Chapter 4.1, Problem 12E

To determine

To find: The rate at which the distance of ships changing at $4:00$ pm.

Answer to Problem 12E

The rate at which the distance of ships changing is $\frac{dC}{dt}=21.39\frac{km}{hr}$

Explanation of Solution

Given information:

The ship A is $150$ km of ship B.

Given speed of ship A is $35$ km/hr to east and ship B is $25$ km/hr to north.

Calculation:

The problem gives us $\frac{dA}{dt}=34\frac{km}{hr},\frac{dB}{dt}=25\frac{km}{hr},t=4$ hours.

  $A\left(0\right)=150$ km, and $B\left(0\right)=0$ km

We are asked to find rate at which the distance between the ships is changing after 4 hours, imagine a line C connecting both of the ships directly; we are finding figure (1)

  

Figure $\left(1\right)$

Looking at the upper figure, we can see we can use the Pythagorean theorem to relate A,B, and C

  $\begin{array}{l}{\left(150-A\left(t\right)\right)}^2+B{\left(t\right)}^2=c{\left(t\right)}^2\\ 22500+A{(}^t-300A\left(t\right)+B{\left(t\right)}^2=C{\left(t\right)}^2\end{array}$

We can also take the derivatives of this equation with respect to time in order to relate the rates of these quantities.

  $\begin{array}{l}2A\left(t\right)\frac{dA}{dt}-300\frac{dA}{dt}+2B\left(t\right)\frac{dB}{dt}=2C\left(t\right)\frac{dC}{dt}\\ 2\left(4*35\right)\left(35\right)-300\left(35\right)+2\left(4*25\right)\left(25\right)=2\left(100.499\right)\frac{dC}{dt}\\ 4300=201\frac{dC}{dt}\\ \frac{dC}{dt}=21.39\frac{km}{hr}\end{array}$

Therefore, the rate at which the distance of ships changing is $\frac{dC}{dt}=21.39\frac{km}{hr}$ .