# linear calculus related rate of boat moving toward a dock

I'm not sure what I did wrong here, but I got 0.217 (rounded to the nearest thousandth) and the answer is supposed to be 0.227.

Problem: "A boat is pulled in to a dock by a rope with one end attached to the front of the boat and the other end passing through a ring attached to the dock at a point 8 ft higher than the front of the boat. The rope is being pulled through the ring attached to the dock at a rate of 0.20 ft/sec.

How fast is the boat approaching the ring attached to the dock when 17 ft of rope is out?"

Here is what I did:
Let $$x$$ be the horizontal distance between the rope's attachment point on the boat and the dock.
Let $$y$$ be the vertical distance between the rope's attachment point on the boat and the ring.
Let $$h$$ be the length of the rope, which is also the hypotenuse of the right triangle formed by $$x$$ and $$y$$.

We want to find the rate at which the boat is approaching the dock, so that's the rate at which $$x$$ is changing with respect to time ($$dx/dt$$).

We are given $$dh/dt$$, so I used the pythagorean theorem and implicit differentation to find an equation for $$dx/dt$$.

$$h=\sqrt{x^{2}+y^{2}}\Rightarrow (x^{2}+y^{2})^{1/2}$$

$$\frac{\mathrm{d} h}{\mathrm{d} t}=\frac{\mathrm{d} }{\mathrm{d} t}\left [ (x^{2}+y^{2})^{1/2} \right ]\Rightarrow \frac{\mathrm{d} h}{\mathrm{d} t}=1/2(x^{2}+y^{2})^{-1/2}(2x)(\frac{\mathrm{d} x}{\mathrm{d} t})$$

Isolate for $$\frac{\mathrm{d} x}{\mathrm{d} t}$$ and I got:
$$\frac{2\frac{\mathrm{d} h}{\mathrm{d} t}\sqrt{x^{2}+y^{2}}}{2x}=\frac{\mathrm{d} x}{\mathrm{d} t}$$

Now, I want to solve this equation when $$h=17$$. Since we have $$h$$ and $$y$$, just use pythagorean theorem again to solve for $$x$$.
$$x=\sqrt{17^{2}+8^{2}}\approx 18.7883$$

Then I just plug in $$x$$ and $$\frac{\mathrm{d} h}{\mathrm{d} t}$$, which we were also given at the beginning, and solve the equation.

$$\frac{2(.02)\sqrt{x^{2}+y^{2}}}{2x}\approx 0.217375538288$$

Where did I go wrong?

• Can you pls type in the question completely? Commented Jun 25, 2021 at 4:12
• It is. That's the entire text. Or did you mean something else? Commented Jun 25, 2021 at 8:51
• Some of it like initial rope length of $17$ etc. is not mentioned in the question. What needs to be found is also not written. Going through your working, I can see some of it but not in the problem statement. Commented Jun 25, 2021 at 8:54
• Oh wow, you're right. Sorry. Fixing it now! Commented Jun 25, 2021 at 10:34

Wait , you mentioned in your question that $$\frac{dh}{dt}=0.2$$ but substituted the value as $$\frac{dh}{dt}=0.02$$

Well one more error is there you have

$$h^2=x^2+y^2$$ therefore $$x=\sqrt{h^2-y^2}$$ and not as you did $$x=\sqrt{h^2+y^2}$$

Plugging in values now gives

$$\displaystyle\frac{2\frac{\mathrm{d} h}{\mathrm{d} t}\sqrt{x^{2}+y^{2}}}{2x}=\frac{\mathrm{d} x}{\mathrm{d} t}$$

$$\displaystyle\frac{2*0.2*17}{2*15}=\frac{\mathrm{d} x}{\mathrm{d} t}$$

$$\frac{\mathrm{d} x}{\mathrm{d} t}=0.227$$