(e ln + ) + e +) d d d 7sin (f) d d (k) y when = and y =, where y 3 3 = + y Compute d sin + d 3 Compute d d tan (e + ) 4 Prove that e > + for 5 Graph y = 3 5/ /3 cos + sin + 4 (h) d ( ) d In the following problems, compute the it, or show that it is undefined sin 3 sin 5 (a) sin (b) ( + h) 5 5 (c) h h 3 4 (d) 4 6 (e) (f)

2 4 (g) Compute the following integrals (a) (e + e 3 ) d (b) + 3 ( + ) 4 d (c) e e + d (3 ln) + (d) d (e) cos (sin ) + sin + d (f) d 3 ( ) sec /3 (g) d (h) /3 ln( + ) + d f () (i) d, if f() = and f() = e f() (j) ( + )4 ( ++5) d 3 (l) d (m) e e (n) d ( + ) d 8 (a) A population of flamingo lawn ornaments grows eponentially in Calvin s yard There are after day and 6 after 4 days How many are there after 6 days? (b) A hot pastrami sandwich with a temperature of 5 is placed in a 7 room to cool After minutes, the temperature of the sandwich is 9 When will the temperature be 8? 9 (a) Find the area of the region in the first quadrant bounded on the left by y =, on the right by + y =, and below by the -ais (b) Find the area of the region between y = and y = from = to = 4 Approimate the area under y = ( sin ) from = 3 to = 5 using rectangles of equal width, and using the midpoints of each subinterval to obtain the rectangles heights

3 Approimate the sum 3 + sin sin sin sin to 5 decimal places Calvin runs south toward Phoebe s house at feet per second Bonzo runs east away from Phoebe s house at 5 feet per second At what rate is the distance between Calvin and Bonzo changing when Calvin is 5 feet from the house and Bonzo is feet from the house? 3 Find the dimensions of the rectangle with the largest possible perimeter that can be inscribed in a semicircle of radius 4 A window is made in the shape of a rectangle with an isosceles right triangle on top s s s h h s (a) Write down an epression for the total area of the window (b) Write down an epression for the perimeter of the window (that is, the length of the outside edge) (c) If the perimeter is given to be 4, what value of s makes the total area a maimum? 5 (a) Find the absolute ma and the absolute min of y = on the interval 5 (b) Find the absolute ma and the absolute min of f() = 3 4 4/3 5 /3 on the interval 8 6 Use a it of a rectangle sum to find the eact area under y = + 3 from = to = 7 Given that f() = , what is (f ) ()? 8 Find the largest interval containing = on which the function f() = has an inverse f 6 9 (a) Compute d + t dt d 4 ( d ) 3 (b) Compute + d d (a) Use the definition of the derivative as a it to prove that d d 4 = ( 4) *(b) Compute h h +h t4 + dt 3

4 Let Is f continuous at =? Why or why not? + 3 if < f() = 6 9 if = 3 if > (a) Suppose that f(3) = 5 and f () = Use differentials to approimate f(99) to 5 places + 6 (b) A differentiable function satisfies dy d = e cos3 and y() = Use differentials to approimate y() 3 Use 3 iterations of Newton s method starting at = to approimate a solution to 4 = e 4 Prove that the function f() = 3 + cos + 5 has eactly one root Solutions to the Review Problems for the Final Compute the following derivatives (a) ( d ) d ( + ) = ( + ) () ()()( + )() ( + ) 4 (b) d ( ln(e + ) + e ln +) = e d e + + ( ) (e ln + ) (c) ( d ln + ln ( + ln)] = d + ln ( + ln) ) ( ) ( ) + ln (d) d d 7sin = (ln 7)(7 sin )(cos) (e) d d sin ( ) ( + = cos + 3 ( )) ( ( + 3)() ( + )() ( + 3) ) 4

