Chapter 6.3, Problem 119AYU

Calculating the Time of a Trip From a parking lot, you want to walk to a house on the beach. The house is located 1500 feet down a paved path that parallels the ocean, which is 500 feet away. See the illustration. Along the path you can walk 300 feet per minute, but in the sand on the beach you can only walk 100 feet per minute.

The time $T$ to get from the parking lot to the beach house expressed as a function of the angle $\theta$ shown in the illustration is

$T\left(\theta \right)\text{=}\text{5}\text{-}\frac{5}{3\tan \theta }\text{+}\frac{5}{\sin \theta }$, $0\text{<}\theta \text{<}\frac{\pi }{2}$

Calculate the time $T$ if you walk directly from the parking lot to the house.

[ Hint: $\tan \theta =\frac{500}{1500}$.]

To determine

To calculate: The time of the trip if you walk directly from the parking lot to the house.

Answer to Problem 119AYU

Solution:

$\text{15}\text{.81}$ min.

Explanation of Solution

Given:

From a parking lot, you want to walk to a house on the beach. The house is located 1500 feet down a paved path that parallels the ocean, which is 500 feet away. See the illustration. Along the path you can walk 300 feet per minute, but in the sand on the beach you can only walk 100 feet per minute.

The time $T$ to get from the parking lot to the beach house expressed as a function of the angle $\theta$ shown in the illustration is,

$T\left(\theta \right)=5-\frac{5}{3\tan \theta }+\frac{5}{\sin \theta },0<\theta <\frac{\pi }{2}$

$\tan \theta =\frac{500}{1500}$

Calculation:

$T\left(\theta \right)=5-\frac{5}{3\tan \theta }+\frac{5}{\sin \theta }$

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

$\tan \theta =\frac{\text{opposite}\text{side}}{\text{Adjacent}\text{side}}=\frac{500}{1500}$

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

$=\sqrt{{500}^2+{1500}^2}=1581.14$

$\sin \theta =\frac{\text{opposite}\text{side}}{\text{hypotenuse}}=\frac{500}{1581.14}$

$T\left(\theta \right)=5-\frac{5}{3\frac{500}{1500}}+\frac{5}{\frac{500}{1581.14}}$

$T\left(\theta \right)=5-\frac{5}{1}+\frac{5}{\frac{500}{1581.14}}=15.81$ min.