Related Rates - Distance between two ships 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?
 A: 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.
A: The solution by Rory Daulton shows the diagram that I would draw.
I would like to add some thoughts about how to draw this diagram in
the first place, however, which go beyond just a comment.
Initially, I would not draw the coordinate axes. I would just try
to make a map showing the location of the two ships $A$ and $B$ at noon,
and the locations $A'$ and $B'$ where we find those ships at 1:00 pm.
Describing the situation of the ships at noon, the problem says,
"Another vessel, 30 km due north of the first vessel ... ."
So if I draw the position of the first vessel at noon
somewhere in the middle of a sheet of paper and label it $A$,
and I follow the convention that "north" is directly "up" on the paper,
I have to draw the position of the second vessel at noon
(labeled $B$) directly "up" from point $A$.
As you can see, this is the relative positions of $A$ and $B$
in Rory's diagram.
Also, from the "30 km due north" we know the distance from $A$ to $B$
is $30$.
Next we find the positions $A'$ and $B'$ where the ships are at 1:00 pm.
Since the first ship sails due north, $A'$ must be directly "up"
from $A$; and since the ship sails $15$ km/hr for an hour, the distance
from $A$ to $A'$ must be $15$. So plot $A'$ at a point $15$ units "up"
from $A$, exactly where Rory's diagram shows it.
For the movement of the second ship, all you need to know is how to
get from $B$ to $B'$, and
"sailing due east at the uniform rate of 20 kilometers per hour"
says how to do that.
Conventionally, due east is directly to the right on the paper,
and sailing at $20$ km/hr four one hour we go $20$ km,
so draw $B'$ exactly $20$ units to the right of $B$.
Now we have everything except the axes.
This is the one part where we can use some judgment to make the
problem easier to solve.
One obvious placement of the axes, since the problem starts by
describing the first ship at noon, is to make the axes cross at $A$.
Rory chose to make the axes cross at $B$ instead.
If you work out the formula for $z$ where the axes cross at $A$,
you should see that Rory's choice of axes results in a simpler formula.
A: here is a vector approach:
Let the first boat be at the origin at noon, and let its position vector at time $t$ be $\underline{a}$. Then $$\underline{a}=\left(\begin {matrix}0\\15\end{matrix}\right)t.$$
Likewise let the second boat have position vector at time $t$ given by $$\underline{b}=\left(\begin {matrix}0\\30\end{matrix}\right)+\left(\begin {matrix}20\\0\end{matrix}\right)t.$$
The displacement of B relative to A is $$\underline{b}-\underline{a}=\left(\begin {matrix}0\\30\end{matrix}\right)+\left(\begin {matrix}20\\-15\end{matrix}\right)t.$$
The distance between them at time $t$ is $$|\underline{b}-\underline{a}|=x$$ and $$x^2=(20t)^2+(30-15t)^2$$
$$\Rightarrow 2x\frac{dx}{dt}=800t+2(30-15t)(-15)$$
Therefore at $t=1$, $$\frac{dx}{dt}=\frac{800+2(15)(-15)}{2\sqrt{20^2+15^2}}=7$$
