This is the question →
I'm having trouble forming an equation for all of this question.
Also do I have only have to use one chain rule or 2 chain rules to find $\frac{\mathrm{d}SA}{\mathrm{d}t}$, I also know we are given $\frac{\mathrm{d}v}{\mathrm{d}t}$… but the part where the top is cut off and a new flat top is added is very confusing… Could someone please guide me!
Best Answer
Imagine that the bottom has not been cut off. Let $y=y(t)$ be the vertical distance at time $t$ from the bottom tip to the top of the water. Then the area $A=A(t)$ of the top of the water is $y^2$.
Let the volume from the bottom tip to the top of the water be $U=U(t)$. We are told that $U$ is increasing at the rate $3000$, so $\frac{dU}{dt}=3000$. Note that $$U=\frac{1}{3}y^3,\tag{1}$$ and therefore $$A=y^2=(3U)^{2/3}.\tag{2}$$ Differentiate both sides with respect tp $t$, and finish by calculting $U$ when $y=35$.
Another way: We could use $U=\frac{1}{3}y^3$ to find $\frac{dy}{dt}$ when $y=35$, and then use $\frac{dA}{dt}=2y\frac{dy}{dt}$ to find the rate of change of area at that time.