Question

A container in the shape of an inverted cone has height 12cm ani
radiuo 6 cm at the top. if it is filled with water at the rate of
12 cm3/sec, what is the rate of change in the height of water
level when the tank is filled 8 cm2​

Answer 1

The rate of change of height of water level is  $\dfrac{3}{64\pi }$   m per sec  .

Step-by-step explanation:

Given as :

A container in the shape of an inverted cone

The height of cone = h = 12 cm

The radius of cone = r = 6 cm

The rate of filling container = $\dfrac{dv}{dt}$  = 12 cm³ per sec           ...........`1

Let The rate of change in the height of water  level = $\dfrac{dh}{dt}$  cm per sec

According to question

Volume of sphere = $\dfrac{1}{3}$ × π × r² × h

where r is radius and h is height

And, The ratio of radius to height = $\dfrac{r}{h}$  = $\dfrac{12}{6}$ = 2

i.e   r = 2 h

So, volume = v =  $\dfrac{1}{3}$ × π × (2 h)² × h

Or,                 v =  $\dfrac{1}{3}$ × π × 4 h³

∵    $\dfrac{dv}{dh}$ = $\dfrac{4}{3}$ × π × $\dfrac{dh^{3} }{dh}$

Or, $\dfrac{dv}{dh}$ = $\dfrac{4}{3}$ × π × 3 h²

Or,  $\dfrac{dv}{dh}$  = 4 π × h²

Rate of change of volume =  4 π × h²

Now,

Rate of change of height = $\dfrac{dh}{dt}$

Or,  $\dfrac{dv}{dh}$ = $\dfrac{dv}{dt}$ × $\dfrac{dt}{dh}$ =  4 π × h²

Or,  12 × $\dfrac{dt}{dh}$ =  4 π × h²                  from eq 1

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

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

Now, at h = 8 cm

i.e  $\dfrac{dh}{dt}$  =  $\dfrac{12}{4\times \Pi \times 8^{2}}$

Or, $\dfrac{dh}{dt}$  = $\dfrac{3}{64\pi }$   m per sec

So, The rate of change of height of water level =  $\dfrac{dh}{dt}$  = $\dfrac{3}{64\pi }$   m per sec

Hence, The rate of change of height of water level is  $\dfrac{3}{64\pi }$   m per sec  . Answer