James is filling an ice cream cone. The cone is 12 cm tall and has a radius of 4 cm. If the ice cream fills the cone evenly at a rate of 1.5 cm3/s, what is the rate of change of the height when the height is 5 cm? (Recall that a cone with radius r and height h as volume V = 1 3 πr2h.)

Answer:[tex]\frac{27}{50\pi}\text{ cm per sec}[/tex] Step-by-step explanation:Since, the volume of the cone,[tex]V =\frac{1}{3} \pi r^2 h----(1)[/tex]Where,r = radius,h = height,Differentiating equation (1) with respect to t ( time ),[tex]\frac{dV}{dt}=\frac{1}{3}\pi \frac{d}{dx}(r^2 h)----(1)[/tex]Since, radius and height of cone must proportional,i.e [tex]\frac{r}{h}=\frac{4}{12}[/tex][tex]\implies r = \frac{1}{3}h[/tex]From equation (1),[tex]\frac{dV}{dt}=\frac{1}{3}\pi \frac{d}{dx}((\frac{1}{3}h)^2 h)[/tex][tex]\frac{dV}{dt}=\frac{1}{27}\pi \frac{d}{dx}(h^3)[/tex][tex]\frac{dV}{dt}=\frac{3}{27}\pi h^2 \frac{dh}{dt}[/tex][tex]\frac{dV}{dt}=\frac{1}{9}\pi h^2 \frac{dh}{dt}[/tex]We have,[tex]\frac{dV}{dt}=1.5\text{ cube cm per sec}, h = 5 cm[/tex][tex]1.5=\frac{1}{9}\pi (5)^2 \frac{dh}{dt}[/tex][tex]\implies \frac{dh}{dt}=\frac{1.5\times 9}{25\pi}=\frac{27}{50\pi}[/tex]Hence, the rate of change of the height would be [tex]\frac{27}{50\pi}[/tex] cm per sec.