Answer to Question #343467 in Calculus for Amy Lee

Question

A ship is 20km west of another ship B. If a sails at 10km/hr. and at the same time B sails north 30km/hr. Find the rate of change of distance between them at the end of half hour.

Answer 1

Let the equation of the motion of ship A is $x(t)=-20+10t.$

Let the equation of the motion of ship B is $y(t)=20t.$

Then the distance between ships will be

$s(t)=\sqrt{(x(t))^2+(y(t))^2}$

$=\sqrt{(-20+10t)^2+(20t)^2}$

$=10\sqrt{4-4t+t^2+4t^2}$

$=10\sqrt{5t^2-4t+4}$

The rate of change of distance between them will be

$ds/dt=\dfrac{10(10t-4)}{2\sqrt{5t^2-4t+4}}$

$=\dfrac{10(5t-2)}{\sqrt{5t^2-4t+4}}$

At the end of half hour

$ds/dt|_{t=1/2}=\dfrac{10(5(1/2)-2)}{\sqrt{5(1/2)^2-4(1/2)+4}}$

$=\dfrac{10}{\sqrt{13}} km/hr$