This is a 60 in 60 Challenge: 60 questions in 60 minutes

THIS IS A 60 IN 60 CHALLENGE:60 QUESTIONS IN 60 MINUTES

• Some are designated as Calc active.

• The expectation is that you’ll be able to get a handle on the solution within a minute, not actually do the question.

• The 60 Focused Questions should create a wealth of other questions. Some of the questions are answered in the next slide.

#1 SOLUTION

2ln2)2/4ln(2

)2ln4(ln)24(

dxdxxx

#2 Find dy/dx for:

yyxyx 23 2

#2 SOLUTION

xyxydxdy

dxdyyy

dxdyxx

Doesn’t this allow you to consider all limit questions a) at a point or b) at the infinities as well as any Indeterminate forms (of which this is one)

x−−>0lim x 3 − 2x 2 + x

#3 SOLUTION Use L’Hopital – answer is 1/2

#4 LET’S TRY IMPLICIT DIFF. AGAIN !!!

The above is non-functional. Could you find

dy/dx ? dx/dy? The value(s) of dy/dx @ y=1?

The name of any horizontal/vertical tangents, if they exist?

y 2x + 3y = 2x 2

#4 SOLUTION dy/dx=(4x – y2)/(2xy – 3)

dy/dx @ y = 1 would require finding the value(s)

for x. When y = 1, then x = 3/2 or -1. Note that for (3/2, 1) the value of dy/dx is undefined (vertical tangency) and for (-1, 1) the value of dy/dx is 1.

4 x[ ]

There is no mystery to the question and the key to its solution is to use the Change of Base Formula. Never forget that all derivatives can be checked in some way or another.

#5 SOLUTION

4 x (ln 4)

#6 (C ACTIVE) Euler and the linear approximation

model go hand in hand……. Find the approximated value of f(4) for

f(x) = ln(2x+1), starting @ x = 1 w/ a step size of 1.5

#6 SOLUTION f(x + dx) ≈ f(x) + f’ (x) dx f(1 + 1.5) ≈ f(1) + f’ (1) (1.5) f(1 + 1.5) ≈ f(1) + f’ (1) (1.5) f(2.5) ≈ ln(3) + (2/3)(1.5) f(2.5) ≈ ln(3) + 1

f(2.5 + 1.5) ≈ f(2.5) + f ‘ (2.5)(1.5) f( 4 ) ≈ [ ln 3 + 1] + (2/6)(1.5) f( 4 ) ≈ [ln 3 + 1] + ½ f( 4 ) ≈ ln 3 + 1.5

#7 Given a graph of the derivative of a

function and an initial condition, then sketch the graph of the function…………

#7 SOLUTION

f ‘ (x)

#8 (C ACTIVE) Calculate the area of the region formed

by the function , the x-axis, the y-axis, and the line x = 2π.

f (x) = esin x − 3cos x

#8 SOLUTION

∫ dx ≈ 13.009

#9 The function g(x) =

Find g(4) Find g ‘ (4)€

2t− 3t 2 − 3t + 5

∫ dt

#9 SOLUTION g(4) = ln 9 – ln 5 by u-substitution

g ‘ (4) =

2• 4 − 34 2 − 3• 4 + 5

#10 If f(x)= then which of the following is

FALSE ?

A) f(0) = 0 B) f is continuous at x for all x ≥ 0. C) f(1) > 0 D) f ‘ (1) =

E) f (-1) > 0

1t 3 + 20

∫ dt

#10 SOLUTON Choice (E)

f (x) = 1

t 3 + 20

∫ dt, f (−1) = 1

t 3 + 20

∫ dt = f (x) = − 1

t 3 + 2−1

∫ dt < 0

f (−1) < 0. So (E) is false

A question that demands your understanding of the definition of the derivative. Of course, L’Hopital’s Rule would also be appropriate !

