BuyFind

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 84E
To determine

Explain that the given curve has two tangent lines at the origin.

Expert Solution

Explanation of Solution

Given:

The given equations are x=sint and y=sin(t+sint) .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt[sin(t+sint)]

Apply chain rule.

  df(a)dx=dfdadadx

Let f=sina,a=t+sint

  dydt=dda(sina)ddt(t+sint)

Apply sum/difference rule.

  (f±g)'=f'±g'

  dydt=dda(sina)[ddt(t)+ddt(sint)]

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

  dydt=cosa[1+cost]

Substitute the value of a=t+sint .

  dydt=cos(t+sint)[1+cost]

  dxdt=sint

Use derivative rules ddx(sinx)=cosx

  dxdt=cost

  dydx(slope)=cos(t+sint)[1+cost]cost

Find the value of the parameter t at the origin.

  sint=0t=[0,π]

At t=0 .

  dydx(slope)=cos(t+sint)[1+cost]cost=cos(0+sin(0))[1+cos(0)]cos(0)=cos(0)[1+1]1=121=2

At t=π .

  dydx(slope)=cos(t+sint)[1+cost]cost=cos(π+sin(π))[1+cos(π)]cos(π)=cos(π+0)[11]1=0

There are two different tangents at the origin because there is two different slope values.

Now use point-slope form for the tangent equations.

  yy1=m(xx1)y1=0,x1=0,m=2y0=2(x0)y=2xandy1=0,x1=0,m=0y0=0(x0)y=0

The graph of the given curve and the tangent lines is given below.

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 3.4, Problem 84E

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