Water is poured in a glass at a rate of $1.9~\text{cm}^3/\text{s}$. How fast does the liquid level rise when the liquid height in the glass is $7.0~\text{cm}$?
I have used uniformity on the triangle to get a second triangle and got it to $7*7$ Not sure how to think to find the equation I need to solve.
We wish to find $\frac{dh}{dt}$, the rate at which the height is changing with respect to time. The volume of a cone is $$V = \frac{1}{3}\pi r^2h$$ In order to find $\frac{dh}{dt}$, we need to write $V$ as a function of $h$, then differentiate the resulting equation with respect to time, so we must eliminate $r$. To do so, we must express $r$ in terms of $h$.
Consider the diagram below:
Since the diameter of the cone is $18~\text{cm}$, its radius is $9~\text{cm}$. By similar triangles, $$\frac{r}{h} = \frac{9~\text{cm}}{18~\text{cm}} = \frac{1}{2} \implies r = \frac{h}{2}$$ Substituting for $r$ in the equation for the volume gives $$V = \frac{1}{3}\pi\left(\frac{h}{2}\right)^2h = \frac{1}{12}\pi h^3$$ Differentiating both sides of the equation implicitly with respect to time yields $$\frac{dV}{dt} = \frac{1}{4}\pi h^2\frac{dh}{dt}$$ Solving for $\frac{dh}{dt}$ gives $$\frac{dh}{dt} = \frac{4}{\pi h^2}\frac{dV}{dt}$$ Now, you just need to substitute the given numbers to complete the problem.