   Chapter 3.4, Problem 82E

Chapter
Section
Textbook Problem

In Example 1.3.4 we arrived at a model for the length of daylight (in hours) in Philadelphia on the t th day of the year: L ( t ) = 12 + 2.8 sin [ 2 π 365 ( t − 80 ) ] Use this model to compare how the number of hours of daylight is increasing in Philadelphia on March 21 and May2l.

To determine

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

Explanation

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

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