   Chapter 3.7, Problem 58E

Chapter
Section
Textbook Problem

# What is the smallest possible area of the triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola y = 4 − x 2 at some point?

To determine

To find:

The smallest possible area of the triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola y=4-x2 at some point

Explanation

1) Concept:

Equation of the tangent line at the point (x1, y1) is given by

y-y1=m(x-x1)

Area of triangle-

A=12·base·height

2) Test:

First derivative test- if f' changes from negative to positive at c then f  has local minimum at c

Second derivative test- Suppose f'' is continuous near c

i. If f'c=0 and f''c>0 then f  has local minimum at c

ii. If f'c=0 and f''c<0 then f  has local maximum at c

3) Given:

The parabola y=4-x2

4) Formula:

5) Calculation:

Given that

y=4-x2

Let (a, 4-a2) be the point of the curve y=4-x2

Differentiate the above equation using the power rule

y'=0-2x2-1

Simplify

y'=-2x

Equation of the tangent line at the point (a, 4-a2) is given by

y-(4-a2)=-2a(x-a)

Simplify

y-4-a2=2a2-2xa

y-4-a2+4-a2=2a2-2xa+4-a2

Simplify

y=a2-2xa+4

By setting x=0 and y=0 we will get endpoints

Put x=0 in y=a2-2xa+4

y=a2-2(0)a+4

Simplify

y=a2+4

This is y- intercept

And Put y=0 in y=a2-2xa+4

0=a2-2xa+4

Simplify

0+ 2xa =a2-2xa+4+ 2xa

Simplify

2xa =a2+4

Divide by 2a

2xa 2a=a2+42a

Simplify

x=a2+42a

This is x- intercept

The area of the triangle is A=12

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