BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.4, Problem 72E
To determine

To find: The number of hours of daylight increases in Philadelphia on March 21 and May 21.

Expert Solution

Answer to Problem 72E

The number of hours of daylight increases in Philadelphia on March 21 and May 21 is approximately 0.0482 and 0.02398.

Explanation of Solution

Given:

The function L(t)=12+2.8sin[2π365(t80)].

Derivative rule:

(1) Constant Multiple Rule: ddx[cf(x)]=cddx[f(x)]

(2) Sum Rule: ddx(f+g)=ddx(f)+ddx(g)

(3) Quotient Rule: ddx(fg)=gddx(f)fddx(g)(g)2

Result used: Chain Rule

If g is differentiable at x and f is differentiable at g(x), then the composite function F=fg defined by F(x)=f(g(x)) is differentiable at x and F is given by the product

F(x)=f(g(x))g(x) (1)

Calculation:

Obtain the derivative of L(t).

L(t)=ddx(L(t))=ddx(12+2.8sin[2π365(t80)])

Apply the sum rule (1) and the constant multiple rule (2),

L(t)=ddx(12+2.8sin(2π365(t80)))=ddt(12)+ddt(2.8sin(2π365(t80)))=0+2.8ddt(sin(2π365(t80)))

L(t)=2.8ddt(sin(2π365(t80)))

Apply the chain rule as shown in equation (1),

Let g(t)=2π365(t80) and f(u)=2.8sinu  where u=g(t).

L(t)=f(g(t))g(t) (3)

The derivative of f(g(t)) is computed as follows,

f(g(t))=f(u)=ddu(f(u))=ddu(2.8sinu)=2.8cosu

Substitute u=2π365(t80) in the above equation,

f(g(t))=2.8cos(2π365(t80))

Thus, the derivative is f(g(t))=2.8cos(2π365(t80)).

The derivative of g(t) is computed as follows,

g(t)=ddt(2π365(t80))=2π365ddt(t80)=2π365(10)=2π365

Thus, the derivative is g(t)=2π365.

Substitute 2.8cos(2π365(t80)) for f(g(t)) and 2π365 for g(t) in equation (3),

L(t)=2.8cos(2π365(t80))(2π365)

Therefore, the derivative is L(t)=2.8cos(2π365(t80))(2π365).

Given that, the rate of the length of daylight is increasing in Philadelphia on March 21 and May 21. That is, L(t)>0.

Obtain the value of t for which L(t)>0.

2.8cos(2π365(t80))(2π365)>0cos(2π365(t80))>0

Note that, cosx>0 when  0 < x < π2 and it can be concluded that 0<2π365(t80)<π2.

0<t80<π23652π0<t80<91.2580<t<91.25+8080<t<171.25

Here, t is from 21st March to 20 th June.

On March 21, the value of t=80.

Substitute t=80 in L(t),

L(80)=2.8cos(2π365(8080))(2π365)=2.8cos(0)(0.0172)=2.8(1)(0.0172)  (Qcos(0)=1)=0.0482

Thus, the number of hours of daylight is increasing in Philadelphia on March 21 is approximately 0.0482.

On May 21, the value of t is t=141 (80+30+21).

Substitute t=141 in above equation,

L(141)=2.8cos(2π365(14180))(2π365)=2.8cos(2π365(61))(2π365)=2.8cos(122π365)(0.0172)=2.8(0.4975)(0.0172)

0.02396

Thus, the number of hours of daylight increases in Philadelphia May 21 is approximately 0.02398.

Note: L(141) is approximately two times the L(80).

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