I have completed a word problem involving related rates, and gone over it myself. However, this is the first relative rates problem I've ever done, and I would appreciate it if people would check my work.

Problem Description:

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1 meter higher than the bow of the boat. If the rope is pulled at a rate of 1 meter per second, how fast is the boat approaching the dock when it is 8 meters from the dock?

Since the bow of the boat, the pulley, and the dock can all form the corners of a triangle, I can solve the problem by thinking of it in terms of triangles.

pulley height (adjacent) $= h = 1$

dock distance (opposite) $= b = 8$

rope length (hypotenuse) $= r = ???$

The rope's rate is $-1$ meters per second.

According to the Pythagorean Theoream, $b^2 + h^2 = r^2$.

So, $2b \frac{db}{dt} + 0 = 2r\frac{dr}{dt}$

$$\frac{dr}{dt} = \frac{2(r)(-1)}{2(8)} = \frac{r(-1)}{8}$$

Using the Pythagorean Theorem again, I know that $1 + 8^2=r^2 \to \sqrt{1 + 64} = r \to \sqrt{65} = r$.

Therefore, since $r = \sqrt{65}$, I know that

$$\frac{dr}{dt}= \frac{\sqrt{65}(-1)}{8}=\frac{-\sqrt{65}}{8}$$

$\frac{-\sqrt{65}}{8}$ refers to meters per second of course.

  • 1
    $\begingroup$ If the boat is 8m from the dock, how is the rope 3m long? $\endgroup$
    – Daryl
    Commented Mar 10, 2019 at 21:14
  • $\begingroup$ @Daryl Um, you've got a point there. Let me see.. $\endgroup$ Commented Mar 10, 2019 at 21:15
  • 1
    $\begingroup$ @Daryl Should I have rewritten $8^2$ as $64$, and so gotten $\sqrt{1 + 64} = r \to \sqrt{65}=r$? $\endgroup$ Commented Mar 10, 2019 at 21:24

1 Answer 1


Your working seems fine.

Just that I think you intended to mean $\frac{db}{dt}$ when you write $\frac{dr}{dt}$ for the last few lines.

Remark: I prefer working with positive numbers, so I would use $\frac{dr}{dt}=1$. But yup, $\frac{dr}{dt}$ and $\frac{db}{dt}$ would share the same sign, i.e. the boat moves towards the direction of pulling. Speed is a scalar, so we can remove the sign after all.

  • $\begingroup$ That's a relief. Thank you! $\endgroup$ Commented Mar 11, 2019 at 1:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .