Calculus I-01Exam 2 – Team Portion

Write solutions completely for maximum credit. Put all work on separatepages and keep this page for your reference. Each of the four problems isworth 16 points.

1. Starting at room temperature 70 at noon, an oven is turned onat 2:30pm and set at 400 . It takes half an hour to reach 400 ,stays at 400 for one hour, and is then turned off and let cool forthe rest of the day.

a. Sketch a possible graph of T, the oven temperature in degreesFahrenheit, as a function of t, the time in hours after noon, for tbetween noon and midnight. State the values of t for which youknow T’(t) is zero, the values of t for which you know T’(t) ispositive, and the values of t for which you know T’(t) isnegative.

45

40

35

30

25

20

15

10

5

-5

2 4 6 8 10 12

Midnight4:003:002:30Noon

400°

70°

T’(t) is zero for 0 < t ≤ 2.5 for 3≤ t ≤ 4 , and for 5 ≤ t <12 . (The graphwas drawn by splicing straight line segments together with 4parabolas, with vertices at (2.5,70), (3,400), (4,400), and (5,70).)By making the curves to be parabolas with vertices where the

straight line segments meet the curves, the whole graph becomessmooth and the temperature function differentiable.

T’(t) is positive for 2.5 < t < 3 (because for these values of t ,T is anincreasing function) and T’(t) is negative for 4 < t < 5 (because forthese values of t, T is a decreasing function).

b. Suppose that someone proposes the function T (t)= 70+ 8910tt 3 + 54as a model for temperature T as a function of time t. Use rulesof differentiation in sections 2.3 - 2.4 to determine T’(t).

Using the differentiation rules for sums, constant multiples, and

quotients, T '(t)= 8910 t3 + 54 − t(3t 2 )(t 3 + 54)2 = 8910−2t

3 + 54(t 3 + 54)2 =17,820 27− t

3

(t 3 + 54)2c. Describe the ways in which the function T proposed to model

oven temperature in part b is similar to, and ways in which itdiffers from, the function you graphed in part a. Consider bothT and its derivative T’ as you make your comparisons.

T(0)=70 (corrected by L.W.) for both functions, but the functiondrawn has a slope, and hence derivative, of 0 for 0<t<2.5, whilethe function in part b has a positive derivative whenever 27− t3 > 0 ;i.e., whenever 0<t<3. So, both functions are increasing for tbetween 2.5 and 3, but only the function in part b is alsoincreasing for t between 0 and 2.5. Since the oven wasn’t turnedon until 2:30, the increasing nature of the function in part b in thisperiod of time is a defect in that function as a model. The functionin part b does reach 400 at 3:00, but is constantly decreasing fort>3. (Since the constant multiple 17,820 is positive and thedenominator of T’(t) is positive for all positive values of t, the signof T’ is determined solely by the numerator of the expression forT’.) As t approaches infinity, T (in part b) approaches roomtemperature, 70, but by midnight, T has dropped to only 130,while the function drawn in part a reaches 70 by 6pm and remainsat that room temperature.

d. Determine a value of the constant k so that the function

C(x)= kx2 − 1x has the property that C '(1)= 0 . (Use rules of

differentiation in sections 2.3 - 2.4 to determine C’(x).)

C '(x)= 2kx− ((−1)x−2 ) = 2kx+ 1x2 , so C '(1)= 2k +1. In order for C '(1) to

equal 0, k must be −12 .

2.

a. Construct (make up) a function by sketching its graph (and notgiving a formula for it) for which all of the following are true:

i. f (0)=1 ii. f '(0)= −2 iii. f '(−1)= −2 iv. f '(1)= −29

Illustrate the values of f ' in parts ii., iii., and iv. by sketchingtangent lines to the graph of f at the appropriate points.

6

4

2

-2

- 4

- 6

- 8

-10

-12

-10 - 5 5 10 15

The graph of f, shown in red, consists of a linesegment with slope -2 for -2<x≤0, a curve whosetangent line at the point (0,1) on the graph containsthe line segment portion of the graph, and a curvethat has a slope of -2/9 at the point on the graphwhere x=1.

