Math 1126 Final Exam Review Solutions Spring 2010

1. If F (x) =∫ x

0

√(1 + t2)(1− t) dt, then F (x) has a local minimum value at the following x-

values:

(i). − 1 (ii). 0 (iii). 1 (iv). 2 (v). None of these

Solution: (v) None of these.

The Fundamental Theorem Part 1 gives F ′(x) :

F ′(x) =√

(1 + x2)(1− x).

F ′(x) is zero when x = 1. In order to classify this critical point, we find the secondderivative:

F ′′(x) =x(1− x)√

(1 + x2)−√

1 + x2

F ′′(1) = −√

2 < 0.

Thus, x = 1 is the location of a local maximum.

2. True/False.

(a)d

dx

∫ x2

4ln(t+ 1) dt = lnx2 + 1

Solution: False. By the Fundamental Theorem of Calculus Part 1 and the chain rule,we have that

ddx

∫ x2

4ln(t+ 1) dt = 2x ln(x2 + 1).

(b) If a particle travels in a straight line with velocity v(t) = t2 + 1, then its displacementfrom t = 0 to t = 3 is the same as its total distance traveled from t = 0 to t = 3.

Solution: True. We have that v(t) = |v(t)|, since t+ 21 > 0. Therefore, the displace-ment, given by

∫ 30 v(t) dt is the same as the speed, which is given by

∫ 30 |v(t)| dt.

(c)∫ 4

04x− x3 dx represents the area between the curve y = 4x and y = x3 from 0 to 4.

Solution: We have that the area between y = 4x and y = x3 between 0 and 4 is givenby ∫ 4

0|4x− x3| dx.

4x > x3 for −2 < x < 2, and 4x ≤ x3 otherwise. Thus, |4x − x3| = 4x − x3 for−2 < x < 2 and |4x− x3| = x3 − 4x otherwise. Thus, the desired area is given by theexpression ∫ 2

04x− x3 dx+

∫ 4

2x3 − 4x dx.

3. A potato at room temperature of 20◦C is put in a 200◦C oven. 30 minutes later the potatohas reached 120◦C. Find the temperature of the potato after one hour. How long will it takefor the temperature of the potato to reach 180◦C? How long will it take for the temperature ofthe potato to reach 200◦C.

Solution: Let X(t) denote the temperature of the potato at time t. We have, by Newton’sLaw of Cooling, that

X(t) = Ta + (X(0)− Ta)ekt

where Ta is the ambient temperature (i.e., that of the oven) and X(0) is the initial temper-ature of the potato. The decay constant k is unknown, but we use the given informationto solve for k:

120 = X(30) = 200− 180e30k

k =ln(4/9)

30≈ −0.027.

If T is when the temperature of the potato is 180◦C, then

X(T ) = 200− 180e−0.027T = 180

T ≈ 81.29 minutes.

According to this model, the temperature of the potato will never reach 200◦C (solving forthis value of T would involve evaluating ln(0).

4. Suppose we know that f ′(x) = x5/2 and that f(4) = 6.

(a) Find a linear approximation of f(x) near a = 4 and use this to approximate f(3.95).

Solution: The linear approximation of f near a = 4 is merely the equation of the linetangent to the curve at a = 4:

L(x) = f(4) + f ′(4)(x− 4) = 6 + 32(x− 4).

We have that f(3.95) ≈ L(3.95) = 6 + 32(−0.05) = 4.4.

(b) Is this approximation an overestimate or and underestimate?

Solution: Since f ′′(x) = 52x

3/2, we have that f ′′(4) = 20 > 0. Therefore f is concaveup at x = 4, and so the tangent line lies under the curve of f(x). From this we havethat the approximation in part (a) is an underestimate.

5. Given the function f(x) = x10 − 10x.

(a) Find all the local minimum and local maximum values on the interval 0 < x < 2.

Solution: First locate critical points: f ′(x) = 10x9 − 10. f ′(x) = 0 only if x = 1.We have that f ′′(x) = 90x8 and so f ′′(1) = 90. Thus, x = 1 is the locus of a localminimum. Since the interval is open and we’re only looking for local minima/maxima,we do not need to consider endpoints. The only local extremum is the local minimumat x = 1.

(b) Find all the absolute minimum and absolute maximum values on the interval 0 ≤ x ≤ 2.

Solution: From part (a), x = 1 is a local minimum. We now check the endpoints:f(0) = 0 and f(2) = 1004. Thus, the absolute maximum and minimum are 1004 and−9 occuring at x = 2 and x = 1 respectively.

