Differentiating the volume with respect to time

Question

A two dimensional polygon B in xy plane starts to move up vertically along the z axis. A size of B is continuously changing. Let $V(t)$ be the volume that B has traced during t seconds and $L(t)$be the distance that $B$ traveled during $t$ seconds. Suppose that $V'(t)$is given by the product of time $t$ and the area of B at $t$.

(a) Find $L(10)$.

(b)Prove that there is $t_{0}\in [0,10]$ such that $V(10)/L(10)$ is the area of B at time $t_{0}$.

Let $A(t)$ be the area of B at $t$. By the condition, $V'(t)=tA(t)$.

By the chain rule, $\frac{dV}{dt}=\frac{dV}{dL}\frac{dL}{dt}$

I'm confused with the expression of volume with respect to $L(t)$ and $A(t)$ using the integral.

Answer 1

For point (a) we have that

• $dV(t)=A(t)dL(t)\implies V'(t)=A(t)L'(t)$

and since

• $V'(t)=tA(t)$

we have that

$$V'(t)=A(t)L'(t)=tA(t)\implies L'(t)=t\implies L(t)=\frac12t^2\implies L(10)=50$$

For point (b) by MVT we have that

$$\frac{V(10)-V(0)}{L(10)-L(0)}=\frac{V(10)}{L(10)}=\frac{dV}{dL}(t_0)=B(t_0)$$