5 (f) d cos + d sin + 4 = ( cos + sin + 4 ) / ( (sin + 4)( sin) (cos + )(cos) (sin + 4) ) ((g) ( d d ( ln)(sin + 37) = ( ln)(cos) + (sin + 37) ) (h) d ( ) d 7 + = ( ( ) 7 + ) / ( 7( ) 6 (3 + 5) + ) (i) dy, where sin y cos y + cosy = d Differentiate implicitly, then solve for y : sin y + y cos y + y sin y y sin y =, y = sin y sin y cosy sin y (j) dy d, where y = (e + ) +3 Use logarithmic differentiation: lny = ln(e + ) +3 = ( + 3) ln(e + ), y y = ln(e + ) + e ( + 3) e, + ( y = (e + ) +3 ln(e + ) + e ( ) + 3) e + (k) y when = and y =, where y 3 3 = + y Differentiate implicitly: Plug in = and y = : 3 y y + y 3 3 = + y 3y + 3 = + y, y =, y = ( ) Now differentiate (*) implicitly: 3 y y + 6 y(y ) + 6y y + 6y y + y 3 6 = y 5

6 Note that the term 3 y y produces three terms when the Product Rule is applied Now set =, y =, and y = : 3y = y, y + 4 =, y = 7 Compute d sin + d d sin + = ( sin + ) / d ( ) + 3 Compute d d tan (e + ) d d tan (e + ) = e + (e + ) 4 Prove that e > + for The result is true, as the picture shows However, a picture is not a proof Let f() = e (+) f measures the distance between the two curves The derivative is f () = e f = at = ; since f () = e and f () = >, the critical point is a local min Since it is the only critical point, it is an absolute min Thus, the minimum vertical distance between the curves occurs at =, when it is f() = Since this is an absolute min, it follows that f() > for that is, e ( + ) > for This is equivalent to what I wanted to show 5 Graph y = 3 5/ /3 The function is defined for all (you can take the cube root of any number) Since y = 3 5/3 ( 8 ), the -intercepts are = and = 8 The y-intercept is y = The derivatives are y = 5 /3 5/3 = /3 (5 ), y = 3 /3 5 3 /3 = 5 3 /3 (In working with these kinds of epressions, it is better to have a sum of terms when you re differentiating On the other hand, it is better to have everything together in factored form when you set up the sign charts Notice how I used these two forms in the derivatives above) 6

7 y = for = and at = 5 y is defined for all f'(-)=6 = f'()=4 =5 f'(8)=- The function increases for 5 and decreases for 5 There is a local ma at = 5, y y = at = y is undefined at = - + f"(-)=-5 = f"()=5/3 = - f"(8)=-5 The graph is concave down for < and for > The graph is concave up for < < = and = are inflection points Since the function is defined for all, there are no vertical asymptotes (3 5/3 38 ) + 8/3 = and (3 5/3 38 ) 8/3 = The graph goes downward on the far left and far right In the following problems, compute the it, or show that it is undefined sin 3 sin 5 (a) sin sin 3 sin 3 sin 5 = sin sin sin 5 sin 3 3 = 3 sin sin = 3 5 = 7

8 (b) = = = 5 3 ( + h) 5 5 (c) h h for f() = 5 Hence, by the Power Rule, ( + h) 5 5 = f (), h h ( + h) 5 5 = 5 49 h h 3 4 (d) ( + )( 4) 4 = 6 4 ( 4)( + 4) = = 5 8 (e) For close to 4 but larger than 4 eg = 4 6 is negative Since the square root of a negative number is undefined, the it is undefined (f) = ( 4)( + ) + 4 ( 4) = 4 4 Plugging in gives 5 Moreover, Hence, the it is undefined + + = + and = 4 (g) = + 3 5( + 3) 5 5( + 3) = ( + 3) 4 = 5( + 3) ( )( + ) = = 5( + 3) 8

