Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

Question

Andrew

1 month ago

7

Salt is poured from a container at 10 cm³ s-¹ and it formed a conical pile whose height at any time is 1/5 the radius of the abo

ve. At what rate is the height of the cone increasing when the height is 2 cm form the base of the cone?

Chemistry

1 answer:

Tems11 [2.7K] · Answer 1 · 2026-04-12T01:39:00+03:00

Answer:

$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

Explanation:

Volume of a cone:

$\displaystyle V=\frac{1}{3} \pi r^2 h$

We have $\displaystyle \frac{dV}{dt} = \frac{10 \ cm^3}{sec}$ and we aim to calculate $\displaystyle \frac{dh}{dt} \Biggr | _{h\ =\ 6}= \?$ when the height is 2 cm.

In our cone volume formula, three variables appear: V, r, and h.

With only dV/dt and dh/dt available, we can express the equation solely in terms of h.

It is given that the height of the cone is 1/5 of the radius at any moment, or 1/5r, thus we can express this as r = 5h.

Substitute this r value into the volume equation:

$\displaystyle V =\frac{1}{3} \pi (5h)^2 h$
$\displaystyle V =\frac{1}{3} \pi \ 25h^3$

Differentiate the equation with respect to time t.

$\displaystyle \frac{dV}{dt} =\frac{25}{3} \pi \ 3h^2 \ \frac{dh}{dt}$
$\displaystyle \frac{dV}{dt} =25 \pi h^2 \ \frac{dh}{dt}$

Insert known values into the equation to find dh/dt.

$\displaystyle 10 = 25 \pi (2)^2 \ \frac{dh}{dt}$
$\displaystyle 10 = 100 \pi \ \frac{dh}{dt}$

Isolate dh/dt by dividing both sides by 100π.

$\displaystyle \frac{10}{100 \pi} = \frac{dh}{dt}$
$\displaystyle \frac{dh}{dt} = \frac{1}{10 \pi}$

The cone's height is rising at a rate of 1/10π cm per second.