Given the following function:
[tex]tan\text{ }\theta=\frac{10}{y}[/tex]
Both θ and y are functions of the time (t)
We will find the derivatives of θ and y with respect of the time (t) as follows:
[tex]sec^2θ*\frac{dθ}{dt}=-\frac{10}{y^2}*\frac{dy}{dt}[/tex]
Now, we will find dy/dt when θ = π/6 and dθ/dt = π/12
First, we need to find the value of y when θ = π/6
[tex]\begin{gathered} tan(\frac{\pi}{6})=\frac{10}{y} \\ \frac{1}{\sqrt{3}}=\frac{10}{y} \\ \\ y=10\sqrt{3} \end{gathered}[/tex]
so, we will substitute the values to find dy/dt as follows:
[tex]\begin{gathered} sec^2(\frac{\pi}{6})*\frac{\pi}{12}=-\frac{10}{(10\sqrt{3})^2}*\frac{dy}{dt} \\ \\ so,\frac{dy}{dt}=-\frac{(10\sqrt{3})^2}{10}*sec^2(\frac{\pi}{6})*\frac{\pi}{12}=-10.4719755 \end{gathered}[/tex]
Rounding to 2 decimal places
So, the answer will be:
[tex]\frac{dy}{dt}=-10.47\text{ feet/hour}[/tex]
