Q. 98 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

(a) The cross-sectional area of the beam is $\frac{8}{9}\sqrt{2}$ if the length of half the base of the beam is One-third of a foot.

(b) The cross-sectional area of the beam is $\sqrt{3}$ if the length of half the base of the beam is One-half of a foot.

(c) The cross-sectional area of the beam is $\frac{8}{9}\sqrt{5}$ if the length of half the base of the beam is Two-thirds of a foot

Step 1. Given Information

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

Part (a) Step 1. The given function is A ( x ) = 4 x 1 - x 2

We have to find the cross-sectional area of the beam if the length of half the base of the beam is One-third of a foot.

So the value of $x=\frac{1}{3}$

Part (a) Step 2. Putting the value of x in the given function.

$$\begin{gathered} A\left(\frac{1}{3}\right)=4\times \frac{1}{3}\sqrt{1-{\left(\frac{1}{3}\right)}^2} \\ A\left(\frac{1}{3}\right)=\frac{4}{3}\sqrt{1-\frac{1}{9}} \\ A\left(\frac{1}{3}\right)=\frac{4}{3}\sqrt{1\times \frac{9}{9}-\frac{1}{9}} \\ A\left(\frac{1}{3}\right)=\frac{4}{3}\sqrt{\frac{9-1}{9}} \\ A\left(\frac{1}{3}\right)=\frac{4}{3}\sqrt{\frac{8}{9}} \\ A\left(\frac{1}{3}\right)=\frac{4}{3}\times \frac{2}{3}\sqrt{2} \\ A\left(\frac{1}{3}\right)=\frac{8}{9}\sqrt{2} \end{gathered}$$

Part (b) Step 1. We have to find the cross-sectional area of the beam if the length of half the base of the beam is One-half of a foot.

So the value of $x=\frac{1}{2}$

Part (b) Step 2. Putting the value of x in the given function.

$$\begin{gathered} A\left(\frac{1}{2}\right)=4\times \frac{1}{2}\sqrt{1-{\left(\frac{1}{2}\right)}^2} \\ A\left(\frac{1}{2}\right)=2\sqrt{1-\frac{1}{4}} \\ A\left(\frac{1}{2}\right)=2\sqrt{1\times \frac{4}{4}-\frac{1}{4}} \\ A\left(\frac{1}{2}\right)=2\sqrt{\frac{4-1}{4}} \\ A\left(\frac{1}{2}\right)=2\sqrt{\frac{3}{4}} \\ A\left(\frac{1}{2}\right)=2\times \frac{1}{2}\sqrt{3} \\ A\left(\frac{1}{2}\right)=\sqrt{3} \end{gathered}$$

Part (b) Step 1. We have to find the cross-sectional area of the beam if the length of half the base of the beam is two-thirds of a foot.

So the value of $x=\frac{2}{3}$

Part (c) Step 2. Putting the value of x in the given function.

$$\begin{gathered} A\left(\frac{2}{3}\right)=4\times \frac{2}{3}\sqrt{1-{\left(\frac{2}{3}\right)}^2} \\ A\left(\frac{2}{3}\right)=\frac{8}{3}\sqrt{1-\frac{4}{9}} \\ A\left(\frac{2}{3}\right)=\frac{8}{3}\sqrt{1\times \frac{9}{9}-\frac{4}{9}} \\ A\left(\frac{2}{3}\right)=\frac{8}{3}\sqrt{\frac{9-4}{9}} \\ A\left(\frac{2}{3}\right)=\frac{8}{3}\sqrt{\frac{5}{9}} \\ A\left(\frac{2}{3}\right)=\frac{8}{3}\times \frac{1}{3}\sqrt{5} \\ A\left(\frac{2}{3}\right)=\frac{8}{9}\sqrt{5} \end{gathered}$$