Question

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 1

Answer:

$\frac{27}{50\pi}\text{ cm per sec}$

Step-by-step explanation:

Since, the volume of the cone,

$V =\frac{1}{3} \pi r^2 h----(1)$

Where,

r = radius,

h = height,

Differentiating equation (1) with respect to t ( time ),

$\frac{dV}{dt}=\frac{1}{3}\pi \frac{d}{dx}(r^2 h)----(1)$

Since, radius and height of cone must proportional,

i.e $\frac{r}{h}=\frac{4}{12}$

$\implies r = \frac{1}{3}h$

From equation (1),

$\frac{dV}{dt}=\frac{1}{3}\pi \frac{d}{dx}((\frac{1}{3}h)^2 h)$

$\frac{dV}{dt}=\frac{1}{27}\pi \frac{d}{dx}(h^3)$

$\frac{dV}{dt}=\frac{3}{27}\pi h^2 \frac{dh}{dt}$

$\frac{dV}{dt}=\frac{1}{9}\pi h^2 \frac{dh}{dt}$

We have,

$\frac{dV}{dt}=1.5\text{ cube cm per sec}, h = 5 cm$

$1.5=\frac{1}{9}\pi (5)^2 \frac{dh}{dt}$

$\implies \frac{dh}{dt}=\frac{1.5\times 9}{25\pi}=\frac{27}{50\pi}$

Hence, the rate of change of the height would be $\frac{27}{50\pi}$ cm per sec.