Related rates of boat being pulled

Question

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.

Answer 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.

Answer 2

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.

Answer 3

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