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

### 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 82E
To determine

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

Expert Solution

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) .

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