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?
Best Answer
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
$\displaystyle\frac{2\frac{\mathrm{d} h}{\mathrm{d} t}\sqrt{x^{2}+y^{2}}}{2x}=\frac{\mathrm{d} x}{\mathrm{d} t}$
$\displaystyle\frac{2*0.2*17}{2*15}=\frac{\mathrm{d} x}{\mathrm{d} t}$
$\frac{\mathrm{d} x}{\mathrm{d} t}=0.227$