Question

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimeters. As you suck on the straw, root beer leaves the cup at the rate of 4 cm3/sec. At what rate is the level of the root beer in the cup changing when the root beer is 10 centimeters deep?

Answer 1

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Step-by-step explanation:

The volume of the cone is :

`V=\frac {1}{3}* \pi* r^2* h`

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

`V=\frac {1}{3}* \frac {22}{7}* ({{{\frac {5}{13}* h}}})^2* h`

`V=\frac {550}{3549}* h^3`

Also differentiating the expression of volume w.r.t. time as:

`\frac {dV}{dt}=\frac {550}{3549}* 3* h^2* \frac {dh}{dt}`

Given:
`\frac {dV}{dt}` = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

`-4=(550)/(3549)* 3* {10}^2* \frac {dh}{dt}`

`(55000)/(1183)* \frac {dh}{dt}=-4`

`\frac {dh}{dt}=-0.08603\ cm/s`

The level of the root beer is dropping at a rate of 0.08603 cm/s.