   Chapter 3.7, Problem 24E

Chapter
Section
Textbook Problem

# Find, correct to two decimal places, the coordinates of the point on the curve y = sin x that is closest to the point ( 4 , 2 ) .

To determine

To find:

The coordinates of the point on the curve y=sinx closest to the point (4,2) correct to two decimal places.

Explanation

1) Concept:

For finding the closest point, use the distance formula and then apply the first derivative test.

2) Test:

First derivative test:

Suppose c is a critical number of a continuous function f defined on an interval.

(a) If f'(x)>0 for all x<c and f'(x)<0 for all x>c, then f(c) is the absolute maximum value of f.

(b) If f'(x)<0 for all x<c and f'(x)>0 for all x>c, then f(c) is the absolute minimum value of f.

3) Formula:

Distance formula: The distance between two points x1,y1 and (x2,y2) is

d=x2-x12+y2-y12

4) Given:

i) The curve y=sinx

ii) The curve y=sinx is closest to the point (4,2)

5) Calculation:

The distance between point (4,2) and point (x,y) is

d=x-42+y-22

If (x,y) lies on the curve y=sinx then x=sin-1y

The expression for d becomes

d=x-42+sinx-22

For easy simplification instead of minimizing d, minimize its square

d2=fx=x-42+sinx-22

There are no restrictions on x,

The domain is all real numbers.

Differentiating fx with respect to x,

f'x=2x-4+2sinx-2cosx

=2x-4+2sinx-2cosx

Therefore,

f'x=2x-4+2sinx-2cosx

f'x=0

2x-4+2sinx-2cosx=0

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