The two tangent lines that just "touch" the graphare shown by dashed lines; the tangent line wherex=-1 is the same as the tangent line where x=0.

b. Apply the rules of differentiation in sections 2.3 and 2.4 todetermine the derivatives of:

i. sinxx ii. 2x2 cos(x)−1

i. Applying the quotient rule, ddx(sin xx ) = xcosx − sin xx2 .

ii. Applying the rule for differences, then the rule for constantmultiples, then the rule for products, and, finally the powerrule and rule for cosine:

ddx(2x2 cos(x)−1) = d

dx(2x2 cos(x))− d

dx(1) = 2 d

dx(x2 cos(x))− 0

= 2[x2 ddx(cos(x))+ cos(x) d

dx(x2 )]= 2[x2(−sin(x))+ cos(x)(2x)]

c. Use the definition of derivative to show that ddx (1

2x +1)=−2

(2x +1)2 .

ddx (

12x+1)= limh→0

( 12[x + h]+1

) − ( 12x +1

)

h= limh→0

2x +1− (2[x + h]+1)h[2[x + h]+1][2x +1]

= limh→0

−2hh[2[x + h]+1][2x +1]

= limh→0

−2[2[x + h]+1][2x +1]

=−2

[2x +1][2x +1]=

−2[2x +1]2

d. Compare and contrast the function g defined by g(x) = 12x +1

with the function you graphed in part a of this problem.Calculate a formula for g ''(x) , and describe what g ''(x) tellsyou about the function g and its derivative.

The function g satisfies g(0)=1, just as the function graphed in

part a is required to do. But g is not defined for x = −12

, while the

function I graphed in part a has the value 2 when x = −12

. The

function g has g’(0)=g’(-1)=-2 and g '(1) = −2[2 +1]2

=−29

, the same as the

function in part a. The function drawn in part a is deceasing for xless than a number that is a little larger than 1 (The number wasby construction 9/8, but that isn’t made clear on the graph.) andincreasing for x greater than that specific number. This is incontrast to the function g whose derivative is always negative,

except where it is undefined, at x = −12

. Thus g is everywhere

decreasing, except at x = −12

. Using the quotient rule,

g"(x) = (g ')'(x) =[2x +1]2 ( d

dx[−2]) − (−2) d

dx[2x +1]2

[2x +1]4=

2 ddx[4x2 + 4x +1]

[2x +1]4=2[8x + 4][2x +1]4

=8[2x +1][2x +1]4

=8

[2x +1]3

The form of g’’ shows that g’’ is negative for x < −12

and positive for

x > −12

. What g’’ tells us about g’ is that g’ is decreasing for x < −12

and increasing for x > −12

. This contrasts with the function graphed

in part a, whose derivative is increasing everywhere.3.

a. Assume that the function r(t) represents the rate at which theworld’s oil is consumed, where t is measured in years startingat t=0 on Jan 1, 2000, and r(t) is the number of barrelsconsumed per year at time t. If r(t) is approximated by:r(t)=109(2t2 − t +1), make an estimate of the number of barrels

of oil consumed in the 5-year period 2000-2004, r(t)dt0

5

∫ , by

using the Evaluation Theorem of section 5.3 to determine theexact value of this integral.

Using the basic properties of integrals,

r(t)dt0

5

∫ = 109(2t2 − t +1)dt0

5

∫ =109[ (2t2dt − tdt0

5

∫ + 1dt0

5

∫0

5

∫ ]

=109[2 t2 − tdt0

5

∫ + 1dt0

5

∫0

5

∫ ]

Then, using the power rule for integrals (and geometry for thethird integral),

r(t)dt

0

5

∫ =109[2 t2 − tdt0

5

∫ + 1dt0

5

∫0

5

∫ ] ==109[2 t3

3|05 −

t 2

2|05 +5]

=109[21253− 252

+ 5] =109[500 − 75 + 306

]=109[4556]

Which is around 76 billion barrels consumed in the 5-year period.

b. Evaluate the integral [sin(x)]dxπ 2

5π4

∫ by using the Evaluation

Theorem and interpret the result as a difference in areas.Illustrate with a sketch.

Since the derivative of the cosine function is the negative of thesine function, the negative of the cosine function has the sinefunction for its derivative, and the Evaluation Theorem implies:

[sin(x)]dxπ 2

5π4

∫ = (−cos(x)) |π 25π4 = −cos(5π4 )− (−cos(

π2)) =

22

+ 0 = 22

6

4

2

-2

- 4

- 6

-10 - 5 5 101π/2 π

5π/4

The integral is the difference between the area of the shadedregion to the left of x =π and the area of the shaded region to theright of x =π . (By the symmetry of the sine function, thisdifference in areas is the same as the area under the sine curve,

above the x-axis, to the right of x = π2

and to the left of x = 3π4

.)

c. Write the Riemann sum for h(t)= 2t , 1≤ t ≤ 4 , with 6 equal

subintervals, taking the sample points to be midpoints. Explain,with a diagram, what the Riemann sum represents.

