# Find the equation of the tangent line usinggiven data.

### 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 80E
To determine

## Find the equation of the tangent line usinggiven data.

Expert Solution

The equation of the tangent line is y=x2+12 .

### Explanation of Solution

Given:

The given curves are x=cosθ+sin2θ and y=sinθ+cos2θ . The parametric value is θ=0 .

Calculation:

Find the x and y value at θ=0 .

x=cosθ+sin2θ=cos(0)+sin(20)=1

y=sinθ+cos2θ=sin(0)+cos(20)=1

Find the slope.

Apply formula dydx=dydtdxdt .

dydθ=ddθ(sinθ+cos2θ)

Apply sum rule (f+g)'=f'+g' .

dydθ=ddθ(sinθ)+ddθ(cos2θ)

Use derivative rule ddx(sinθ)=cosθ and ddx(cosθ)=sinθ .

dydθ=cosθsin2θ2dydθ=cosθ2sin2θ

dxdθ=ddθ(cosθ+sin2θ)

Apply sum rule (f+g)'=f'+g' .

dxdθ=ddθ(cosθ)+ddθ(sin2θ)

Use derivative rule ddx(sinθ)=cosθ and ddx(cosθ)=sinθ .

dxdθ=sinθ+2cos2θ

dydx(slope)=cosθ2sin2θsinθ+2cos2θ

At the value θ=0 .

dydx(slope)=cosθ2sin2θsinθ+2cos2θ=cos(0)2sin(20)sin(0)+2cos(20)=12

Use point-slope form for the equation of tangent line.

yy1=m(xx1)y1=1,x1=1,m=12y1=12(x1)y1=x212y1+1=x212+1y=x2+12

Hence theequation of the tangent line is y=x2+12 .

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