Chapter 6.3, Problem 120AYU

Calculating the Time of a Trip Two oceanfront homes are located 8 miles apart on a straight stretch of beach, each a distance of 1 mile from a paved path that parallels the ocean. Sally can jog 8 miles per hour on the paved path, but only 3 miles per hour in the sand on the beach. Because a river flows directly between the two houses, it is necessary to jog in the sand to the road, continue on the path, and then jog directly back in the sand to get from one house to the other. See the illustration. The time $T$ to get from one house to the other as a function of the angle $\theta$ shown in the illustration is

$T\left(\theta \right)=\text{ 1 + }\frac{2}{3\sin \theta }\text{ - }\frac{1}{4\tan \theta }$, $0\text{ < }\theta \text{ < }\frac{\pi }{2}$

(a) Calculate the time $T$ for $\text{tan}\theta \text{ = }\frac{1}{4}$.

(b) Describe the path taken.

(c) Explain why $\theta$ must be larger than ${14}^{\circ }$.

To determine

To calculate: The time of the trip.

Answer to Problem 120AYU

Solution:

$15.81$ min.

Explanation of Solution

Given:

Two oceanfront homes are located 8 miles apart on a straight stretch of beach, each a distance of 1 mile from a paved path that parallels the ocean. Sally can jog 8 miles per hour on the paved path, but only 3 miles per hour in the sand on the beach. Because a river flows directly between the two houses, it is necessary to jog in the sand to the road, continue on the path, and then jog directly back in the sand to get from one house to the other. See the illustration. The time $T$ to get from one house to the other as a function of the angle $\theta$ shown in the illustration is,

$T\left(\theta \right)=1+\frac{2}{3\sin \theta }-\frac{1}{4\text{tan}\theta },0<\theta <\frac{\pi }{2}$

$\tan \theta =\frac{1}{4}$

Calculation:

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

$\sin \theta =\frac{\text{opposite}\text{side}}{\text{hypotenuse}}$

$\tan \theta =\frac{\text{opposite}\text{side}}{\text{Adjacent side}}=\frac{1}{4}$

$\text{hypotenuse}=\sqrt{{\left(\text{opposite}\text{side}\right)}^2+{\left(\text{Adjacent}\text{side}\right)}^2}$

$=\sqrt{1^2+4^2}=4.12$

$\sin \theta =\frac{\text{opposite}\text{side}}{\text{hypotenuse}}=\frac{1}{4.12}$

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

$T\left(\theta \right)=1+\frac{2}{3\times \frac{1}{4.12}}-\frac{1}{4\times \frac{1}{4}}$

$T\left(\theta \right)=1-2.75-1=2.75$ hours.