The interval [1,4] is partitioned into 6 equal-width subintervals:

[1,1.5], [1.5,2], [2,2.5], [2.5,3], [3,3.5], [3.5,4] and the midpoints ofeach subinterval: 1.25, 1.75, 2.25, 2.75, 3.25, 3.75 are chosen assample points. The sample points determine the height of the 6

rectangles: 254=85

, 274=87

, 294=89

, 2114=811

, 2134=813

, and 2154=815

as

shown below:2

1.5

1

0.5

-0.5

- 1

-1.5

- 4 - 2 2 4 6

y=2

x

x[6]* = 3.75

x[5]* = 3.25x[4]* = 2.75

x[3]* = 2.25

x[2]* = 1.75

x[1]* = 1.25

The Riemann sum is then the sum of the areas of the rectangles:

(8 5)(12) + (8 7)(

12) + (8 9)(

12) + (811)(

12) + (813)(

12) + (815)(

12) , or

4[(15) + (17) + (

19) + (

111) + (

113) + (

115)] = 4

12i + 3i=1

6

∑d. Use the Theorem 4 of section 5.2 to express the exact value of

the integral in part a as a limit, as n approaches infinity, of aRiemann sum. You need not simplify or evaluate this limit.

Proceeding as in part c, except with n subintervals instead of 6,and right endpoints instead of midpoints for sample points, the

Riemann sum is ( 21+ 3n

+2

1+ 6n

+2

1+ 9n

+ ...+ 21+ 3in

+ ...+ 21+ 3nn

)( 3n) = 2

1+ 3in

i=1

n

∑ ( 3n) .

The integral in part c (Oops, this problem was supposed to beabout part a, not part c!!) is the limit, as n approaches infinity, of

this Riemann sum: limn→∞2

1+ 3in

i=1

n

∑ ( 3n) .

As for the integral in part a: The interval [0,5] is divided into n

equal subintervals, each 5n

in width and sample points are chosen

to be the right endpoints, 5n

, 10n

, 15n

, ..., 5in

, ..., 5nn

. The function

r(t)=109(2t2 − t +1) evaluated at those endpoints determinesthe “height of the rectangles” in the Riemann sum:

109(2(5in)2 − 5i

n+1)

i=1

n

∑ (5n), and the limit, as n approaches infinity, of this

Riemann sum: limn→∞ 109(2(5in)2 − 5i

n+1)

i=1

n

∑ (5n) , is the exact value of the

number of barrels of oil consumed in the 5-year period.

4.

a. Evaluate the integral (x+ | x |)dx−1

3

∫ by using basic properties of

the integral and geometry. Illustrate your use of geometry witha diagram.

Since |x|=-x for negative values of x and |x|=x for positive values

of x, (x+ | x |)dx−1

3

∫ = (x+ | x |)dx0

3

∫ = 2xdx0

3

∫ = 2 xdx0

3

∫ , which is twice the

area of the triangle drawn below:

6

4

2

-2

- 4

- 6

-10 - 5 5 10

So, (x+ | x |)dx−1

3

∫ is the number 9.

b. Use the Evaluation Theorem to evaluate:

i. 2 x + 3x3x dx

1

4

∫ ii. [1− cos(x)]dxπ 2

π

∫

i. 2 x + 3x3x dx

1

4

∫ =2 xx dx+ 3x3

x dx1

4

∫1

4

∫ =2xdx+ 3x2dx

1

4

∫1

4

∫

= 2 x−12dx+ 3 x2dx

1

4

∫1

4

∫ = 2 x12

12|14 +x3 |14= 8 − 4 + 64 −1 = 67

ii. [1− cos(x)]dx

π2

π

∫ = 1dxπ2

π

∫ − cos(x)dxπ2

π

∫ = x |π 2π − sin(x) |π 2

π = π − π 2 − (0 −1)

= π2 +1

c. Express the sum 1− 32 + 5

3 −74 + 9

5 − 116 + ... + 141

71 using the sigma

notation. You need not evaluate the sum.

The sum alternates, which is accomplished by inserting a factor(−1)i+1; its numerator increases by 2 in each successive term,which is accomplished by inserting the factor 2i−1; and itsdenominator increases by 1 in each successive term, which is

accomplished by inserting the factor 1i . Putting the three pieces

together, (−1)i+1(2i−1i )i=1

71∑ expresses the sum using the sigma

notation.d. Suppose that Si(x) is a function whose derivative is the function

f defined by f (x)= sinxx when x is not zero and f (0)=1. If

Si(3)=A and Si(1)=B, what is sinxx dx1

3

∫ in terms of A and B?

Since Si’(x)=f(x), sinxx dx1

3

∫ = Si(x) |13= Si(3)− Si(1)= A− B by the

Evaluation Theorem.