# Find the equation of the tangent lines at the given point. ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 3.4, Problem 86E

a.

To determine

## Find the equation of the tangent lines at the given point.

Expert Solution

The equation of tangent line is y=3xr(3π32) .

### Explanation of Solution

Given:

The given cycloid x=r(θsinθ) and y=r(1cosθ) , where θ=π3 .

Calculation:

Find the slope.

Apply formula dydx=dydθdxdθ .

dydθ=ddθ[r(1cosθ)]

Apply difference rule (fg)'=f'g' .

dydθ=r[ddθ(1)ddθ(cosθ)]

Use derivative rule ddx(constant)=0 and ddx(cosx)=sinx .

dydθ=r[0(sinθ)]dydθ=rsinθ

dxdθ=ddθ[r(θsinθ)]

Apply difference rule (fg)'=f'g' .

dxdθ=r[ddθ(θ)dxdθ(sinθ)]

Use derivative rule ddx(xn)=nxn1 and ddx(sinx)=cosx .

dxdθ=r[1cosθ]

dydx(slope)=rsinθr[1cosθ]=sinθ1cosθ

At θ=π3 .

dydx(slope)=sinθ1cosθ=sinπ31cosπ3=32112=3212=3

Find the value of value of x and y at θ=π3 .

x=r(π3sinπ3)=r(π332)

y=r(1cos(π3))=r(112)=r2

Now use point-slope form for the tangent equations.

yy1=m(xx1)y1=r2,x1=r(π332),m=3yr2=3[xr(π332)]

yr2=3xr(3π332)y=3xr(3π332)+r2y=3xr(3π33212)y=3xr(3π32)

Hence the equation of tangent line is y=3xr(3π32) .

b.

To determine

### Find the horizontal and vertical tangent line points on the curve.

Expert Solution

The tangent is horizontal at the point (r(π+2nπ),2r) and the vertical at the point (2rnπ,0) .

### Explanation of Solution

Given:

The given cycloid x=r(θsinθ) and y=r(1cosθ) , where θ=π3 .

Calculation:

Find the slope.

Apply formula dydx=dydθdxdθ .

dydθ=ddθ[r(1cosθ)]

Apply difference rule (fg)'=f'g' .

dydθ=r[ddθ(1)ddθ(cosθ)]

Use derivative rule ddx(constant)=0 and ddx(cosx)=sinx .

dydθ=r[0(sinθ)]dydθ=rsinθ

dxdθ=ddθ[r(θsinθ)]

Apply difference rule (fg)'=f'g' .

dxdθ=r[ddθ(θ)dxdθ(sinθ)]

Use derivative rule ddx(xn)=nxn1 and ddx(sinx)=cosx .

dxdθ=r[1cosθ]

dydx(slope)=rsinθr[1cosθ]=sinθ1cosθ

For the horizontal tangent line.

dydx=0sinθ1cosθ=0sinθ=0θ=nπcosθ1θ2nπθ=π+2nπ

Substitute θ=π+2nπ in the given equations.

x=r[π+2nπsin(π+2nπ)]=r[π+2nπ0]=r[π+2nπ]

y=r[1cos(π+2nπ)]=r[1(1)]=2r

For vertical tangent line dxdθ=0 .

r[1cosθ]=01cosθ=0cosθ=1θ=2nπ

Substitute θ=2nπ in the given equations.

x=r[2nπsin(2nπ)]=r[2nπ0]=2rnπ

y=r[1cos(2nπ)]=r=0

Hence the tangent is horizontal at the point (r(π+2nπ),2r) and the vertical at the point (2rnπ,0) .

c.

To determine

Expert Solution

### Explanation of Solution

Given:

The given cycloid x=r(θsinθ) and y=r(1cosθ) , where θ=π3 . The given value of r=1 .

Calculation:

At r=1

Theequation of tangent line is y=3xr(3π32)=3x(3π32) .

Hence the graph of the cycloid is given below. The tangents are x=2π,x=0andy=2 .

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