9 5( + 3) ( )( + ) = 5( + 3)( + ) = 7 Compute the following integrals (a) (e + e 3 ) d (e + e 3 ) d = (e 4 + e 5 + e 6) d = 4 e4 + 5 e5 + 6 e6 + C (b) + 3 ( + ) 4 d ( + 3 (u ) + 3 u + ( + ) 4 d = u 4 du = u 4 du = u 3 + ) u 4 du = u = +, du = d, = u ] u 3u 3 + C = ( + ) 3( + ) 3 + C (c) e e + d e e e + d = u du e = du u = ln u + C = ln e + + C u = e +, du = e d, d = du ] e (3 ln) + (d) d (3 ln) + 3u + d = du = (3u + ) du = u 3 + u + C = (ln) 3 + ln + C u = ln, du = d ], d = du (e) cos (sin ) + sin + d cos (sin ) + sin + d = cos u + u + du cos = u = sin, du = cosd, d = du ] cos du u + u + = 9

10 du (u + ) = u + + C = sin + + C (f) d d = + du 3 3 u 6( + ) = 6 u /3 du = ] u = 3 3, du = ( 6 6 du ) d, d = 6( + ) 6 3 u/3 + C = 4 ( 3 3 ) /3 + C ( ) sec /3 (g) d /3 ( ) sec /3 (sec u) d = 3 /3 du = 3 (secu) du = 3 tanu + C = 3 tan /3 + C /3 /3 u = /3, du = ] 3 /3 d, d = 3 /3 du (h) ln( + ) + d ln( + ) + d = lnu u du = u = +, du = d, d = du ; =, u = ; =, u = lnu u du = ln w u u dw = ln w dw = ] ln w = 4 (ln ) w = lnu, dw = duu ], du = u dw; u =, w = ; u =, w = ln ] (i) f () d, if f() = and f() = e f() f () e f() d = f () du e u f () = du u = ln u ]e = u = f(), du = f () d, d = du ] f ; =, u = f() = ; =, u = f() = e ()

11 (j) ( + )4 ( ++5) d ( + )4 ( ++5) d = ( + )4 u du ( + ) = 4 u du = ] u = du + + 5, du = ( + ) d = ( + ) d, d = ( + ) ln 4 4u + C = ln4 4( ++5) + C 3 (l) d 3 d = ( ) 3 d = 5 d = ln c (m) e d e e d = e e u ( e du) = du = u sin u + C = sin e + C u = e, du = e d, d = e du ] (n) ( + ) d d = ( + ) ( + u ) du = u =, du = ] d, d = du + u du = tan u + C = tan + C 8 (a) A population of flamingo lawn ornaments grows eponentially in Calvin s yard There are after day and 6 after 4 days How many are there after 6 days? Let F be the number of flamingos after t days Then When t =, F = : When t = 4, F = 6: Divide 6 = F e 4k by = F e k and solve for k: Plug this back into = F e k : F = F e kt = F e k 6 = F e 4k 6 = F e 4k F e k, 3 = e3k, ln 3 = lne 3k = 3k, k = ln 3 3 = F e (ln 3)/3, F = e (ln3)/3

12 Hence, When t = 6, F = e (ln 3)/3 e(t ln 3)/3 F = e (ln 3)/3 e ln flamingos (b) A hot pastrami sandwich with a temperature of 5 is placed in a 7 room to cool After minutes, the temperature of the sandwich is 9 When will the temperature be 8? Let T be the temperature at time t The initial temperature is T = 5, and the room s temperature is 7, so T = 7 + (5 7)e kt or T = 7 + 8e kt When t =, T = 9: Hence, Set T = 8 and solve for t: 9 = 7 + 8e k, = 8e k, ln 4 = k, k = ln 4 T = 7 + 8e t(/) ln(/4) 8 = 7 + 8e t(/) ln(/4), = 8e t(/) ln(/4), 4 = ek, ln 4 = lnek, ln 8 = t ln ln 4, t = 8 ln 5 minutes 4 8 = et(/) ln(/4), ln 8 = lnet(/) ln(/4), 9 (a) Find the area of the region in the first quadrant bounded on the left by y =, on the right by + y =, and below by the -ais y +y= y= The curves intersect at the point =, y = (You can see this by solving y = and + y = simultaneously) Divide the region up into horizontal rectangles A typical rectangle has width dy The right end of a rectangle is on = y; the left end of a rectangle is on = y (ie y = ) Therefore, the length of a rectangle is y y, and the area of a rectangle is ( y y) dy