h−−>0lim

+ h ⎛ ⎝ ⎜

⎞ ⎠ ⎟8

− 8 12 ⎛ ⎝ ⎜

⎞ ⎠ ⎟8

#11 SOLUTION f(x) = 8x8 This limit is the definition of the

derivative of the above f(x) function evaluated @ x = ½

The answer is f ‘(1/2) = 64 (1/2) 7 = 1/2

If h(x) = f 2 (x)− g2 (x), f ' (x) = −g(x), & g ' (x) = f (x),

then h ' (x) = ______________________________

#12 SOLUTION If Then

h(x) = f 2 (x)− g2 (x), f ' (x) = −g(x), & g ' (x) = f (x),

h'(x) = 2 f (x)• f ' (x)− 2g(x)• g' (x)h'(x) = 2g' (x)• −g(x)[ ] − 2g(x)• g' (x)

h'(x) = −4g(x)• g' (x) ORh'(x) = −4 f (x)• g(x)

#13 Let’s see what you learned from the last

question………

If ddx

f (x)( ) = g(x) & ddx

g(x)( ) = f (x 2 ),

then ddx

f (x)• g(x)[ ] = ______________________________

#13 SOLUTION

If ddx

f (x)( ) = g(x) & ddx

g(x)( ) = f (x 2 ),

then ddx

f (x)• g(x)[ ] = f (x)• ddx

g(x)[ ] + g(x) f ' (x)

then ddx

f (x)• g(x)[ ] = f (x)• f (x 2 ) + f ' (x) f (x 2 )

= f (x 2 )• f (x) + f (x 2 )[ ]

#14 (C ACTIVE) What is the volume of the solid formed when

the region bounded by y = ex, y = ln(1/x), x = .1, & y = 0 is rotated about the x-axis?

Sketch the region and store the intersection point. This will require two integrations. Be efficient with your use of FnInt.

#14 SOLUTION

π • ex( )2

.26987

∫ dx + π • ln(1/ x)( )2

.26987

∫ dx ≈ 1.688

NOTE: ln(1/x) is easier to “see” and work with by simplifyingto ln (x-1) which is equivalent to - ln(x)

NOTE: A boundary line of x = .1 is necessary because if the y-axis is used (w/out the x-axis boundary), then animproper integral results !!

#15 This looks harmless………

cos xsin x

π / 4

π / 2

∫ dx

#15 SOLUTION (ln2)/2

#16 What is the average value of the

function 3t3 – t2 over the interval -1 ≤ t ≤ 2 ?

#16 SOLUTION 2.75

#17 A region in the plane is bounded by the graph of

y = 1/x, the x-axis, the line x = m, and the line x = 2m, m > 0. The area of this region

A) is independent of m B) increases as m increases C) decreases as m increases D) decreases as m increases when m < ½;

increases as m increases when m > ½ E) increases as m increases when m < ½;

decreases as m increases when m > ½

#17 SOLUTION A)

∫ dx = ln x |m2m = ln(2m)− ln(m) = ln 2

The 4th degree Taylor polynomial for f about

x = 1 is given by T(x) = 5 - 10(x -1) + 15(x – 1)2 – 20(x – 1)3 + 25(x – 1)4

Calculate f(1) + f ‘ (1) + f iv (1)

#18 SOLUTION T(x) = 5 - 10(x -1) + 15(x – 1)2 – 20(x –

1)3 + 25(x – 1)4

f(1) = 5 f ‘ (1) = -10 f ‘’ (1) = 30 f ‘’’ (1) = -120 f iv (1) = 600 So, f(1) + f ‘ (1) + f iv (1) = 595

#19 Create the 3rd degree Maclaurin

polynomial for f(x) = ln(1 – x) Use this polynomial to approximate

#19 SOLUTION From previous experience (or memorization) we know

ln (1 + x) =

So, ln (1 – x) =

And ln (1 – (-2) ) = ln (3) ≈

NOTE: ln 3 ≈1.0986, so the 3rd degree polynomial is not a great approximation

x − x 2

2+ x 3

3− x 4

4+....(−1)n−1 x n

−x − x 2

2− x 3

3− x 4

4−....(−1) x

−(−2)− (−2)2

2− (−2)3

3≈ 8

Which of the following series converge?

