# Find the equation of the tangent line usinggiven data.

### 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 90E
To determine

## Find the equation of the tangent line usinggiven data.

Expert Solution

The equation of the tangent line is y=x1 .

### Explanation of Solution

Given:

The given curves are x=3t2+1 and y=2t3+1 . The tangent passes through the point (4,3) .

Calculation:

Find the slope.

Apply formula dydx=dydtdxdt .

dydt=ddt(2t3+1)

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

dydt=ddt(2t3)+ddt(1)

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

dydt=6t2+0dydt=6t2

dxdt=ddt(3t2+1)

Apply sum rule (f+g)'=f'+g' .

dxdt=ddt(3t2)+ddt(1)

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

dxdt=6t+0dxdt=6t

dydx(slope)=6t26t=t

Substitute the value of x=4 and y=3 in the given equations and find the value of t .

3t2+1=43t2+11=413t2=333t2=33t2=1(i)2t3+1=32t3+11=312t3=222t3=22t3=1(ii)

Divide equation (ii) by (i) .

t3t2=11t=1

At the value t=1 .

dydx(slope)=t=1

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

yy1=m(xx1)y1=3,x1=4,m=1y3=1(x4)y3=x4y3+3=x4+3y=x1

Hence theequation of the tangent line is y=x1 .

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