# Related rates of boat being pulled

I have the following related rates I am helping a student on, however, another tutor got $$.9923$$ and I get $$1.007$$.

Here is my attempt and someone please just point out the silly mistake I might be making:

Let $$a$$ be the height above the water of dock which is $$1$$, let $$b$$ be horizontal distance between boat and dock which at our instance is $$8$$, this makes the hypotenuse $$c$$ being the rope which at this instant is $$\sqrt{65}$$. Then we have a right triangle $$a^2+b^2=c^2$$ differentiating with respect to $$t$$ $$aa'+bb'=cc'$$ But $$a'=0$$ as height does not change. Then I plug in $$c'=-1, b=8, c=\sqrt{65}, a'=0$$ and $$b'=1.007$$, but I am unsure wether it's correct or not. Thanks in advance.

Your method looks correct to me (I did not check the numbers). The other tutor's answer cannot be correct since the boat must be approaching the dock at a faster rate than the rope is.

• oh my gosh that's what I said too! ok so I WAS right, thanks :) Sep 30, 2021 at 21:16

Your approach is correct. Your colleague's answer confuses velocity and acceleration. This is probably how they arrived at it: $$(1 \ \mathrm{m}\!\cdot\!\mathrm{s}^{-1} \ \cos( \tan^{-1}(\frac{1 \ \mathrm{m}}{8 \ \mathrm{m}}) ) = 0.9923 \ \mathrm{m}\!\cdot\!\mathrm{s}^{-1}$$

So it turns out, I was correct. Let $$a,b,c$$ denote the height, horizontal distance, and rope length respectively. Then if $$a=1,b=8$$ we get that $$c=\sqrt{65}.$$ By the Pythagorean theorem however, we know that since this triangle is right, $$a^2+b^2=c^2.$$ Differentiating this with respect to the time variable $$t$$ we obtain $$2a \frac{da}{dt} + 2b \frac{db}{dt}=2c \frac{dc}{dt}$$ we know however that the height is not changing at all which gives us $$\frac{da}{dt}=0$$ Then substituting and cancelling $$2$$'s out we obtain $$8 \frac{db}{dt}=-\sqrt{65}$$ and thus $$\frac{db}{dt}=\frac{-\sqrt{65}}{8}$$ as needed.