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 81E
To determine

Find the equation of the tangent line usinggiven data.

Expert Solution

Answer to Problem 81E

The equation of the tangent line is y=2xe+3 .

Explanation of Solution

Given:

The given curves are x=et and y=tlnt2 . The parametric value is t=1 .

Calculation:

Find the x and y value at t=1 .

  x=et=e1=e

  y=tlnt2=1ln(12)=1ln(1)=1

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt(tlnt2)

Applydifferencerule (fg)'=f'g' .

  dydt=ddt(t)ddt(lnt2)

Use derivative rule ddx(xn)=nxn1 and ddx(ln(x))=1x .

  dydt=11t22t=12t

  dxdt=ddt(et)

Use derivative rule ddx(ex)=ex and ddx(x)=12x

  dxdt=et12t

  dydx(slope)=12tet12t

At the value t=1 .

  dydx(slope)=121e1121=1e2=2e

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

  yy1=m(xx1)y1=1,x1=e,m=2ey1=2e(xe)y1=2xe+2y1+1=2xe+2+1y=2xe+3

Hence theequation of the tangent line is y=2xe+3 .

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