Chapter 4.7, Problem 15E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Minimum Distance In Exercises 15 and 16, find the points on the graph of the function that are closest to the given point. y = x 2 , ( 0 , 3 )

To determine

To calculate: The point on the graph for the function y=x2 that lie closest to the point (0,3).

Explanation

Given:

The function y=x2 and the point (0,3).

Formula Used:

For a function f that is twice differentiable on an open interval I, if f'(c)=0 for some c, then,

If f''(c)>0 the function f has relative minima at c if f''(c)<0 the function f has relative maxima at c. If f''(c)=0, the test fails.

Calculation:

Consider the desired point to be (x,y).

Then the quantity to be minimized or the primary equation is:

d=(0âˆ’x)2+(3âˆ’y)2

The desired point satisfies the function y=x2. This implies:

(x,y)=(x,x2)

Substitute this back in the primary equation to obtain:

d=(0âˆ’x)2+(3âˆ’x2)2=x4âˆ’5x2+9

Here as x is the x-coordinate of a point on the function, the feasible domain is:

âˆ’âˆžâ‰¤xâ‰¤âˆž

Differentiate the equation obtained and equate it to zero to obtain,

12x4âˆ’5x2+9(4x3âˆ’10x)=02x(2x2âˆ’5)=0x2=52x=Â±52

Now, replace this in the second derivative and check the sign.

d''(x)=2x6âˆ’15x4+54x2âˆ’45(x4âˆ’5x2+9)32d''(52)=2(52)6âˆ’15(52)4+54(52)2âˆ’45((52)4âˆ’5(52)2+9)32=27

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