13 The area is A = ( y y) dy = y y 3 y3/ ] = (b) Find the area of the region between y = and y = from = to = Find the intersection point: =, + =, ( + 4)( 3) =, = 4 or = 3 = 3 is the intersection point between and 4 Use vertical rectangles Between = and = 3, the top curve is y = and the bottom curve is y = Between = 3 and = 4, the top curve is y = and the bottom curve is y = The area is 3 ( ) d+ 4 3 ( ( )) d = 3 3 ] ] = Approimate the area under y = ( sin ) from = 3 to = 5 using rectangles of equal width, and using the midpoints of each subinterval to obtain the rectangles heights The width of each rectangle is = 5 3 = is The midpoints start at 35 and go to 495 in steps of size If f() = ( sin ), the rectangle sum 9 f(35 + k) 4477 k= Approimate the sum 3 + sin sin sin sin to 5 decimal places 3 + sin sin sin sin = 3 + sin(n + ) n n n= 3

14 Calvin runs south toward Phoebe s house at feet per second Bonzo runs east away from Phoebe s house at 5 feet per second At what rate is the distance between Calvin and Bonzo changing when Calvin is 5 feet from the house and Bonzo is feet from the house? Let be the distance from Bonzo to the house, let y be the distance from Calvin to the house, and let s be the distance between Calvin and Bonzo Calvin y s Bonzo Phoebe's house By Pythagoras, s = + y Differentiate with respect to t: s ds dt = d dt + ydy dt, sds dt = d dt + ydy dt d dy = 5 and dt dt y = 5, s = 3 So = (negative, because his distance from the house is decreasing) When = and 3 ds dt = ()(5) + (5)( ), ds dt = feet per second 3 3 Find the dimensions of the rectangle with the largest possible perimeter that can be inscribed in a semicircle of radius Y X X The height of the rectangle is y; the width is The perimeter of the rectangle is p = 4 + y 4

15 By Pythagoras, + y =, so y = Therefore, p = 4 + = gives a flat rectangle lying along the diameter of the semicircle = gives a thin rectangle lying along the vertical radius dp d = 4 dp, so d = for = (The negative root does not lie in the interval ) 5 5 p 5 4 ma = 5 gives y = 5 The dimensions of the rectangle with the largest perimeter are = 4 5 and y = 5 ; the maimum perimeter is p = A window is made in the shape of a rectangle with an isosceles right triangle on top s s s h h (a) Write down an epression for the total area of the window s A = s + sh (b) Write down an epression for the perimeter of the window (that is, the length of the outside edge) p = h + s + s (c) If the perimeter is given to be 4, what value of s makes the total area a maimum? Since 4 = p = h + s + s, Hence, A = s + s ( s h = s s The etreme cases are s = and h =, which gives s = ) s = s + s s s = s s s 5 4 +

16 The derivative is Find the critical point: Plug the critical point and the endpoints into A: When s =, the area is a maimum + da ds = s s = s s, s = + s A (a) Find the absolute ma and the absolute min of y = on the interval 5 The derivative is y = 3 = 3( 4) = 3( )( + ) y = for = and = ; however, only = is in the interval 5 y is defined for all 5 y 5 7 The absolute ma is at = 5; the absolute min is at = (b) Find the absolute ma and the absolute min of f() = 3 4 4/3 5 /3 on the interval 8 f () = /3 5 /3 = 5 /3 f = for = 5 and f is undefined for = Both points are in the interval 8 5 f() The absolute ma is at = and the absolute min is at = 5 6 Use a it of a rectangle sum to find the eact area under y = + 3 from = to = Divide the interval up into n equal subintervals Each subinterval has length = n I ll evaluate the function at the right-hand endpoints of the subintervals, which are The function values are ( ) + 3 n ( ), n n, n, 3 n,, n n, n n ( ) + 3 n 6 ( ) ( n ) ( n, + 3 n n n)

