High School

**Modeling with Mathematics**

The time [tex]t[/tex] (in seconds) it takes a trapeze artist to swing back and forth is represented by the function [tex]t=2 \pi \sqrt{\frac{r}{32}}[/tex], where [tex]r[/tex] is the rope length (in feet).

It takes the trapeze artist 6 seconds to swing back and forth. Is this rope [tex]\frac{3}{2}[/tex] as long as the rope used when it takes the trapeze artist 4 seconds to swing back and forth? Explain your reasoning.

Answer :

Let's solve the problem step by step to determine if the rope used for a 6-second swing is [tex]\(\frac{3}{2}\)[/tex] times as long as the rope used for a 4-second swing.

1. Understand the Function:
The time [tex]\( t \)[/tex] it takes for a trapeze artist to swing back and forth is given by:
[tex]\[
t = 2\pi \sqrt{\frac{r}{32}}
\][/tex]
where [tex]\( r \)[/tex] is the rope length in feet.

2. Find Rope Length for t = 6 seconds:
- Given [tex]\( t = 6 \)[/tex] seconds, substitute into the formula:
[tex]\[
6 = 2\pi \sqrt{\frac{r}{32}}
\][/tex]
- Solve for [tex]\( r \)[/tex]:
[tex]\[
\sqrt{\frac{r}{32}} = \frac{6}{2\pi}
\][/tex]
[tex]\[
\frac{r}{32} = \left(\frac{6}{2\pi}\right)^2
\][/tex]
[tex]\[
r = 32 \cdot \left(\frac{6}{2\pi}\right)^2
\][/tex]
- This calculation gives a rope length [tex]\( r_1 \approx 29.18 \)[/tex] feet for [tex]\( t = 6 \)[/tex] seconds.

3. Find Rope Length for t = 4 seconds:
- Given [tex]\( t = 4 \)[/tex] seconds, substitute into the formula:
[tex]\[
4 = 2\pi \sqrt{\frac{r}{32}}
\][/tex]
- Solve for [tex]\( r \)[/tex]:
[tex]\[
\sqrt{\frac{r}{32}} = \frac{4}{2\pi}
\][/tex]
[tex]\[
\frac{r}{32} = \left(\frac{4}{2\pi}\right)^2
\][/tex]
[tex]\[
r = 32 \cdot \left(\frac{4}{2\pi}\right)^2
\][/tex]
- This calculation gives a rope length [tex]\( r_2 \approx 12.97 \)[/tex] feet for [tex]\( t = 4 \)[/tex] seconds.

4. Compare the Rope Lengths:
- Check if [tex]\( r_1 \)[/tex] is [tex]\(\frac{3}{2}\)[/tex] times [tex]\( r_2 \)[/tex]:
[tex]\[
\frac{3}{2} \times r_2 = \frac{3}{2} \times 12.97 \approx 19.455
\][/tex]
- Compare with [tex]\( r_1 \approx 29.18 \)[/tex].

Since [tex]\( 29.18 \)[/tex] is not equal to [tex]\( 19.455 \)[/tex], the rope used for the 6-second swing is not [tex]\(\frac{3}{2}\)[/tex] times as long as the rope used for the 4-second swing. Thus, the answer is No.