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?

  • $\begingroup$ Is it $25kmph$ at end of hour $\endgroup$ – Archis Welankar Jan 9 '16 at 14:23

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.

Ships on a Cartesian plane

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):


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




At the relevant time 1:00 p.m.,

$$x=20$$ $$y=-15$$


Differentiate the expression for $z$ then substitute the given values for $x,\ y,\ \dfrac{dx}{dt},\ \dfrac{dy}{dt}$.


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.

  • $\begingroup$ Based on the diagram. Does this mean I'll only consider the part from (0, y) up to point B? $\endgroup$ – Jem Celespara Jan 11 '16 at 12:02
  • $\begingroup$ @JemCelespara: Vertically, yes, but you also need to consider the horizontal dimension from point $B$ to $B'$. The problem only considers the two ships at the two times noon and 1:00 pm, and my diagram covers the two ships at the two times. There is no need to consider other times or places. $\endgroup$ – Rory Daulton Jan 11 '16 at 18:41

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.

  • 1
    $\begingroup$ Good analysis, +1! I did place the origin at point $A$ at first but I later saw that placing it at $B$ would simplify the formula for $z$, so I moved the origin. $\endgroup$ – Rory Daulton Jan 9 '16 at 15:47
  • $\begingroup$ @RoryDaulton Ah, yes, the other way to deal with axes is to draw them right away, but if you later see a better choice of axes, change the axes. As long as all the formulas you finally use are based on the new axes (which they are, of course, in your answer) then all is fine. $\endgroup$ – David K Jan 9 '16 at 16:08

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 $t=1$, $$\frac{dx}{dt}=\frac{800+2(15)(-15)}{2\sqrt{20^2+15^2}}=7$$

  • $\begingroup$ I dont think its right $\endgroup$ – Archis Welankar Jan 9 '16 at 15:25
  • $\begingroup$ Any particular reason? @ArchisWelankar $\endgroup$ – David Quinn Jan 9 '16 at 15:28
  • $\begingroup$ @RoryDaulton..yikes! Thanks for that, edited accordingly. I hope we are now in agreement. I hesitated about entering an answer when I saw your perfectly-formed offering, but I was half-way through typing it, somewhat laboriously, so I decided to persevere... $\endgroup$ – David Quinn Jan 9 '16 at 15:45
  • $\begingroup$ I think it is good to give different approaches to a question, and finishing my answer after another answer has appeared happens often to me. I think your work here is good. Plus, you continued to the final answer, which I decided not to do (since the OP did not ask for it). I removed the negative sign in front of the final answer, since it seems that you meant to do that. I hope you don't mind. Put it back if you want it there. $\endgroup$ – Rory Daulton Jan 9 '16 at 15:50
  • $\begingroup$ That's strange. I though I removed it myself! Thanks anyway. $\endgroup$ – David Quinn Jan 9 '16 at 16:41

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