# Thread: related rates

1. ## related rates

Coffee is poured at a uniform rate of 20 cm^3/s into a cup whose inside is shaped like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm and the height of the cup is 4 cm, how fast will the coffee level be rising when the coffee is halfway up? [ hint: Extend the cup downward to form a cone.]

i know that v = (1/3)(r^2)(h)
i dont know how to set r as a function of h or vice versa.

2. Originally Posted by larryboi7
Coffee is poured at a uniform rate of 20 cm^3/s into a cup whose inside is shaped like a truncated cone. If the upper and lower radii of the cup are 4 cm and 2 cm and the height of the cup is 4 cm, how fast will the coffee level be rising when the coffee is halfway up? [ hint: Extend the cup downward to form a cone.]

i know that v = (1/3)(r^2)(h)
i dont know how to set r as a function of h or vice versa.
following the hint ...

top radius = 4 cm

height of the extended cone = 8 cm

$\frac{r}{h} = \frac{4}{8} = \frac{1}{2}$

$r = 2h$

now work with the cone, assuming that it is filled to a height $h = 4$ cm at $t = 0$ ... the remaining void is the truncated cone (a frustrum).

to answer the question, note that $h = 6$ cm from the cone's bottom when the liquid is halfway up the frustrum.