6. (a) State Rolle’s Theorem, the Mean Value Theorem, and the Fundamental Theorem.

Solution:Rolle’s Theorem: If f is continuous on [a, b] and differentiable on (a, b) and f(a) =f(b), then there is some number c in (a, b) so that f ′(c) = 0.

Mean Value Theorem: If f is continuous on [a, b] and differentiable on (a, b), thenthere is some number c in (a, b) so that f(b)− f(a) = f ′(c)(b− a).

Fundamental Theorem: Part 1: If f is continuous on [a, b] and g is defined by

g(x) =∫ x

af(t) dt,

for a ≤ x ≤ b, then g is also continuous on [a, b], and

g′(x) = f(x).

Part 2: If f is continuous on [a, b] and F is any antiderivative of f (i.e., F ′ = f), then∫ b

af(x) dx = F (b)− F (a).

(b) A tank containing 200 gallons of water develops a leak, and the water completely drainsout in 8 hours. Use the Mean Value Theorem to show that at some point in this 8 hourtime interval, the water was leaking out at exactly 25 gallons per hour.

Solution: Let f(t) be the amount of water in the tank at time t (in hours). We aregiven that f(0) = 200 gallons and f(8) = 0 gallons. Using the Mean Value Theoremon the interval [0, 8], we have that there is some time t = c in the interval (0, 8) sothat

(8)− f(0) = f ′(c)(8− 0)0− 200 = f ′(c) · 8

−2008

= f ′(c)

−25 = f ′(c)

This shows that there is some point t = c during the 8 hour time interval in whichf ′(c) = 25 gallons per hour, i.e., the water is leaking from the tank at a rate of 25gallons per hour.

7. Let f be a continuous function given by the following graph. Suppose F is an antiderivative off ; that is, F ′(t) = f(t), and F (0) = 5.

(a) At what values of x does F (x) have a local minimum value?

Solution: Notice that F ′(x) = f(x) is zero when x = −4,−2, 0, and 4. Create a signchart for F ′ = f :Using the first derivative test, we see that x = −4 and x = 0 correspond to localminimum values of F .

(b) On what intervals is the graph of F (x) concave down?

Solution: Look at the picture of f to determine the intervals of increase and decreaseof f . When f is increasing, f ′ = F ′′ > 0, and when f is decreasing, f ′ = F ′′ < 0.

From the graph, f ′ changes sign at x = −3,−1, and 2. We could make a sign chartfor F ′′ = f ′:From the chart, we see that F is concave down on (−3,−1) ∪ (2, 6).

(c) If F (0) = 5 then use the Fundamental Theorem to determine F (6).

Solution:

F (6)− F (0) =∫ 6

0f(x) dx

F (6)− 5 =12· 4 · 2− 1

2· 2 · 2

F (6)− 5 = 4− 2F (6) = 7.

(d) On what intervals is the graph concave up?

Solution: : Use the chart from part (b).We see that F is concave up on (−5,−3) ∪ (−1, 2).

8. Evaluate limx→∞(1 + 1

x

)x.

Solution: As x → ∞, 1x → 0, so 1 + 1

x → 1. Then this limit is of the indeterminate form1∞.

Let y =(1 + 1

x

)x. Then

ln y = ln((

1 +1x

)x)= x ln

(1 +

1x

),

and as x→∞, x ln(1 + 1

x

)is of the indeterminate form∞·0. Rewrite as a fraction in order

to use L’Hospital’s rule. (It’s easier to take the derivative of 1/x than 1/ ln(1 + 1/x), sowe’ll rewrite the fraction by moving x to the denominator as 1/x and leaving the ln(1+1/x)term alone.)

x ln(

1 +1x

)=

ln(1 + 1

x

)1x

Now this of the indeterminate form 0/0, so we can apply L’Hospital’s rule:

limx→∞

ln y = limx→∞

ln(1 + 1

x

)1/x

(L) = limx→∞

−1/x2

1+1/x

−1/x2

= limx→∞

−1/x2

1 + 1/x· x

2

−1

= limx→∞

11 + 1/x

=1

1 + 0= 1.

Since ln y → 1, y = eln y → e1 = e.

9. Given f(x) = 15x2+90xx−2 , f ′(x) = 15(x−4)(x+3)

(x−2)2, f ′′(x) = 30(x−2)4(15−x)

(x−2)3, find

(a) Vertical asymptotes

Solution: x = 2 is a vertical asymptote of f since

limx→2+

15x2 + 90xx− 2

=∞.

(b) Critical numbers

Solution: f ′(x) = 0 when x = 4 and x = −3. f ′(x) does not exist when x = 2.However, x = 2 is also not in the domain of f . Therefore, the only critical numbersare x = 4 and x = −3.

