Math 110 Fall 2014 Final Exam Solutions, Summaries of Differential Equations

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Math 110

Fall 2014

Final Exam

Solutions

1. Graph these functions. The main thing is to get the general shape, but please also mark a few points and any maxima, minima, asymptotes or discontinuities.

(a) y =

ex x

Solution: The limit at −∞ is zero (use L’Hˆopital’s rule). The x in the deniminator produces a vertical asymptote at x = 0, with the function positive to the right and negative to the left. There is just one minimum, which may be found by diferentiating.

d dx

ex x = ex

x

x^2

which vanishes at x = 1.

2. Compute is the trapezoidal approximation to

1

ex x dx using just one trapezoid. Leave the expression in exact form here.

The one trapezoid has base 1/2 and heights e^1 and

e^3 /^2. Therefore the trapezoidal approximation to the integral is

1 4

e +

e^3 /^2

Then select which of these is closest to the value you have written down. (i) 0. (ii) 0. (iii) 1. (iv) 1. (v) 2.

To evaluate numerically, we can use e ≈ 2 .7 but then we must come up with a reasonable value for e^3 /^2. Here are two ways to do that. (1) e^3 /^2 ≈ 101.^5 /^2.^3 ≈ 100.^65 which should be about midway between 10^0.^6 ≈ 4 and 10^0.^7 ≈ 5. So we say e^3 /^2 ≈ 4 .5. (2) Use the Taylor series: e^3 /^2 = e ∗ e^1 /^2 ≈ e ∗ (1 + 1/2 + (1/2)^2 /2) = e ∗ 13 / 8 ≈ 2. 7 ∗ 13 /8 = 35. 1 / 8 ≈ 4 .5. In either case, we get about

= a little bigger than 1.4, so we choose answer (iv).

3. (a) Use linearization to estimate log 10 1 .069.

Solution: d dx log 10 x at x = 1 is equal to

ln 10 , therefore L(x) =

ln 10 (x − 1). Plugging in x = 1.069 gives

log 10 1. 069 ≈

ln 10

(b) Use this to estimate log 10

Solution: log 10

= 30 log 10 1. 069 ≈ 30 · 0 .03 = 0.9.

(c) Use this to estimate 1. 06930.

Solution: In the middle we use the fact that 10^0.^3 is approximately 2 (it’s on the log cheatsheet):

1. 06930 = 10log^10 (1.^069

(^30) )

≈ 100.^9 =

100.^3

5. Uncle Sam has a debt of 18 trillion dollars (as of January 1, 2015). Senator Paul decides he needs to pay it off, and proposes an installment scheme, under which Uncle Sam will make payments of one trillion dollars on December 31, 2015 and every December 31 thereafter. Unfortunately, the debt increases by 5% during the course of every year.

(a) Write expressions for the amount owed by Uncle Sam on January 1 of 2015, 2016 and

Solution: Each year the new debt is the old debt multiplied by 1.05, with one trillion subtracted. This give the following table.

date debt in trillions of dollars Jan 1, 2015 18 Jan 1, 2016 18 × 1. 05 − 1 Jan 1, 2015 18 × 1. 052 − 1. 05 − 1

(b) Write an expression using Sigma notation for how much Uncle Sam owes on January 1 of year n, counting 2015 as year 0.

Solution: After year n, Uncle Sam’s debt is given by

18 × 1. 05 n^ −

∑^ n−^1

k=

1. 05 k^.

(c) Evaluate the sum to get an algebraic expression.

Solution: The geometric series has first term 1 and r = 1.05. Plugging into the formula for a geometric series gives

∑^ n−^1

k=

1. 05 k^ =
2. 05 n^ − 1
3. 05 − 1 = 20 (1. 05 n^ − 1).

Therefore, on January 1 of year n, Uncle Sam owes

18 × 1. 05 n^ − 20 (1. 05 n^ − 1) = 20 − 2 × 1. 05 n^ trillion dollars.

(d) Solve for the number of years, n, in which the debt will be paid off. Please leave this as an exact expression (it is OK if this leads to a value which is not a whole number).

Solution: From 20 − 2 · 1. 05 n^ = 0 we get 1. 05 n^ = 10. We can write this as

n = log 1. 05 10.

It may be somewhat easier to continue to simplify. An alternative expression is

n = ln 10 ln 1. 05

(e) Estimate the numerical value of n to the nearest whole number.

Solution: Estimate the value of ln 1.05 by the linearization of ln x near 1 whicih is x − 1. Thus ln 1. 05 ≈ 0 .05. Using the log cheatsheet, ln 10 ≈ 2 .3, therefore ln 10 ln 1. 05

In other words, it will take Uncle Sam about 46 years to pay off this debt according to Senator Paul’s schedule.

7. Sketch the region and evaluate the integral.

∫ (^9)

4

∫ √x

0

ey/

√x dy dx

Solution:

The indefinite integral of ey/

√x dy may be evaluated by the substitution u = y/

x; it is equal to

xey/

√x

. Therefore the inner integral comes out to ∫ √x

0

ey/

√x dy =

xey/

√x^ ∣∣ ∣

√x y=

x(e − 1).

The outer integral then evaluates to ∫ (^9)

4

x(e − 1) dx = (e − 1)

x^3 /^2

9

4

(27 − 8)(e − 1) =

(e − 1).

8. The price of a turkey is proportional to its weight and inversely proportional to the square of its age. Jack’s mother gives him enough money to buy a 10 pound turkey that is one year old. When Jack gets to the fair, he realizes that he needs a turkey that is slightly bigger. How much older will the turkey have to be per extra pound in weight? Please begin by writing down an equation satisfied by price, weight and age, giving the interpretation and units for all varialbles and constants used.

