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 82E
To determine

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

Expert Solution

Answer to Problem 82E

The tangent are horizontal at the points (0,1) and (13,2) . The tangent are vertical at the points (20,3) and (7,6) .

Explanation of Solution

Given:

The given curves are x=2t3+3t212t and y=2t3+3t2+1 .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

  dydt=ddt(2t3+3t2+1)

Applysumrule (f+g)'=f'+g' .

  dydt=ddt(2t3)+ddt(3t2)+ddt(1)

Use derivative rule ddx(xn)=nxn1 and ddx(constant)=0 .

  dydt=23t2+32t+0dydt=6t2+6tdydt=6t(t+1)

  dxdt=ddt(2t3+3t212t)

Apply sum/difference rule.

  dxdt=ddt(2t3)+ddt(3t2)ddt(12t)

Use derivative rule ddx(xn)=nxn1 .

  dxdt=23t2+32t12dxdt=6t2+6t12dxdt=6(t2+t2)

  dydx(slope)=6t(t+1)6(t2+t2)=t(t+1)(t2+t2)

For the horizontal tangent line.

  dydx=0t(t+1)(t2+t2)=0t(t+1)=0t=0,1

Substitute t=0 in the given equations.

  x=2t3+3t212t=203+302120=0y=2t3+3t2+1=203+302+1=1(x,y)=(0,1)

Substitute t=1 in the given equations.

  x=2t3+3t212t=2(1)3+3(1)212(1)=13y=2t3+3t2+1=2(1)3+3(1)2+1=2(x,y)=(13,2)

For vertical tangent line dxdt=0 .

  dxdt=06(t2+t2)=0t2+t2=0t2+2tt2=0t(t+2)(t+2)=0(t+2)(t1)=0t=2,1

Substitute t=2 in the given equations.

  y=2t3+3t2+1=2(2)3+3(2)2+1=3y=2t3+3t2+1=2(2)3+3(2)212(2)=20(x,y)=(20,3)

Substitute t=1 in the given equations.

  x=2t3+3t212t=2(1)3+3(1)212(1)=7y=2t3+3t2+1=2(1)3+3(1)2+1=6(x,y)=(7,6)

Hence thetangent are horizontal at the points (0,1) and (13,2) . The tangent are vertical at the points (20,3) and (7,6) .

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!