A) I only B) II only C) III only D) II and III E) I and II€

I. 2n 2 +1n=0

∑ II . 2n −13n +1n=1

∑ III . n4

#20 SOLUTION Choice E Item I may be broken up into to summations;

one from 0 to 1 which is a finite series that must converge and the other from 1 to ∞ which when compared to the p-series with p = 2 must converge.

Item II closely resembles a geometric series with r = 2/3. Convergence must occur

Item III is a p-series with p = ¾ < 1, so it diverges.

#21 Use the definition of the derivative

to find the instantaneous

rate of change for f(x) = sin(2x) €

f ' (x) =h−−>0lim

f (x + h)− f (x)h

#21 SOLUTION

f ' (x) =h−−>0lim

f (x + h)− f (x)h

with f (x) = sin(2x)

f ' (x) =h−−>0lim

sin(2(x + h))−sin(2x)h

f ' (x) =h−−>0lim

sin(2x + 2h)−sin(2x)h

f ' (x) =h−−>0lim

sin(2x)• cos(2h) + cos(2x)• sin(2h)−sin(2x)h

f ' (x) =h−−>0lim

sin(2x) cos(2h)−1[ ]h

+h−−>0lim

cos(2x)• sin(2h)h

f ' (x) =h−−>0lim sin(2x)•

h−−>0lim

cos(2h)−1[ ]h

+h−−>0lim cos(2x)•

h−−>0lim

sin(2h)h

f ' (x) = sin(2x) • 0 + cos(2x) • 2f ' (x) = 2cos(2x)

#22 Which of the following is an equation of

the normal line to y = sin x + cos x at x = π ?

A) y = - x +π – 1 B) y = x – π + 1 C) y = x – π – 1 D) y = x + π + 1 E) y = x + π - 1

#23 Which of the following gives dy/dx for

the parametric curve y = 3 sin t, x = 2 cos t ?

A) -3/2 cot t B) 3/2 cot t C) -2/3 tan t D) 2/3 tan t E) tan t

Choice A

#24 Which of the following gives dy/dx if

y = sin -1 (2x) ?

A) −2

1− 4 x 2 B) −1

1− 4 x 2 C ) 2

1− 4 x 2

1− 4 x 2 E) 2x

1+ 4 x 2

#24 SOLUTION Choice C Consider all the inverse trig derivatives, including tan -1 (f (x) )

and the associated antiderivatives that are sure to come up such as:

636+ 4x 2∫ dx (÷ each term by 36)

11+ x 2 /9∫ dx

− − − > Let u2 = x 2

9− − − − > u = x

du = (1/ 3)dx 3du = dx

AND 16

11+ x 2 /9∫ dx = 1

1+ u2∫ = 12

tan−1 u +C

tan−1 x3 ⎛ ⎝ ⎜

⎞ ⎠ ⎟ +C

#25 A slope field for the differential

equation dy/dx = 42 – y will show

A) A line with slope -1 and y-intercept of 42

B) A vertical asymptote at x = 42C) A horizontal asymptote at y = 42D) A family of parabolas opening

downwardE) A family of parabolas opening to the

#25 SOLUTION A family of exponential functions result

that by transformational geometry result in horizontal asymptotes at y = 42, regardless of the constant C that you use in the general solution of the differential equation.

So, choice C

#26 If

A) 7 + a B) 7 C) 7 – a D) a – 7 E) -7€

f (x− a) dx3

5∫ = 7, then f (x) dx

5−a∫ =

#26 SOLUTION If you shift the limits of integration a

units to the left (5 to 5 – a and 3 to 3 – a), then that will compensate for the integrand’s shift from f(x – a) to f(x) which also would have been a units to the left…….CHOICE B

A good numerical example might help:

(x −1) dx3

∫ = (x) dx2

#27 The graph of y = f(x) conforms to the slope field for

the differential equation dy/dx = 4x ln x, as shown below. Which of the following could be f(x)?

