I need help solving an issue of mine.
The question goes like this:
Q) $A$ is a lighthouse and $\overline{BC}$ is the shoreline. It is a right triangle and the lighthouse $A$ is perpendicular to the shoreline. The lighthouse turns once a second so $\dfrac{d\theta}{dt}=2\pi$ radians/second. Let the distance from the lighthouse to shoreline at $B$ be the perpendicular distance $100$ and the light move up from $B$ to $C$ and the distance $\overline{BC}=y$. And finally the hypotenuse is $z$. Find a relationship between the rate of change of $z$ to $y$ with respect to time.
My solution was to use the Pythagorean theorem So $$z^2 = y^2 + 100^2$$ differentiating with respect to time implicity: $$2z\frac{dz}{dt} = 2y\frac{dy}{dt}$$ so $$\frac{dz}{dt} = \frac{y}{z}\times \frac{dy}{dt}$$ as $\frac{y}{z} = \sin\theta$ $$\frac{dz}{dt} = \sin\theta\frac{dy}{dt}$$ This is the right answer. I tried another approach I do not know why this method does not work. I tried the following $$y = z\sin\theta$$ $$\frac{d}{dt}(y) = \frac{d}{dt}(z\sin\theta)$$ From here if I use product rule on the RHS then I do not get the right answer. I do not know why this approach is not working. I will appreciate any insight as to why this does not work or if I am doing this question wrong.
Thank you and stay safe!!!!
From the first picture (left), you've got $$\begin{split}y^2+100^2&=z^2\\ \frac{dz}{dt}&=\frac{y}{z}\frac{dy}{dt}\end{split}$$
Now suppose you are given the additional information that $\theta$ is $\angle A$ (right). You can indeed find the relationship between dz/dt and dy/dt from this, you just need 2 equations now.
$$\begin{split}\sin\theta&=\frac{y}{z}\\ \cos\theta\frac{d\theta}{dt}&=-\frac{y}{z^2}\frac{dz}{dt}+\frac{1}{z}\frac{dy}{dt}\end{split}$$
and
$$\begin{split}\cos\theta&=\frac{100}{z}\\ -\sin\theta\frac{d\theta}{dt}&=-\frac{100}{z^2}\frac{dz}{dt}\\ \Rightarrow\frac{d\theta}{dt}&=\frac{100}{z^2\sin\theta}\frac{dz}{dt}\end{split}$$
Go back to above
$$\begin{split}\frac{100\cot \theta}{z^2}\frac{dz}{dt}&=-\frac{y}{z^2}\frac{dz}{dt}+\frac{1}{z}\frac{dy}{dt}\\ \left(\frac{100^2}{yz^2}+\frac{y}{z^2}\right)\frac{dz}{dt}&=\frac{1}{z}\frac{dy}{dt}\\ \underbrace{\left(\frac{100^2+y^2}{z^2}\right)}_{=\frac{z^2}{z^2}=1}\frac{dz}{dt}&=\frac{y}{z}\frac{dy}{dt}\end{split}$$