Solution: The wording of the problem may be translated into the equation

p = k

w a^2 where w is the weight (the problem suggests units of pounds), a is its age (units of years), and p is the price of the turkey (units of in dollars or your favorite medieval currency). The proportionality constant k then has units of dollars times years squared per pound.

The problem calls for us to evaluate da/dw along the level curve where p is constant, throught the point (w, a) = (10, 1). We use the formula da dw

pw pa

. The partial deriva- tives of p(w, a) are given by ∂p ∂w (w, a) = k a^2 ∂p ∂a (w, a) = − 2 kw a^3

Therefore, da dw

pw pa

k/a^2 2 kw/a^3

a 2 w

Evaluating at the point (10, 1) gives 1/20. Thus, for each pound more tha Jack needs, he must be willing to accept a turkey 1/20 year older and tougher. [Note that the units are in years per pound, which is consistent with computation we are trying to compute how mthat it represents how many more years the turkey will have per ound of extra weight.]

10. Uncle Sam owes 18 trillion dollars (this is a new problem, really!). Interest accmulates continuously at the rate of 5%/year. Suppose that Congress forces Uncle Sam to pay back the debt continuously at a rate of one trillion dollars per year. (a) Write a differential equation for the amount owed by Uncle Sam at time t. Please give the meaning and units of all variables.

Solution: Let A(t) be the amount owed at time t, measured in trillions of dollars. Then dA(t) dt

= 0. 05 A(t) − 1 where the constant 0.05 has units of inverse years and the constant 1 has units of trillions of dollars per year.

(b) Find the general solution of this differential equation.

Solution: The equation A′^ − 0. 05 A = −1 is first order linear with P and Q both constant: P (t) = − 0 .05 and Q(t) = −1. Multiplying by the integrating factor e−^0.^05 t and integrating yields e−^0.^05 tA(t) = 20e−^0.^05 t^ + C and therefore A(t) = 20 + Ce^0.^05 t^.

(c) State the initial condition and give the solution to the initial value problem.

Solution: The initial condition is A(0) = 18. Solving for C gives C = −2 therefore A(t) = 20 − 2 et/^20.

(d) How long will it take for Uncle Sam’s debt to reach zero?

Solution: Setting A(t) = 0 gives et/^20 = 10 hence t = 20 ln 10. Numerically, this comes out to about 46 years. [Note: this is similar to Problem 4 except that this version does not require you to approximate ln 1.05.]

11. (a) Compute the indefinite integral

x−^3 ln x dx.

Solution: This is a straightforward integration by parts. Let u = ln x and dv = x−^3 dx so that v = −(1/2)x−^2 dx and dv = x−^1 dx. Then ∫ x−^3 ln x dx = −

x−^2 ln x +

x−^2 (x−^1 dx) = −

x−^2 ln x −

x−^2.

Alternative solution: You can do this as a substitution with u = ln x and du = dxx. The integral then becomes

x−^2 ln x(dx/x) =

e−^2 uu du. Now you can integrate by parts to get ∫ ue−^2 u^ du =

ue−^2 u^ +

e−^2 u 2 du =

ue−^2 u^ −

e−^2 u

and plugging in u = ln x produces the same answer as before.

(b) Write the improper integral

1

x−^3 ln x dx as a limit and evaluate it.

Solution: The integrand is continous so it is a type-I improper integral. Using the previous result, ∫ (^) ∞

1

x−^3 ln x dx = (^) Mlim →∞

∫ M

1

x−^3 ln x dx

= (^) Mlim →∞

x−^2 ln x −

x−^2

M

1

= (^) Mlim →∞

M −^2 ln M −

M −^2

It is obvious that the term (1/4)M −^2 goes to zero; to see why (1/2)M −^2 ln M also goes to zero note that ln M grows much slower than any power of M.

13. A swimming pool holding 300 cubic meters of water is determined to contain 1/100 of a cubic meter of toxins. Immediately a drain is opened and pool water starts flowing out at 3 cubic meters per minute. Also a hose is inserted to pump in fresh water at the rate of 1 cubic meter per minute. Assuming the fresh water mixes rapidly with the pool water, which of these differential equations models the total amount P of poison in the pool at time t minutes after the toxicity is discovered? You need only circle the correct number from (i) to (v).

(i) P ′(t) = − 3 P (t) 300 − 2 t

(ii) P ′(t) = −1 + P (t) 3

(iii) P ′(t) P (t)

(iv) P ′(t) − P (t) = −

300 − t

(v) None of the above

Solution: The pool is filling at rate 1 m^3 / min and draining at rate 3 m^3 / min. Therefore, the amount of water in the pool at time t is 300 − 2 t cubic meters. If P (t) is the total amount of poison in the pool, the concentration of poison is P (t)/(300 − 2 t). The amount of poison leaving the pool is the concentration times the total outflow, which is 3m^3 / min times the concentration. Because no poison enters the pool, this is the rate of change of P (t) and it is negative. Thus the correct answer is (i).

Logarithm Cheat Sheet

These values are accurate to within 1%:

e ≈ 2. 7 ln(2) ≈ 0. 7 ln(10) ≈ 2. 3 log 10 (2) ≈ 0. 3 log 10 (3) ≈ 0. 48

Some other useful quantities to with 1%:

π ≈

22

√^7 10 ≈ π √ 2 ≈ 1. 4 √ 1 / 2 ≈ 0. 7

(ok so technically

√ 2 is about 1.005% greater than 1.4 and 0.7 is about 1.005% less than

√ 1 /2)

1