A) 2x2(ln x)2 + 3 B) x3ln x + 3 C) 2x2ln x – x2 + 3 D) (2x2 + 3)ln x – 1 E) 2x(ln x)2 – (4/3)(ln x)3 + 3

[0,2] by [0,5]

#27 SOLUTION Choice C done through Integration by

Parts Obviously, this is tied to the initial

condition which is imbedded within the window of [0,2] by [0,5]

You need the general solution after IBP and “window” observation to see this solution !!!

#28 A GOOD SUBSTITUTION PROBLEM

If the substitution is made in the integrand of , the resulting integral is

x = sin y

x1− x0

1/ 2∫ dx

A) sin 2 y dy0

1/ 2∫ B) 2 sin 2 y

cosy dy

1/ 2∫

C ) 2 sin 2 y dy0

π / 4∫ D) sin 2 y dy

π / 4∫

E) 2 sin 2 y dy0

π / 6∫

#28 SOLUTION (THIS “TRIG SUB” IS BEYOND WHAT YOU NEED TO STUDY, BUT AN INTERESTING BY PRODUCT OF “U-SUB” WORK ! )

is to be converted to a “y” integral

Let , so

x1− x0

1/ 2∫ dx

x = sin y

y = sin−1 x anddydx

1− x( )2

• 12 x

dy = 11− x

• dx2 x

−− > 2 x • dy = dx1− x

= 2sin y • dy

x1− x0

1/ 2∫ dx

2sin y • sin y • dy0

π / 4∫

#29 (C ACTIVE) Let v(t) be the velocity, in feet per second,

of a skydiver at time t seconds, t ≥ 0. After her parachute opens, her velocity satisfies the differential equation dv/dt = -2(v + 17), with the initial condition v(0) = -47.

It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?

#29 SOLUTION 1.15

#30 (C ACTIVE) Pollution is being removed from a lake at a

rate modeled by the function y = 20e-0.5t

tons/yr, where t is the number of years since 1995. Estimate the amount of pollution removed from the lake between 1995 and 2005. Round your answer to the nearest ton.

A) 40 B) 47 C) 56 D) 61 E) 71

#30 SOLUTION Choice A) 40

#31 Let f and g be the functions given by

f(x) = ex and g(x) = 1/x. Which of the following gives the area of the region enclosed by the graphs of f and g between x = 1 and x = 2?

A) e2 – e – ln 2 B) ln 2 - e2 + e C) e2 – ½ D) e2 - e – ½ E) 1/e – ln 2

#31 SOLUTION

A simple setup once you know that x = 1 is the right of the point of intersection of the two curves

CHOICE A

#32 (C ACTIVE) Let R be the region enclosed by the

graphs of y = e –x, y = ex , and the line x = 1. Which of the following gives the volume of the solid generated when R is revolved about the x-axis?

A) (ex − e− x

1∫ ) dx B) (e2x − e−2x

1∫ ) dx C ) (ex − e− x

1∫ )2 dx

D) π (e2x − e−2x

1∫ ) dx E) π (ex − e− x

1∫ )2 dx

#32 SOLUTION This question, while calculator active, is

calculator neutral. CHOICE D

#33 (C ACTIVE) The base of a solid S is the region enclosed

by the graph of y = ln x, the line x = e, and the x-axis. If the cross sections of S perpendicular to the x-axis are squares, which of the following gives the best approximation of the volume of S?

A) 0.718 B) 1.718 C) 2.718 D) 3.171 E) 7.338

#33 SOLUTION

ln x( )2

∫ dx

CHOICE A

#34 (C ACTIVE) A developing country consumes oil at a

rate given by r(t) = 20e 0.2t million barrels per year, where t is the time measured in years, for 0 ≤ t ≤ 10. Which of the following expressions gives the amount of oil consumed by the country during the time interval 0 ≤ t ≤ 10.

