CALCULUS I HOMEWORK 6 Due October 31 Solutionpicard/hwk6_soln.pdf · CALCULUS I HOMEWORK 6 Due...

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CALCULUS I HOMEWORK 6 Due October 31 1. Find the linearization L(x) of f (x) = (sin x) tan x at the point x = π 4 . Solution: We write f (x)= e ln(sin x) tan x . Differentiating using the chain rule f 0 (x)= e ln(sin x) tan x d dx ln(sin x) tan x = e ln(sin x) tan x d dx tan x ln(sin x). Using the product rule, f 0 (x) = (sin x) tan x ( d dx tan x) ln(sin x) + tan x d dx ln(sin x) . Taking the derivatives f 0 (x) = (sin x) tan x ln(sin x) cos 2 x + tan x 1 sin x cos x . Evaluating the expression at x = π 4 gives f 0 ( π 4 )=( 1 2 ) 1 ln 1 2 1 2 +1 = 1 2 (1 - ln 2). Evaluating f (x) at π 4 gives f ( π 4 )= 1 2 . Therefore the linearization L(x)= f (a)+ f 0 (a)(x - a) is L(x)= 1 2 + 1 2 (1 - ln 2)(x - π 4 ). 2. A ladder is leaning against a wall. The ladder starts to slide down, and when the bottom of the ladder is 4m from the wall, the bottom is sliding away at 1 4 m/s and the top is sliding down at 1 2 m/s. How long is the ladder? Solution: Denote by x the distance between the bottom of the ladder and the wall, and by y the distance between the top of the ladder and the floor. Let L denote the length of the ladder. By the Pythagorean theorem, x 2 + y 2 = L 2 . Differentiating both sides of the equation gives 2x dx dt +2y dy dt =0, 1

Transcript of CALCULUS I HOMEWORK 6 Due October 31 Solutionpicard/hwk6_soln.pdf · CALCULUS I HOMEWORK 6 Due...

Page 1: CALCULUS I HOMEWORK 6 Due October 31 Solutionpicard/hwk6_soln.pdf · CALCULUS I HOMEWORK 6 Due October 31 1. Find the linearization L(x) of f(x) = (sinx)tanxat the point x= ˇ 4.

CALCULUS I HOMEWORK 6

Due October 31

1. Find the linearization L(x) of f(x) = (sinx)tanx at the point x = π4 .

Solution: We write f(x) = eln(sinx)tan x

. Differentiating using the chainrule

f ′(x) = eln(sinx)tan x d

dxln(sinx)tanx = eln(sinx)

tan x d

dxtanx ln(sinx).

Using the product rule,

f ′(x) = (sinx)tanx(

(d

dxtanx) ln(sinx) + tanx

d

dxln(sinx)

).

Taking the derivatives

f ′(x) = (sinx)tanx(

ln(sinx)

cos2 x+ tanx

1

sinxcosx

).

Evaluating the expression at x = π4 gives

f ′(π

4) = (

1√2

)1( ln 1√

212

+ 1

)=

1√2

(1− ln 2).

Evaluating f(x) at π4 gives

f(π

4) =

1√2.

Therefore the linearization L(x) = f(a) + f ′(a)(x− a) is

L(x) =1√2

+1√2

(1− ln 2)(x− π

4).

2. A ladder is leaning against a wall. The ladder starts to slide down, andwhen the bottom of the ladder is 4m from the wall, the bottom is slidingaway at 1

4m/s and the top is sliding down at 12 m/s. How long is the ladder?

Solution: Denote by x the distance between the bottom of the ladder andthe wall, and by y the distance between the top of the ladder and the floor.Let L denote the length of the ladder. By the Pythagorean theorem,

x2 + y2 = L2.

Differentiating both sides of the equation gives

2xdx

dt+ 2y

dy

dt= 0,

1

Page 2: CALCULUS I HOMEWORK 6 Due October 31 Solutionpicard/hwk6_soln.pdf · CALCULUS I HOMEWORK 6 Due October 31 1. Find the linearization L(x) of f(x) = (sinx)tanxat the point x= ˇ 4.

2 CALCULUS I HOMEWORK 6

since dLdt = 0, because the length of the ladder is not changing. Substituting

the information given in the equation, at a certain time we have

2(4)1

4− 2y

1

2= 0.

Solving for y gives y = 2. Therefore,

42 + 22 = L2,

and so L =√

20.

3. Suppose you have a can of a drink which you prefer to drink cold, butyou bought unrefrigerated. The temperature of the drink is currently at21C and you would like to drink it at 8C. You are living in Canada, and youplace the can in a snow pile so that the ambient temperature around it is−5C. After 15 minutes, you are impatient and check the temperature of thedrink, but it is still too warm at 18C. Assuming Newton’s Law of Cooling,how much longer do you need to wait until the drink reaches 8C?

Solution: Let t denote the time in minutes since you placed the can inthe snow, T denote the temperature of the drink, and y = T + 5. Newton’sLaw of Cooling says that

y = y(0)ekt.

At t = 0, we have y(0) = T (0) + 5 = 21 + 5 = 26.

y = 26ekt.

At t = 15, we have(18 + 5) = 26e15k.

Solving for k,

23

26= e15k, ln

23

26= 15k, k =

1

15ln

23

26≈ −0.00817.

We are looking for the time when y = 8 + 5 = 13. We must solve

13 = 26e−0.00817t.

Solving for t gives

e−0.00817t =13

26, −0.00817t = − ln

1

2, t = 84.8.

Therefore, after waiting 15 minutes, you must wait an extra 69.8 minutesbefore drinking.

4. Suppose you run a credit company and you want to exploit your cus-tomers as much as possible. Your annual interest rate is 21.99%. You areconsider two options for accumulating interest. Option A: daily compound-ing. This means that at the end of each day (365 times per year), the currentbalance is multiplied by

1 +21.99

100

1

365.

Page 3: CALCULUS I HOMEWORK 6 Due October 31 Solutionpicard/hwk6_soln.pdf · CALCULUS I HOMEWORK 6 Due October 31 1. Find the linearization L(x) of f(x) = (sinx)tanxat the point x= ˇ 4.

CALCULUS I HOMEWORK 6 3

For example, if the balance starts as A0 dollars, after one day the balancebecomes

A0

(1 +

21.99

100

1

365

).

After two days, the balance becomes

A0

(1 +

21.99

100

1

365

)2

,

and so on.

You are also considering Option B: continuously compounded interest.The balance after one year is mathematically defined as the limit as n→∞of multiplying the balance by

1 +21.99

100

1

nn times during the year. This means that if the balance starts as A0 dollars,after one year the balance becomes

limn→∞

A0

(1 +

21.99

100

1

n

)n.

A customer asks for a 3500$ loan. Unfortunately for the customer, theyblew all their money gambling in Vegas and are unable to pay back thedebt for 3 years. How much does the customer owe you after 3 years underOption A and Option B?

Solution: Under option A, the customer owes you

3500

(1 +

21.99

100

1

365

)365(1 +

21.99

100

1

365

)365(1 +

21.99

100

1

365

)365

.

This can be evaluated to 6768.40$. Under option B, the customer owes you

3500e0.2199e0.2199e0.2199.

This can be evaluated to 6769.74$.