# 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 83E
To determine

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

Expert Solution

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) .

### 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!