A) r(t) B) r(10) − r(0) C ) r ' (t) dt0

D) r (t) dt0

10∫ E) 10• r(10)

#34 SOLUTION CHOICE D

#35 (C ACTIVE) Which of the following gives the length

of the path described by the parametric equations x = (1/4)t4 and y = t3, where 0 ≤ t ≤ 2 ?

A) t 6 + 9t 4 dt B) 0

2∫ t 6 +1 dt C ) 1+ 9t 4 dt

D) t 6 + 9t 4 dt E) 0

2∫ t 3 + 3t 2 dt

#35 SOLUTION CHOICE D – Length is

dxdt ⎛ ⎝ ⎜ ⎞

⎠ ⎟2

+ dydt ⎛ ⎝ ⎜ ⎞

⎠ ⎟2

∫ dt

#36 (BC) Which of the following is the limit of the

sequence with nth term

an = nsin 3πn

⎛ ⎝ ⎜

⎞ ⎠ ⎟ ?

A) 1 B) π C ) 2π D) 3π E) 4π

#36 SOLUTION

n−>∞lim an =

n−>∞lim n sin 3π

n ⎛ ⎝ ⎜

⎞ ⎠ ⎟

=n−>∞lim

sin 3πn

⎛ ⎝ ⎜ ⎞

⎠ ⎟

(apply L'Hopital b /c this is 0 /0)

= n−>∞lim

3π • −1n 2 • cos 3π

n ⎛ ⎝ ⎜ ⎞

⎠ ⎟

−1n 2

=n−>∞lim

3π • cos 3πn

⎛ ⎝ ⎜ ⎞

⎠ ⎟

1= 3π

CHOICE D

#37 Which of the following gives the value of

x−−>2lim

sin t dt2

x 2 − 4 ?

A) −sin 24

B) sin 24

C ) − sin 22

D) sin 22

E) DNE

#37 SOLUTION Choice B; This is a L’Hopital Rule

example

#38 Which of the following gives the area

under the curve y = 1/(x2 + 1) in the 1st quadrant ?

A) π4

B) 1 C ) π2

D) π E) diverges

#38 SOLUTION Choice C

#39 Which of the following is the coefficient

of x4 in the Maclaurin series generated by cos(3x) ?

A) 27/8 B) 9 C) 1/24 D) 0 E) -27/8

#39 SOLUTION Choice A

#40 Which of the following is the

fourth order Taylor polynomial generated by f(x) at x = π/2 ?

A) (x− π /2)− (x− π /2)2

2!+ (x− π /2)4

B) 1+ (x− π /2)2

2!+ (x− π /2)4

C ) 1− (x− π /2)2

2!+ (x− π /2)4

4!D) 1− (x − π /2)2 + (x− π /2)4

E) 1+ (x− π /2)2 + (x− π /2)4

#41 Which of the following is the sum of the

series

e2nn=0

∞∑ ?

A) ee− π

B) ππ − e

C ) ππ − e2

e2 − π E) The series diverges.

#41 SOLUTION The sum of an infinite series will only be

asked if the series is geometric or directly related to one of the basic series. . In this case the ratio is easy to see >>>>>

Since r = π/e2 and r < 1 , then €

e2n = πe2

⎛ ⎝ ⎜

⎞ ⎠ ⎟n

n−−∞lim Sn = a

1− r=

e(2•0)

1− πe2

e2 − π ; choice D

#42 (C ACTIVE) Which of the following is equal to the

area of the region inside the curve r = 2 cos θ and outside the curve r = cos θ ?

