Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

Question

12 days ago

15

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

ers. 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?

Physics

1 answer:

ValentinkaMS [1.1K] · Answer 1 · 2026-02-21T22:23:29+03:00

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

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

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

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

$V=\frac {550}{3549}\times h^3$

Additionally, we differentiate the volume expression in relation to time:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given that $\frac {dV}{dt}$ = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

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

The rate at which the root beer level is decreasing is 0.08603 cm/s.