Written Homework 11 Solutions

MATH 1310 - CSM

Solutions to: pp 257-260; problems: 4, 15, 16. pp 269; problems: 1 - 5. Pp 286-289;problems: 1 (abcegijmqr), 2 (bcdeijk), 5, 9, 10, 11.

PP 257-260

4. Find the derivatives of the following functions.Solution:

(a)d

dxln(3x) = 1/x

(b)d

dx17 ln(x) = 17/x

(c)d

dwln(ew) = 1

(d)d

dtln(2t) = ln(2)

(e)d

dsπ ln(3e4s) = 4π

15. Find a solution (using lnx) to the differential equation

f ′(x) = 3/x satisfying f(1) = 2

Using the hint provided, let’s first guess that f(x) = lnx. If f(x) = lnx, then f ′(x) = 1/x,but we want f ′(x) = 3/x, so let’s multiply our fist guess of f(x) = lnx by 3 to getf(x) = 3 lnx. Then f ′(x) = 3/x. Now let’s consider that we want f(1) = 2. Currentlyf(1) = ln 1 = 0, so let’s add 2 to our f(x) equation to get f(x) = 3 lnx + 2. Nowf ′(x) = 3/x and f(1) = 2.

16. (a) Find a formula using the natural logarithm function giving the solution of y′ = a/xwith y(1) = b.Solution: In problem 15 we saw that if y′ = 3/x and y(1) = 2, then y(x) = 3 lnx+2.So similarly if y′ = a/x with y(1) = b, then y = a lnx+ b.

(b) Solve P ′ = 2/t with P (1) = 5.Solution: Using the general equation y = a ln t+ b for y′ = a/t and y(1) = b, in thisproblem we have a = 2 and b = 5, so y = 2 ln t+ 5.

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PP 269

1. Find a formula y = F (t) for a solution to the differential equation dy/dt = f(t) when f(t)is:

(a) 5t− 3

Solution: F (t) =5t2

2− 3t.

(b) t6 − 8t5 + 22π3

Solution: F (t) =t7

7− 8t6

6+ 22π3t.

(c) 5et − 3 sin tSolution: F (t) = 5et + 3 cos t.

(d) 12√t

Solution: F (t) =12t3/2

3/2= 8t3/2.

(e) 2t + 7/t9

Solution: F (t) =2t

ln 2− 7

8t8.

(f) 5e4t − 1/t

Solution: F (t) =5e4t

4− ln t.

2. Find G(5) if y = G(x) is a solution to the initial value problem:

dy

dx=

1

x2y(2) = 3.

Solution:We take the antiderivative of dy/dx = 1/x2 to find that y = −1/x+ C.But we’re given that y(2) = −1/2 + C = 3, which implies that C = 7/2.So G(x) = y = −1/x+ 7/2, thus G(5) = −1/5 + 7/2 = 33/10.

3. Find F (2) if y = F (x) is a solution to the initial value problem:

dy

dx=

1

xy(1) = 5.

Solution:We take the antiderivative of dy/dx = 1/x to find that y = lnx+ C.But we’re given that y(1) = ln 1 + C = 5, which implies that C = 5.So F (x) = y = lnx+ 5, thus F (2) = ln 2 + 5 ≈ 5.69.

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4. Find H(3) if y = H(x) is a solution to the initial value problem:

dy

dx= x3 − 7x2 + 19 y(−1) = 5.

Solution:We take the antiderivative of dy/dx = x3− 7x2 + 19 to find that y = 1

4x4− 7

3x3 + 19x+C.

But we’re given that y(−1) = 5 = 14(−1)4 − 7

3(−1)3 − 19 + C, so C = −257/12.Thus H(x) = y = 1

4x4− 7

3x3 +19x− 257

12 , and so H(3) = 14(3)4− 7

3(3)3 +19(3)− 25712 = 58.5.

5. Find L(−2) if y = L(x) is a solution to the initial value problem:

dy

dx= e3x y(1) = 6.

Solution:We take the antiderivative of dy/dx = e3x to find that y = 1

3e3x + C.

But we’re given that y(1) = 6 = 13e

(3)(1) + C, so C ≈ −0.695.

Thus L(x) = y = 13e

3x − 0.695, and so L(−2) = 13e

(3)(−2) − 0.695 ≈ −0.694.

PP 286-289

1. Solutions:

(a)d

dx(3x5 − 10x2 + 8) = 15x4 − 20x.

(b)d

dx((5x12 + 2)(π − π2x4)) = 60x11(π − π2x4) + (5x12 + 2)(−4π2x3).

(c)d

du(√u− 3/u3 + 2u7 =

−1

2√u

+ 9/u4 + 14u6.

(e)d

dx(0.5 sinx+ 3

√x+ π2) = 0.5 cosx+

1

3x3/2.