17 The sum of the rectangle area is n (k ) ( ) ] k + 3 n n n = n n 3 k + 3 n k= The eact area is n k= n k= k = n(n + )(n + ) + 3 n(n + ) n3 6 n ( n(n + )(n + ) + 3 ) n(n + ) = n3 6 n = 6 7 Given that f() = , what is (f ) ()? First, f () = Then (f ) () = f (f ()) I need to find f () Suppose f () = Then f() =, so =, and = I can t solve this equation algebraically, but I can guess a solution by drawing the graph (of y = ): It looks like = is a solution Check: = Thus, f() =, so f () = Hence, (f ) () = f (f ()) = f () = 7 8 Find the largest interval containing = on which the function f() = has an inverse f The derivative is f () = 3 48 = ( 4) f () = for = and = 4 f () is defined for all Here is the sign chart for y : f'(-)=-6 = f'()=-36 =4 f'(5)=3 7

18 f decreases for 4, and this is the largest interval containing = on which f is always increasing or always decreasing Therefore, the largest interval containing = on which the function f() = has an inverse is 4 9 (a) Compute d 6 + t dt d 4 d 6 + t dt = d6 d 6 + t d 4 d d 6 dt = (6 5 ) + ( 6 ) = ( d ) 3 (b) Compute + d d ( d ) 3 + d = 3 d + + C (a) Use the definition of the derivative as a it to prove that d d 4 = ( 4) f f( + h) f() () = = h h h h *(b) Compute h h ( 4) ( + h 4) ( 4)( + h 4) h +h h t4 + dt + h 4 4 h = h = h h ( 4)( + h 4) h ( 4)( + h 4) = ( 4) 4 ( 4)( + h 4) + h 4 ( 4)( + h 4) h h = h h( 4)( + h 4) = The it as h goes to, the, and the its and +h remind me of the definition of the derivative h So I make a guess that the it in the problem is actually the derivative of a function The problem is to figure out what function is being differentiated Let f() = t4 + dt Then f(+h) f() = Therefore, h h +h +h t4 + dt t4 + dt = +h +h t4 + dt+ t4 + dt = t4 + dt f( + h) f() t4 + dt = (f( + h) f()) = = f () = h h h h d d t4 + dt = 4 + = 8

19 Let Is f continuous at =? Why or why not? + 3 if < f() = 6 9 if = 3 if > The left and right-hand its agree Therefore, + 3 f() = 6 = 4 5 = 8 f() = ) = 3 = 8 + +(3 f() = 8 However, f() = 9, so f() f() Hence, f is not continuous at = (a) Suppose that f(3) = 5 and f () = Use differentials to approimate f(99) to 5 places + 6 Use f( + d) f() + f () d d = 99 3 = and f (3) = 9 = 36 Therefore, 5 f(99) f(3) + f (3) d = 5 + (36)( ) = (b) A differentiable function satisfies dy d = e cos3 and y() = Use differentials to approimate y() d = = ; when =, dy d = e cos = Therefore, dy = dy d = ()() = d Hence, y() y() + dy = + = 3 Use 3 iterations of Newton s method starting at = to approimate a solution to 4 = e Write the equation as 4 e = Set f() = 4 e, so f () = e f() f ()

20 The solution is approimately 58 4 Prove that the function f() = 3 + cos + 5 has eactly one root First, f() = 4 > and f ( π ) = 7π 8 π + 5 < By the Intermediate Value Theorem, f has a root between π and Suppose f has more than one root Then f has at least two roots, so let a and b be two roots of f By Rolle s theorem, f has a critical point between a and b However, f () = sin Since sin, + sin But 3, so f () = sin + = > Thus, f has no critical points Therefore, it can t have more than one root It follows that f has eactly one root The best thing for being sad is to learn something - Merlyn, in T H White s The Once and Future King c 8 by Bruce Ikenaga

Review Problems for Test 3

Review Problems for Test 3 Math 6-3/6 7 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the