Q. 119 - Precalculus Enhanced with Graphing Utilities

Question

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 can be expressed as a function of the angle u shown in

the illustration and is

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

Calculate the time T if you walk directly from the parking lot

to the house.

Step-by-Step Solution

Answer

The time taken is 15.8 minutes.

Step 1. Given Information

Expression to calculate time

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

Speed = 300 feet per minute

Horizontal distance = 1500 feet

Vertical distance = 500 feet

Step 2. Calc ulation

$$\begin{gathered} From the figure \\ \tan \theta =\frac{y}{x} \\ \tan \theta =\frac{500}{1500} \\ \tan \theta =\frac{1}{3} \\ Now we will calculate Sin\theta \\ from the figure \\ Sin\theta =\frac{y}{r} \\ where, \\ r=\sqrt{x^2+y^2} \\ therefore, \\ \sin \theta =\frac{y}{\sqrt{x^2+y^2}} \\ and \sqrt{x^2+y^2}=\sqrt{3^2+1^2} \\ =\sqrt{10} \\ \sin \theta =\frac{1}{\sqrt{10}} \\ substituting this in the equation for time we get \\ T\left(\theta \right)=5-\frac{5}{3\tan \theta }+\frac{5}{\sin \theta }, 0<\theta <\frac{\mathrm{\pi }}{2} \\ T\left(\theta \right)=5-\frac{5}{3\times {\displaystyle \frac{1}{3}}}+\frac{5}{{\displaystyle \frac{1}{\sqrt{10}}}} \\ =5-5+5\sqrt{10} \\ =15.8 minutes \end{gathered}$$