At noon, a vessel is sailing due north at the uniform rate of $15$ kilometers per hour. Another vessel, $30$ km due north of the first vessel, is sailing due east at the uniform rate of $20$ kilometers per hour. At what rate is the distance between the vessels changing at the end of an hour?
Here's what I thought.
$$dy/dt = 15 \;\text{km/hr}$$
$$dx/dt = 20 \;\text{km/hr}$$
$$x = 30 \;\text{km}$$
$y$ is unknown
$z$ is unknown
The first ship purely sailed from south to north since it's stated "due north" on the problem.
The second one, was a tough nut to crack for me. I was hoping that the second ship sailed as well from south to north but changed its direction from north to east.
My problem is depicting the diagram. How would it look like? By the way, I let z be the distance that would serve as the main subject for the problem. Am I also wrong on citing my analysis on the problem?
Best Answer
Here is a diagram for your situation. The first ship is at point $A$ at noon and at point $A'$ one hour later. The second ship is at point $B$ at noon and at point $B'$ one hour later.
I teach my calculus class that in related rates problems you should separate the "general" information, which is always true, from the "snapshot" information, which is true only at the relevant moment in time. In your case we have (leaving out the units):
GENERAL INFO:
The first ship is at position $(0,y)$ while the second is at position $(x,0)$.
The distance between them is $z=\sqrt{x^2+y^2}$.
The ship's speeds are given by
$$\frac{dy}{dt}=15$$
$$\frac{dx}{dt}=20$$
SNAPSHOT INFO:
At the relevant time 1:00 p.m.,
$$x=20$$ $$y=-15$$
SOLUTION:
Differentiate the expression for $z$ then substitute the given values for $x,\ y,\ \dfrac{dx}{dt},\ \dfrac{dy}{dt}$.
NOTES:
I use those particular coordinates for the ships in my diagram to get a simple expression for $z$. It should now be clear where your analysis went wrong, but ask if you need details.