(c) Intervals of increase

Solution: Make a sign chart for f ′:We see that f is increasing on (−3, 2) ∪ (2, 4).

(d) Intervals of decrease

Solution: From the chart in part (c), we see that f is decreasing on (−∞,−3) and(4,∞).

(e) Local extrema

Solution: Using the first derivative test with the chart from part (c), we see that fhas a local maximum at x = 4 and a local minimum at x = −3.

(f) Concavity

Solution: f ′′(x) = 0 when x = 15. f ′′(x) does not exist when x = 2, but this is alsonot in the domain of f . Make a sign chart for f ′′:We see that f is concave up on (2, 15) and concave down on (−∞, 2) ∪ (15,∞).

10. A box with a square base and open top must have a volume of 24 cm3. Find the dimensions ofthe box that minimize the amount of material used.

Solution: Let x be the length of each side of the base and y be the height of the container.We wish to minimize the amount of material used, i.e., the total surface area A.

A = (area of base) · (total area of four sides)

A = x2 + 4xy

Also, the volume of the container is 24 cm3. We can use this to solve for y in terms of x:

V = x2y

24 = x2y

24x2

= y

Then

A = x2 + 4x(

24x2

)= x2 +

96x

Find the critical number(s) of A:

A′ = 2x− 96x2

0 = 2x− 96x2

2x =96x2

2x3 = 96

x3 = 48

x = 3√

48

We can use the first derivative test for absolute min/max to check that x = 3√

48 is anabsolute minimum of A. Notice that A′(x) < 0 for x < 3

√48 and A′(x) > 0 for x > 3

√48,

so x = 3√

48 is an absolute minimum of A. The dimension are

x = 3√

48 cm and y =24x2

=24

( 3√

48)2cm.

11. One tower is 50 ft high and another tower is 30 ft high. The towers are 150 ft apart. A guywire is run from the top of each tower to point A on the ground between the two towers. Usecalculus to locate point A so that the total length of the guy wire is minimal.

42−3

⊕⊕f ′

f

Solution: Label the figure as above. By the Pythagorean Theorem,

z21 = x2 + 302

z1 =√x2 + 900

z22 = (150− x)2 + 502

z2 =√

(150− x)2 + 2500

Let ` be the total length of the wire. We wish to minimize `.

` = z1 + z2

` =√x2 + 900 +

√(150− x)2 + 2500

Find the critical number(s) of `:

`′ =2x

2√x2 + 900

+−2(150− x)

2√

(150− x)2 + 2500

`′ =x√

x2 + 900− (150− x)√

(150− x)2 + 2500

0 =x√

x2 + 900− (150− x)√

(150− x)2 + 2500(150− x)√

(150− x)2 + 2500=

x√x2 + 900(

(150− x)√(150− x)2 + 2500

)2

=(

x√x2 + 900

)2

(150− x)2

(150− x)2 + 2500=

x2

x2 + 900(150− x)2(x2 + 900) = x2((150− x)2 + 2500)

(150− x)2(x2 + 900) = (150− x)2x2 + 2500x2

(150− x)2(x2 + 900− x2) = 2500x2

(150− x)2(900) = 2500x2

9(150− x)2 = 25x2

9(22500− 300x+ x2) = 25x2

202500− 2700x+ 9x2 = 25x2

0 = 16x2 + 2700x− 2025000 = 4(x+ 225)(4x− 225)

x =2254,−225

Since 0 ≤ x ≤ 150, the only critical number is x = 225/4 = 56.25, and we can use theclosed interval method to find the absolute minimum of `:

`(225/4) = 170

`(0) ≈ 188.11

`(150) ≈ 202.97

Therefore, the length of the wire is minimal when A is 56.25 ft from the 30 ft tower and150− 56.25 = 93.75 ft from the 50 ft tower.

12. A car is traveling at 25m/s when the driver sees an accident 60m ahead and slams on thebrakes. What constant deceleration is required to stop the car in time to avoid a pileup?

Solution: Let A denote the acceleration. We have that v(t) = At+C1. The car is initiallytraveling at a rate of 25m/s and so C1 = 25. Therefore v(t) = At + 25. We have thatx(t) = A

2 t2 + 25t + C2. Let x(0) denote the position of the car at the moment that the

driver sees the accident. Then x(0) = C2 = 0. Thus x(t) = A2 t

2 + 25t. The car (ideally)will come to a complete stop when x = 60. Let T denote the time for the car to stop. Wehave that x(T ) = 60 and V (T ) = 0. From this we get two equations:

A

2T 2 + 25T = 60

AT + 25 = 0.

