   Chapter 2.3, Problem 56E

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# Find equations of the tangent line and normal line to the curve at the given point. y 2 = x 3 ,   ( 1 , 1 )

To determine

To find:

Equation of tangent line and normal line to the curve y2= x3 at (1, 1)

Explanation

Concept:

Differentiate the given function to find the slope of tangent line and then find its negative reciprocal to find slope of normal line. Then, using the given point and these two slopes, find the equation of tangent line and normal line.

Formula used:

Power rule:  ddxxn=nxn-1

Slope point formula:  y-y1=m  x- x1

Given:

We have y2=x3

Calculation:

Solving for y: y= ± x3

This can be rewritten as y= ±x32

According to the given point (1, 1), the required equation of the curve is only for the positive part of the function.

So, y=x32

Then y'=32x12

Substituting x = 1, y'=32112=32

So this is the slope of tangent line at x = 1

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