Chapter 6.2, Problem 127AYU

a.

To determine

Calculate the time $T$ for $\theta =30°$

  $=\boxed{0.57\min utes}$.

Explanation of Solution

Given information: Two oceanfront homes are located $8$ miles apart on a straight stretch of beach, each a distance of $1$ mile from a paved road that parallels the ocean.

See the figure.

  

Sally can jog $8$ miles per hour along the paved road, but only $3$ miles per hour in the sand on the beach. Because of a river directly between the two houses, it is necessary to jog in the sand to the road, continue on the road, and then jog directly back in the sand to get from one house to the other. The time $T$ to get from one house to the other as a function of the angle $\theta$ shown in the illustration is

  $T\left(\theta \right)=1+\frac{2}{3\sin \theta }-\frac{1}{4\tan \theta \text{'}}$

  $0°<\theta <90°$

Calculate the time $T$ for $\theta =30°$ . How long is sally on the paved road?

Calculation: The two oceanfront houses, $8$ miles apart on the beach, that are $1$ mile apart from a paved path are shown in the below figure. Further one can only jog on the sand to the path because of river in between two houses.

  

If $\theta =30$ degree, then

  $T\left(30\right)=1+\frac{2}{3\sin \theta }-\frac{1}{4\tan \theta }$

  $=1+\frac{2}{3\times \frac{1}{2}}-\frac{1}{4\times 0.5773}$

  $=1+\frac{4}{3}-\frac{1}{2.309}$

  $=1+1.3333-0.4331$

  $=2.3333-1.4331$

  $\boxed{1.9hours}$

Mr.S remained on paved path to cover a total distance of $DF=2DE$ i.e., the time taken to cover $DF$ is double of time taken to cover $DE$ or Mr.S remained on both for.

  $=2\times \left(\frac{1}{2}-\frac{1}{8\tan 30}\right)$

  $=1-\frac{2}{8\times 0.5773}$

  $=1-0.433$

Hence, the time taken is

  $=\boxed{0.57\min utes}$

b.

To determine

Calculate the time $T$ for $\theta =45°$ .

Answer to Problem 127AYU

  $\boxed{0.75\min utes}$

Explanation of Solution

Given information:

Two oceanfront homes are located $8$ miles apart on a straight stretch of beach, each a distance of $1$ mile from a paved road that parallels the ocean.

See the figure.

  

Sally can jog $8$ miles per hour along the paved road, but only $3$ miles per hour in the sand on the beach. Because of a river directly between the two houses, it is necessary to jog in the sand to the road, continue on the road, and then jog directly back in the sand to get from one house to the other. The time $T$ to get from one house to the other as a function of the angle $\theta$ shown in the illustration is

  $T\left(\theta \right)=1+\frac{2}{3\sin \theta }-\frac{1}{4\tan \theta \text{'}}$

  $0°<\theta <90°$

Calculate the time $T$ for $\theta =45°$ . How long is sally on the paved road?

Calculation: The two oceanfront houses, $8$ miles apart on the beach, that are $1$ mile apart from a paved path are shown in the below figure. Further one can only jog on the sand to the path because of river in between two houses.

  

If $\theta =45$ degree, then

  $T(45)=1+\frac{2}{3\sin 45}-\frac{1}{4\tan 45}$

  $=1+\frac{2}{3\times 0.7071}-\frac{1}{4\times 1}$

  $=1+\frac{2}{2,1213}-0.25$

  $=1+0.9428-0.25$

  $=1.9428-0.25$

  $=\boxed{1.69hours}$

Hence, Mr.S remained on path for:

  $=2\times \left(\frac{1}{2}-\frac{1}{8\tan 45}\right)$

  $=1-\frac{2}{8\times 1}$

  $=1-0.25$

  $=\boxed{0.75\min utes}$

c.

To determine

Calculate the time $T$ for $\theta =60°$ .

Answer to Problem 127AYU

  $\boxed{0.86\min utes}$

Explanation of Solution

Given information:

Two oceanfront homes are located $8$ miles apart on a straight stretch of beach, each a distance of $1$ mile from a paved road that parallels the ocean.

See the figure.

  

Sally can jog $8$ miles per hour along the paved road, but only $3$ miles per hour in the sand on the beach. Because of a river directly between the two houses, it is necessary to jog in the sand to the road, continue on the road, and then jog directly back in the sand to get from one house to the other. The time $T$ to get from one house to the other as a function of the angle $\theta$ shown in the illustration is

  $T\left(\theta \right)=1+\frac{2}{3\sin \theta }-\frac{1}{4\tan \theta \text{'}}$

  $0°<\theta <90°$

Calculate the time $T$ for $\theta =60°$ . How long is sally on the paved road?

Calculation: The two oceanfront houses, $8$ miles apart on the beach, that are $1$ mile apart from a paved path are shown in the below figure. Further one can only jog on the sand to the path because of river in between two houses.

  

If $\theta =60$ degree, then

  $T(60)=1+\frac{2}{3\sin 60}-\frac{1}{4\tan 60}$

  $=1+\frac{2}{3\times 0.866}-\frac{1}{4\times 1.732}$

  $=1+\frac{2}{2.598}-\frac{1}{6.928}$

  $=1+0.7698-0.1443$

  $=\boxed{1.63hours}$

Hence, Mr.S remained on path for:

  $=2\times \left(\frac{1}{2}-\frac{1}{8\tan 60°}\right)$

  $=1-\frac{2}{8\times 1.732}$

  $=1-0.1443$

  $=\boxed{0.86\min utes}$

d.

To determine

Calculate the time $T$ for $\theta =90°$ .

Answer to Problem 127AYU

  $\boxed{1.67hours}$

Explanation of Solution

Given information:

Two oceanfront homes are located $8$ miles apart on a straight stretch of beach, each a distance of $1$ mile from a paved road that parallels the ocean.

See the figure.

  

Sally can jog $8$ miles per hour along the paved road, but only $3$ miles per hour in the sand on the beach. Because of a river directly between the two houses, it is necessary to jog in the sand to the road, continue on the road, and then jog directly back in the sand to get from one house to the other. The time $T$ to get from one house to the other as a function of the angle $\theta$ shown in the illustration is

  $T\left(\theta \right)=1+\frac{2}{3\sin \theta }-\frac{1}{4\tan \theta \text{'}}$

  $0°<\theta <90°$

Calculate the time $T$ for $\theta =90°$ . How long is sally on the paved road?

Calculation: The two oceanfront houses, $8$ miles apart on the beach, that are $1$ mile apart from a paved path are shown in the below figure. Further one can only jog on the sand to the path because of river in between two houses.

  

If $\theta =90$ degree, then

  $T(90)=1+\frac{2}{3\sin 90}-\frac{1}{4\tan 90}$

  $=1+\frac{2}{3\times 1}-\frac{1}{4\times 0}$

  $=1+\frac{2}{3}-\frac{1}{0}$

  $=1+0.67$

  $=\boxed{1.67hours}$

Hence, at $90$ degree, Mr.S will jog first through $AC$ , then $CG$ and then covering the distance $GB$ , he will reach to second house or he will not cover any slant path.