Question

A conical paper cup is 10 cm tall with a radius of 10 cm. The bottom of the cup ispunctured so that the water leaks out at a rate of 16p3 cm³/sec. At what rate is the waterlevel changing when the water level is 3 cm?

Answer 1

Answer:

0.59 cm

Explanation:

height of cup (h) = 10 cm

radius of cup (r) = 10 cm

rate of change of water level (V') = 16π/3 $cm^{3} /sec$

rate of change of height (h) = 3 cm

the ratio of the height to radius (h:r) = 10:10

h/r = 10/10

h/r = 1

r=h

formula for volume of a volume (v) = $\frac{1}{3}$.π$r^{2}$h

substituting r=h into the formula above we have

v = $\frac{1}{3}$.π$h^{2}$. h = $\frac{1}{3}$.π$h^{3}$

differentiating the above we have

v' = 3($\frac{1}{3}$π$h^{2}$)h' = π$h^{2}$.h'

rearranging the above we have

h' = v' / (π$h^{2}$)

where

• v' = rate of change of water level
• h' = rate of change of water level
• h = water level

h' = ((16 x π) / 3) ÷ ( π x $3^{2}$)

h' = ((16 x 3.142) / 3) ÷ ( 3.142 x $3^{2}$)

h' = 16.76 / 28.28

h' = 0.59 cm