   Chapter 7.6, Problem 25E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding Distance In Exercises 23–28, find the minimum distance from the curve or surface to the given point. (Hint: Start by minimizing the square of the distance.)Parabola: y = x 2 ,   ( − 3 ,   0 ) Minimize d 2 = ( x + 3 ) 2 + y 2

To determine

To calculation: The minimum distance from the parabola y=x2 to the point (3,0) and minimize d2=(x+3)2+y2

Explanation

Given Information:

The equation for the parabola y=x2 at the point (3,0) and minimum distance d2=(x+3)2+y2

Formula Used:

If f has a relative extreme at (x0,y0) in an open region R and the xy – plane. And the first partial derivatives of f exist in R, then

fx(x0,y0)=0fy(x0,y0)=0

Calculation:

Consider the given equation of parabola.

y=x2

Now, differentiate the above equation with respect to x.

dydx=2x

Consider the given equation of distance.

d2=(x+3)2+y2

Now, differentiate the above equation with respect to x.

d(d2)dx=ddx((x+3)2+y2)d(d2)dx=2(x+3)+2ydydx

Substitute the value of dydx and y in above equation.

d(d2)dx=2(x+3)+22xx2d(d2)dx=4x3+2x+6F=4x3+2x+6F=12x2+2

Consider the primary equation and equate above expression to zero

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