(g)d

dx

(2√x− 1√

x

)=

1√x

+1

2x3/2.

(i)d

dx

(sinx

x2

)=x2 cosx− 2x sinx

x4.

(j)d

dx(x2ex) = 2xex + x2ex.

(m)d

dx(ex lnx) = ex lnx+

ex

x.

(q)d

dxln(x2 + xex) =

2x+ ex + xex

x2 + xex.

(r)d

dx

(5x2 + lnx

7√x+ 5

)=

(7√x+ 5)(10x+ 1

x)− (5x2 + lnx)( 72√x)

(7√x+ 5)2

.

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2. Suppose f and g are functions that we are given

f(2) = 3, g(2) = 4, g(3) = 2,

f ′(2) = 2, g′(2) = −1, g′(3) = 17.

Evaluate the derivative of each of the following functions at t = 2.Solutions:

(b)d

dt(5f(t)−2g(t)) = 5f ′(t)−2g′(t); evaluate at t=2: 5f ′(2)−2g′(2) = 5·2−2·(−1) = 12.

(c)d

dt(f(t)g(t)) = f ′(t)g(t) + f(t)g′(t); evaluate at t = 2: f ′(2)g(2) + f(2)g′(2) = 5.

(d)d

dt

(f(t)

g(t)

)=g(t)f ′(t)− f(t)g′(t)

g2(t); evaluate at t = 2:

g(2)f ′(2)− f(2)g′(2)

g2(2)=

11

16.

(e)d

dtg(f(t)) = g′(f(t)) · f ′(t); evaluate at t = 2: g′(f(2)) · f ′(2) = 34.

(i)d

dt

(1

f(t)

)=−1

f2(t); evaluate at t = 2:

−1

f2(2)= −1

9.

(j)d

dtf(3t− (g(1 + t))2) = f ′(3t− (g(1 + t))2) · [3− 2g(1 + t)] · g′(1 + t); evaluate at t = 2:

f ′((3)(2)− (g(1 + 2))2) · [3− 2g(1 + 2)] · g′(1 + 2) = f ′(6− (g(3))2) · (3− 2g(3)) · g′(6− 2g(3))

= f ′(2) · (−1) · g′(2)

= (2)(−1)(−1)

= 2.

(k) What additional piece of information would you need to calculate the derivative off(g(t)) at t = 1?

Solution: Sinced

dtf(g(t)) = f ′(g(t)) · g′(t). When we evaluate at t = 2 we get:

f ′(g(2)) · g′(2) = −f ′(4), so we’d need to additionally know f ′(4).

5. Which of the following functions has a derivative which is always positive (except at x = 0,where neither the function nor its derivative is defined)?

1/x − 1/x 1/x2 − 1/x2

Solution:d

dx(1/x) = −1/x2, which is not positive for all x 6= 0.

d

dx(−1/x) = 1/x2, which is positive for all x 6= 0.

d

dx(1/x2) = −2/x3, which is not positive for all x 6= 0.

d

dx(−1/x2) = 2/x3, which is not positive for all x 6= 0.

Hence onlyd

dx(−1/x) = 1/x2 is positive for all x 6= 0.

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9. The population of a particular country is 15,000,000 people and is growing at the rate of10,000 people per year. In the same country the per capita yearly expenditure for energy is$1,000 per person and is growing at the rate of $8 per year. What is the country’s currenttotal yearly energy expenditure? How fast is the country’s total yearly energy expendituregrowing?Solution: Let P (t) represent the population over time, C(t) represent the per capitayearly expenditure for energy, and E(t) represent the total yearly energy expenditure.Then,

E(t) = P (t) · C(t) (1)

E′(t) = P ′(t) · C(t) + P (t) · C ′(t) (2)

Using equation 1, we see that the current energy expenditure is E = (15, 000, 000)(1, 000) =$15, 000, 000, 000. Then using equation 1, we see that the energy expenditure is growingat a rate of E′ = (10, 000)(1, 000) + (15, 000, 000)(8) = $130, 000, 000 per year.

10. The population of a particular country is 30 million and is rising at the rate of 4,000 peopleper year. The total yearly personal income in the country is 20 billion dollars, and it isrising at the rate of 500 million dollars per year. What is the current per capita personalincome? Is it rising or falling? By how much?Solution:Let P (t) represent the population over time, I(t) represent the total yearly personal in-come, and C(t) represent the per capita personal income. Then,

C(t) =I(t)

P (t)(3)

C ′(t) =P (t) · I ′(t)− P ′(t) · I(t)

P (t)2(4)

Using equation 3 we see that the current per capita personal income is C =20 billion

30 million=

$666.67. Using equation 4, we see that the current per capita personal income is growing

at a rate of C ′ =(30 million)(500 million)− (4, 000)(20billion)

(30 million)2= $16.76 per year.

11. Not graded

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