Chapter 5.6, Problem 35E

To determine

To find: the ships moving apart when $\text{OA}=5$ and $\text{OB}=3$ nautical miles.

Answer to Problem 35E

The value of $\frac{dz}{dx}$ is $29.5\text{ knots}$

Explanation of Solution

Given information :

  $x=5$

  $y=3$

  $\frac{dx}{dt}=14\text{knots}$

  $\frac{dy}{dt}=21\text{knots}$

All variables are differentiable functions of t .

Calculation:

We have to find the ships moving apart when $\text{OA}=5$ and $\text{OB}=3$ nautical miles.

Let’s suppose OA = x , OB = y and AB = z .

Therefore,

  $z^2=x^2+y^2-2xy\cos \theta$

Putting the value of $x=5$ , $y=3$ , $\theta ={120}^{\circ }$ in above equation,

  $\begin{array}{l}{z}^2={\left(5\right)}^2+{\left(3\right)}^2-\left(2\right)\left(5\right)\left(3\right)\cos \left({120}^{\circ }\right)\\ z^2=25+9-30\left(-\frac{1}{2}\right)\\ z^2=25+9+15\\ z=\sqrt{49}\\ z=7\end{array}$

Again,

  $z^2=x^2+y^2-2xy\cos \theta$

Putting $\theta ={120}^{\circ }$ in above equation,

  $\begin{array}{l}{z}^2=x^2+y^2-2xy\left(-\frac{1}{2}\right)\\ z^2=x^2+y^2+xy\end{array}$

Differentiate with respect to t using product rule.

  $2z\frac{dz}{dx}=2x\frac{dx}{dt}+2y\frac{dy}{dt}+y\frac{dx}{dt}+x\frac{dy}{dt}$

Putting the value of $x=5$ , $y=3$ , $z=7$ , $\frac{dx}{dt}=14$ and $\frac{dy}{dt}=21$ in above equation,

  $\begin{array}{l}2\left(7\right)\frac{dz}{dx}=2\left(5\right)\left(14\right)+2\left(3\right)\left(21\right)+\left(3\right)\left(14\right)+\left(5\right)\left(21\right)\\ 14\frac{dz}{dx}=140+126+42+105\\ 14\frac{dz}{dx}=413\\ \frac{dz}{dx}=\frac{413}{14}\\ \frac{dz}{dx}=29.5\text{ knots}\end{array}$

Hence, the value of $\frac{dz}{dx}$ is $29.5\text{ knots}$