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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 86E

a.

To determine

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

Expert Solution

Answer to Problem 86E

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

Answer to Problem 86E

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[11]=0

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

c.

To determine

Draw a graph for the cycloid for the given value of r .

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.

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.4, Problem 86E

The tangents are x=2π,x=0andy=2 .

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