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.
Best Answer
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.