A) 3 cos2

π / 2∫ θ dθ B) 3 cos2

π∫ θ dθ C ) 3

2 cos2

π / 2∫ θ dθ

D) 3 cos θ0

π / 2∫ dθ E) 3 cos

π∫ θ dθ

#43 For -1 < x < 1 if f(x) =

(−1)n+1 x 2n−1

2n−1n=1

∞∑ , then f ' (x) =

A) (−1)n+1 x 2n−2

∞∑ B) (−1)n x 2n−2

∞∑

C ) (−1)2n x 2n

∞∑ D) (−1)n x 2n

∞∑

E) (−1)n+1 x 2n

∞∑

#44 If n is a positive integer, then

n−−>∞lim

1n ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

+ 2n ⎛ ⎝ ⎜

⎞ ⎠ ⎟2

+.... + 3nn ⎛ ⎝ ⎜

⎞ ⎠ ⎟2 ⎡

⎣ ⎢ ⎢

⎦ ⎥ ⎥ can be expressed as

A) 1x 2

1∫ dx B) 3 1

1∫ dx C ) 1

3∫ dx

D) x 2

3∫ dx E) 3 x 2

3∫ dx

#44 SOLUTION Choice D

#45 For x > 0,

∫ ⎛

⎝ ⎜ ⎜

⎠ ⎟ ⎟∫ dx =

A) 1x 3 +C B) 8

x 4 − 2x 2 +C C ) ln (ln x) + C

D) ln x 2( )

2+C E)

ln x( )2

#45 SOLUTION A u-substitution example………

∫ ⎛ ⎝ ⎜

⎞ ⎠ ⎟∫ dx =

1x ⎡ ⎣ ⎢

⎤ ⎦ ⎥• ln u∫

1x ⎡ ⎣ ⎢

⎤ ⎦ ⎥∫ • ln x dx =

u du∫ = u2

ln x( )2

#46 Which of the following is the slope of

the inverse of y = ln (x – 3) at x = 2 ?

A) ½ B) 1/e2 C) -1

D) e2 – 3 E) e2

#46 SOLUTION If y = ln (x – 3) , then f -1 (x): x = ln (y – 3) y = ex + 3 Therefore dy/dx = ex and @ x = 2 dy/dx = e2 choice E

You’ll have more difficult inverse derivative questions than this one. A strong suggestion is to graph, when possible, both the given function and its inverse.

#47 Which of the following gives y “ for

y = cos x + tan x ?

A) – cos x + 2 sec 2 x tan xB) cos x + 2 sec 2 x tan xC) - sin x + sec 2 xD) - cos x + sec 2 x tan xE) cos x + sec 2 x tan x

#48 If f is a function such that f ‘ (x) exists

for all x and f(x) > 0 for all x, which of the following is NOT necessarily true?

A) f (x)dx > 0−1

B) 2 f (x)dx = 2 f (x)dx −1

∫−1

C ) f (x)dx = 2 f (x)dx0

∫−1

D) f (x)dx = − f (x)dx1

∫−1

E) f (x)dx = f (x)dx + f (x)dx0

∫−1

#49 What is the volume of the solid

generated by rotating about the x-axis the region enclosed by the curve y = sec x and the lines x = 0, y = 0, and x = π/3 ?

A) π3

B) π C ) π 3

D) 8π3

E) π ln 12

+ 3 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

#50 (FINALLY !!!) (C ACTIVE) A rectangle is inscribed between the parabolas y = 4x2

and y = 30 – x2 as shown.

What is the maximum area of such a rectangle ?

A) 20 2 B) 40 C ) 30 2

D) 50 E) 40 2

[-3,3] by [-2, 40]

#50 SOLUTION A(x) = 2x(30 – x2 -4x2 ) A ‘ (x) = 60 – 30x2 A ‘ (x) = 0 @ x = A “( ) < 0, so a maximum occurs Maximum Area is Choice E€

#51 If f is a continuous function for all x, the

maximum # of horizontal asymptotes that f can have is

A) 0 B) 1 C) 2 D) 3E) there is no maximum #

51 SOLUTION Choice C; there can be a horizontal

asymptote at -∞ and a different one at +∞

x−−>π / 4lim

tan(5x) −1x −π /4

A) 0 B) 2 C) 5/2D) 10 E) d.n.e.

#53 If f is continuous on [2,5] and differentiable

on (2,5) with f(2) = -4 and f(5) =14 which is true?

I. f(x) = 6 has a solution in (2, 5) II. f ‘ (x) = 6 has a solution in (2, 5) III. f ‘’ (x) = 6 has a solution in (2, 5)

A) I only B) II only C) I & II only D) I & III only E) I, II, & III

#53 SOLUTION Both the Intermediate Value Theorem

and the Mean Value Theorem are used in the problem

Choice C

#54 If w = 2x, then

f (2x) dx0

A) f (w) dw0

∫ B) 12

f (w) dw0

∫ C) 12

f (w) dw0

D) f (w) dw0

∫ E ) 2 f ' (w) dw0

#55 If g(x) = use the table below to

compute g ‘ (3).

A) -4 B) 5 C) 10 D) 11 E) 14

f (t)dt0

X 0 3 6f(x) 1 5 7f ‘ (x) 9 11 -4

#55 SOLUTION Choice E

#56 Let F(x) = Then F “ (1) =

A) -3/4 B) -1/4 C) 0 D) ½ E) 3/4

1t 3 +11

∫ dt.

#57 If

A) -10 B) -2 C) -1D) 2 E) 10

f (x)dx = 6 & 0

∫ f (x)dx = 4 0

then f (x)dx = 3

#57 SOLUTION Choice B

#58 Let g(x) =

A) What is the domain of g ? B) Find the interval(s) on which g is

increasing. C) What is g “ (0) ?

t 3 + 640

∫ dt

#58 SOLUTION A) The domain is must include all values of x

such that the antiderivative exists. Therefore this will include all reals. [NOTE: If the upper limit had been just x, then the domain is restricted to be x ≥ -4]

B) > 0 for x > 0. Therefore, g(x) increases for x > 0

C) g” (0) = 0 ; NOTE: Without actually find the second derivative you should sense that each factor of g ‘(x) will include a factor of x in the 2nd derivative.

g'(x) = 2x x 2( )3+ 64

#59 A solid has base given by the triangle with vertices (-

4,0), (0, 8), and (4, 0). Cross sections perpendicular to the y-axis are semi-circles with diameter in the plane. The volume of the solid is given by

A) π 8 − 2x( )2

∫ dx B) π2

2x+ 8( )2

∫ dx+ 8 − 2x( )2

∫ dx ⎛ ⎝ ⎜

⎞ ⎠ ⎟

C) π8

8 − y( )2

∫ dy D) π4

8 − y( )2

∫ dy E) π2

8 − y( )2

∫ dy

#59 SOLUTION Choice D The lines forming the triangle are y =

2x + 8 and y = -2x + 8, and y = 0. Since the diameter of the semicircles are perpendicular to the y-axis, then you need to represent the distance in terms of y………each diameter is

…….remember radius is squared, not diameter !!!!

y − 8−2

⎛ ⎝ ⎜ ⎞

⎠ ⎟− y − 82

⎛ ⎝ ⎜ ⎞

⎠ ⎟= 8 − y

V = π 8 − y2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

#60 Circle City has a population density of

p(r) = where r is the distance in miles from the center of the city, and p (r) is in thousands of people/square mile. Which of the following gives the total population of Circle City?

4 + r 2 for 0 ≤ r ≤ 3,

A) 2πr 4 + r 2dr B) 0

∫ 2πr 4 + r 2dr C) π 4 + r 2dr −3

∫ −3

D) 2πr 4 + r 2 dr E 4 + r 2dr −3

#60 p ( r ) times square miles will give total

population For a small amount of area (dA) we still

multiply by the population density function p ( r )

Since A = , then dA = 2πr Consequently, Total Pop. = the

accumulation from 0 to 3 miles away from the center of the city of p ( r ) times 2πr…………Choice A

This is a 60 in 60 Challenge: 60 questions in 60 minutes

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