Chapter 4.1, Problem 41E

To determine

To find: The rate is the distance from the plane to the radar station increasing a minute later.

Answer to Problem 41E

The rate $\approx 296$ km/h is the distance from the plane to the radar station increasing a minute later.

Explanation of Solution

Given information:

The given constant speed is $300$ km/h, the angle of elevation is ${30}^o$ .

Calculation:

Consider the figure.

  

Figure $\left(1\right)$

In the figure,

The total angle at $A$ does not change while the plane climbing at ${30}^o$ .

  $A={30}^o+{90}^o={120}^o$

Convert the speed into per minute.

  $\frac{dc}{dt}=300\frac{km}{h}\cdot \frac{h}{60\min }=5\frac{km}{\min }$

Thus, $c=5$ .

Use the law cosines to find $a$ at $1$ minute.

  $\begin{array}{c}{a}^2=b^2+c^2-2bc\cos A\\ a^2=1^2+c^2-2\left(1\right)\left(c\right)\cos {120}^o\\ a^2=1+c^2-2c\cdot \left(-\frac{1}{2}\right)\\ a^2=1+c^2+c\\ a=\sqrt{1+25+5}\\ a=\sqrt{31}\end{array}$

Differentiate with respect to $t$ an equation to find $\frac{da}{dt}$ at $1$ minute.

  $\begin{array}{c}{a}^2=1+c^2+c\\ 2a\frac{da}{dt}=0+2c\frac{dc}{dt}+\frac{dc}{dt}\\ \frac{da}{dt}=\frac{\left(2c+1\right)\frac{dc}{dt}}{2a}\\ =\frac{\left(2\left(5\right)+1\right)5}{2\sqrt{31}}\\ =\frac{5}{2\sqrt{31}}\frac{km}{\min }\end{array}$

Convert into km/h.

  $\frac{55}{2\sqrt{31}}\frac{km}{\min }\cdot \frac{60\min }{h}\approx 296km/h$

Therefore, the rate $\approx 296$ km/h is the distance from the plane to the radar station increasing a minute later.