Question

sand is leaking through a small hole at the bottom of a conical funnel at the rate of 12cm3/s . if the radius of the funnel is 8 cm and the altitude of the funnel is 16 cm, find

a) the depth of the salt water after 30 second

b)the rate of change of its depth at that instant

Answer 1

Answer:

h ≅ 16.29 cm

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$

Step-by-step explanation:

From the conical funnel;

Let the adjacent line at the base of the cone be radius (r) and the opposite ( vertical length) be the height (h)

Then;

$tan \theta= \dfrac{r}{h}$

$tan \theta= \dfrac{8}{16}$

$tan \theta= \dfrac{1}{2}$

∴

$\dfrac{r}{h} = \dfrac{1}{2}$

$r = \dfrac{h}{2}$

The volume (V) of the cone is expressed as:

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

$V = \dfrac{1}{3}\pi (\dfrac{h}{2})^2 h$

$V = \dfrac{\pi h^3}{3\times 4}$

$\dfrac{dv}{dt} = \dfrac{3 \pi h^2}{3 \times 4} \dfrac{dh}{dt}$

$\dfrac{dv}{dt} = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

Given that:

$\dfrac{dv}{dt} = 12 \pi$

Then:

$-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$- \int 48 \ dt = \int h^2 \ dh$

$\dfrac{h^3}{3}= -48 t + c$

$At \ t = 0 \ ; h = 0$

∴

$\dfrac{0^3}{3} = -48(0) + c$

c = 0

So;

$\dfrac{h^3}{3}= -48 t + 0$

$\dfrac{h^3}{3}= -48 t$

t = 30

$\implies h = (-48(30))^{1/3}$

h = 16.2865

h ≅ 16.29 cm

Thus;

from $-12 \pi = \dfrac{ \pi h^2}{4} \dfrac{dh}{dt}$

$\dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} } =\dfrac{dh}{dt}$

$\dfrac{dh}{dt} = \dfrac{-12 \pi} { \dfrac{ \pi h^2}{4} }$

$\dfrac{dh}{dt}=\dfrac{-12 \times 4} { {16.29^2} }$

$\mathbf{\dfrac{dh}{dt}= -0.1809 \ cm/s}$