1. Rates Of Change

Water is poured into an upside down open cone at a constant rate of 24cm^3 per second. Depth Of water at any time t seconds is hcm. What is the rate of increase of the area of the surface area of the surface S of the liquid when the depth is 16cm.

Surface Area S is the circukar area inside the cone. the radius of the base is 30cm and the heightof the cone is 30cm

2. Originally Posted by MC Squidge
Water is poured into an upside down open cone at a constant rate of 24cm^3 per second. Depth Of water at any time t seconds is hcm. What is the rate of increase of the area of the surface area of the surface S of the liquid when the depth is 16cm.

Surface Area S is the circukar area inside the cone. the radius of the base is 30cm and the heightof the cone is 30cm

As the volume of water is:

$V(t)=\frac{1}{3} \pi (h(t))^3$

we have:

$\frac{dV}{dt}=\pi (h(t))^2 \frac{dh}{dt}=24$

so:

$\frac{dh}{dt}=\frac{24}{\pi h^2}$

At any time we have:

$S(t)=\pi r^2=\pi h^2$

so:

$
\frac{dS}{dt}= 2 \pi h \frac{dh}{dt}=\frac{48}{h}
$

RonL