Chapter 2.1, Problem 98AYU

Cross-sectional Area The cross-sectional area of a beam cut from a log with radius 1 foot is given by the function $A\left(x\right)=4x\sqrt{1-x^2}$, where $x$ represents the length, in feet, of half the base of the beam. See the figure. Determine the cross-sectional area of the beam if the length of half the base of the beam is as follows:

a. One-third of a foot

b. One-half of a foot

c. Two-thirds of a foot

To determine

To find: The cross sectional area of a beam cut from a log with radius 1 foot is given by the function $A\left(x\right) = 4x\sqrt{1 - x^2}$ where $x$ represents the length, in feet, of the base of the beam is

1. One - third of a foot
2. One - half of a foot
3. Two - thirds of a foot

Answer to Problem 98AYU

a) $\frac{8\sqrt{2}}{9}$; b) $\sqrt{3}$; c) $\frac{8\sqrt{5}}{9}$;

Explanation of Solution

Given:

$A\left(x\right) = 4x\sqrt{1 - x^2}$ where $x$ is the length of the base of the beam (in feet)

Calculation:

$A\left(x\right)=4x\sqrt{1 - x^2}$

a) when $x = \text{one - third of a foot}$

$A\left(x\right)=4x\sqrt{1 - x^2}$

$= 4 \times \frac{1}{3}\sqrt{1 - \frac{1}{9}}$

$= \frac{4}{3} \times \sqrt{\frac{9 - 1}{9}}$

$= \frac{4}{3} \times \sqrt{\frac{8}{9}}$

$= \frac{4}{3} \times \frac{2\sqrt{2}}{3}$

$= \frac{8\sqrt{2}}{9}$

b) when $x = \text{one - half of a foot}$

$A\left(x\right)=4x\sqrt{1 - x^2}$

$= 4 \times \frac{1}{2} \times \sqrt{1 - \frac{1}{4}}$

$= 2 \times \frac{\sqrt{3}}{2}$

$= \sqrt{3}$

c) when $x = \text{two - thirds of a foot}$

$A\left(x\right)=4x\sqrt{1 - x^2}$

$= 4 \times \frac{2}{3} \times \sqrt{1 - \frac{4}{9}}$

$= \frac{8}{3} \times \sqrt{\frac{9 - 4}{9}}$

$= \frac{8}{3} \times \frac{\sqrt{5}}{3}$

$= \frac{8\sqrt{5}}{9}$