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 83E
To determine

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

Expert Solution

Answer to Problem 83E

The tangents are horizontal at the points (6,16) and (6,16) . The tangent is vertical at the point (10,0) and (7,6) .

Explanation of Solution

Given:

The given curves are x=10t2 and y=t312t .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt(t312t)

Applydifferencerule (fg)'=f'g' .

  dydt=ddt(t3)ddt(12t)

Use derivative rule ddx(xn)=nxn1 .

  dydt=3t212dydt=3(t24)

  dxdt=ddt(10t2)

Apply sum/difference rule.

  dxdt=ddt(10)ddt(t2)

Use derivative rules ddx(xn)=nxn1 and ddx(constant)=0

  dxdt=02tdxdt=2t

  dydx(slope)=3t2122t=123t22t

For the horizontal tangent line.

  dydx=0123t22t=0123t2=03(4t2)=04t2=0t2=4t=±4t=2,2

Substitute t=2 in the given equations.

  x=10t2=10(2)2=6y=t312t=(2)312(2)=16(x,y)=(6,16)

Substitute t=2 in the given equations.

  x=10t2=10(2)2=6y=t312t=(2)3122=16(x,y)=(6,16)

For vertical tangent line dxdt=0 .

  dxdt=02t=0t=0

Substitute t=0 in the given equations.

  x=10t2=1002=10y=t312t=(0)3120=0(x,y)=(10,0)

Hence thetangents are horizontal at the points (6,16) and (6,16) . The tangent is vertical at the point (10,0) and (7,6) .

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