Use the second equation and solve for A:

A = −25T

Plug this into the first equation:

A

2T 2 + 25T =

−25T

2T 2 + 25T =

−252T + 25T =

252T = 60

T =12025

= 24/5.

Plug this into the first equation, and we have that

A = −125/24.

Thus, the necessary deceleration is 125/24 m/s2.

13. Water flow into and out of a storage tank. A graph of the rate of change r(t) of the volume ofwater in the tank, in liters per day, is shown. If the amount of water in the tank at time t = 0is 25, 000 L, use the Midpoint Rule with n = 4 to estimate the amount of water in the tankfour days later.

Solution:v(4)− v(0) ≈ (−1000)1 + (1000)1 + (3000)1 + (2000)1

v(4) ≈ 5000 + 25000 = 30000.

14. If∫ 6

8f(x) dx = −3, and

∫ 4

8f(x) dx = 5, find

∫ 6

4f(x) dx.

Solution: We have that∫ 8

4f(x) dx =

∫ 6

4f(x) dx+

∫ 8

6f(x) dx,

and so ∫ 6

4f(x) dx =

∫ 8

4f(x) dx−

∫ 8

6f(x) dx

= −∫ 4

8f(x) dx+

∫ 6

8f(x) dx = −8.

15. Oil is pumped continuously from a well at a rate proportional to the amount of oil left in thewell. Initially ther were 10 million barrels of oil in the well and six years later there were only2 million barels remaining.

(a) Find a formula for the amount of oil remaining in the well after t years.

Solution: Let V (t) denote the amount of oil in the well (in millions of barrels). Wehave that oil is pumped out at a rate proportional to V , and so

dVdt

= kV.

We know that the solution to this differential equation is V (t) = V (0)ekt. We havethat V (0) = 10. We can use the other given information to solve for k:

2 = V (6) = 10e6k

k = − ln(5)6≈ −0.2682.

Therefore, the desired formula is given as

V (t) = 10e−0.2682t.

(b) When will there be only 0.25 million barrels remaining?

Solution: Set V (t) = 0.25 and solve for t:

10e−0.2682t = 0.25

t ≈ 13.75 years.

16. If you jump out of an airplane and your parachute fails to open, your downward velocity tseconds after the jump is approximated by v(t) = 49(1− e−0.2t) m/s.

(a) Write an expression for the distance you fall during the first 3 seconds.

Solution: d =∫ 3

049(1− e−0.2t) dt.

(b) Use the Fundamental Theorem to find the exact distance you fall during the first 3 seconds.

Solution: We have that the function x(t) = 49(t + 52e−0.2t) is an antiderivative of

v(t). Therefore ∫ 3

049(1− e−0.2t) dt = 49(t+

52e−0.2t)

∣∣∣30≈ 91.73 m.

17. Evaluate

(a)∫

sec6(x) tan(x) dx

Solution: Let u = sec(x), then du = sec(x) tan(x) dx. In this case, the integralbecomes ∫

sec6(x) tan(x) dx =∫u5 du =

u6

6+ C

=sec6(x)

6+ C.

(b)∫

tan2(θ) sec2(θ) dθ

Solution: Let u = tan(θ), then du = sec2(θ) dθ. In this case, the integral becomes∫tan2(θ) sec2(θ) dθ =

∫u2 du =

u3

3+ C

=tan3(θ)

3+ C.

18. Find the area of the region bounded by the parabola y = 5x2, the tangent line to this parabolaat (4, 80) and the y-axis.

Solution: First we will find the equation of the tangent line. The slope is 10x|x=4 = 40.We use point slope:

y − 80 = 40(x− 4)

y = 40x− 80

The region enclosed looks like the following:

In this case, it is easier to integrate with respect to y. Written in terms of y, the parabolahas the equation x =

√y5 and the line has the equation x = y

40 + 2. The integral whichgives the area, then, is∫ 80

0

( y40

+ 2)−(√

y

5

)dy =

y2

80+ 2y − 10

3

(y5

)3/2 ∣∣∣80

0

= 80 + 160− 6403

=803.

19. Consider the region bounded by the graphs of y = x3 and y = 2x2. A solid is formed when theregion is rotated about the x-axis. Find the exact integrals that must be evaluted to get thevolume of this solid, using method of discs.

Solution: The intersection of these curves occurs at x = 2, and the resulting solid hasannular cross-sections:

The desired integral is ∫ 2

0π(2x2)2 − π(x3)2 dx.