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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270336
Textbook Problem

Find equations of the tangents to the curve x = 3t2 + 1, y = 2t3 + 1 that pass through the point (4, 3).

To determine

To find: The equation of the tangent to the curve for the parametric equations x=3t2+1 and y=2t3+1 .

Explanation

Given:

The parametric equation for the variable x is as follows.

x=3t2+1 (1)

The parametric equation for the variable y is as follows.

y=2t3+1 (2)

Calculation:

Differentiate the parametric equation x with respect to t .

x=3t2+1

dxdt=6t

Differentiate the parametric equation y with respect to t .

y=2t3+1dydt=6t2

Write the chain rule for dydx .

dydx=dydtdxdt

Substitute (6t2) for dydt and (6t) for dxdt in the above equation.

dydx=(6t2)(6t)=t

The tangent is horizontal when the condition dydt=0 is true.

dydt=02t=0t=0

The point corresponding to the value t with an equation of the tangent line is

[y(2t3+1)]=t[x(3t2+1)]

The tangent line passes through (4,3) .

Substitute 4 for x and 3 for y in the expression [y(2t3+1)]=t[x(3t2+1)] .

[y(2t3+1)]=t[x(3t2+1)](3)(2t3+1)=4t3t3tt33t+2=0(t+1)2(t+2)=0t=2,t=1

Substitute -2 for t in equation (1).

x=3t2+1=3(2)2+1=12+1=13

Substitute -2 for t in equation (2)

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