   Chapter 4.5, Problem 37E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use the guidelines of this section to sketch the curve. y = sin x + 3 cos x , −2π ≤ x ≤ 2π

To determine

To Sketch: The curve by plotting x and y coordinates using guidelines.

Explanation

Given:

The Cartesian equation is as below.

y=f(x)=sinx+3cosx (1)

Calculation:

a)

Calculate the domain.

The function is defined for the given interval 2π<x<2π . So the domain is (2π,2π) .

b)

Calculate the intercepts.

Calculate the value of x -intercept.

Substitute 0 for y in the equation (1).

sinx+3cosx=0

Divide both sides by cosx in the above equation.

sinx+3cosxcosx=0cosxsinxcosx+3cosxcosx=0

Substitute tanx for sinxcosx in the above equation.

tanx+3=0tanx=3x=tan1(3)x=2π3+nπ

Substitute -1 for n in the above equation.

x=2π3+π=2π3π3x=π3

Substitute -2 for n in the above equation.

x=2π3+2π=2π6π3x=4π3

Substitute 0 for n in the above equation.

x=2π3+0x=2π3

Substitute 1 for n in the above equation.

x=2π3+(1)πx=2π+3π3=5π3

Hence x -intercept points are (π3,0) , (4π3,0) , (2π3,0) and (5π3,0) for the domain (2π,2π) .

Calculate the y -intercept.

Substitute 0 for x in the equation (1).

f(0)=sin(0)+3cos(0)=3

Therefore, the y -intercept is (0,3) .

c)

The function f is periodic with period 2π .

d)

Calculate asymptotes.

Apply limit of x tends to (x) in the equation (1).

limxy=

Apply limit of x tends to (x) in the equation (1).

limxy=

Here, the value of limit gets infinity; this implies there is no horizontal asymptote or vertical asymptote.

e)

Calculate the intervals.

Differentiate the equation (1) with respect to x .

f'(x)=cosx3sinx (2)

Substitute 0 for f'(x) in the equation (8).

cosx3sinx=03sinx=cosxsinxcosx=13tanx=13x=tan1(13)x=11π6,5π6,π6or7π6

Take the interval (11π6,5π6) .

Substitute π for x in the equation (2).

f'(π)=cos(π)3sin(π)=10=1

Here the condition f'(x)<0 is true, hence the function f is decreasing on (11π6,5π6) .

Take the interval (π6,7π6) .

Substitute π for x in the equation (2).

f'(π)=cos(π)3sin(π)=1

Here the condition f'(x)<0 is true, hence the function f is decreasing on (π6,7π6) .

Take the interval (2π,11π6) .

Substitute 23π12 for x in the equation (2).

f'(23π12)=cos(23π12)3sin(23π12)=6+24(6+324)=6+2+6324=622=3.172

Here the condition f'(x)>0 is true, hence the function f is increasing on (2π,11π6)

Thus the function f is increasing in the interval (2π,11π6) and decreasing in the interval (11π6,5π6) and (π4,π4) .

f)

Calculate the local maximum and minimum value.

Refer the increasing and decreasing information from the part (e).

Calculate the local maximum.

Substitute 11π6 for x in the equation (1).

f(11π6)=sin(11π6)+3cos(11π6)=12+3(32)=12+32=2

Therefore local maximum occurs at (11π6,2) .

Calculate the local minimum.

Substitute 5π6 for x in the equation (1).

f(5π6)=sin(5π6)+3cos(5π6)=12+3(32)=1232=2

Therefore local minimum occurs at (5π6,2) .

g)

Calculate the concavity.

Differentiate the equation (2) with respect to x .

f''(x)=sinx3cosx (3)

Substitute 0 for f''(x) in the equation (9).

sinx3cosx=0sinx=3cosxsinxcosx=3tanx=3x=tan1(3)x=4π3,π3,2π3or5π3

Take the interval (4π3,π3)

Substitute π in the equation (3).

f''(π)=sin(π)3cos(π)=0+3=1.732

Here, the function f''(x)>0 in the interval (4π3,π3) is true

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