A funnel in the shape of an inverted cone is 30 cm deep and has a diameter across the top of 20 cm.

Question

Liquid is flowing out of the funnel at the rate of 12 cm^3/sec. At what rate is the height of the liquid decreasing at the instant when the liquid in the funnel is 20 cm deep?    Related Rates of AP Calculus!!!!Please help!!!

Answer 1

Hi Gabriele,

 

The funnel looks like

 

radius r

--------

 \      /

  \    / h

   \  /

    \/

 

 

We know that the original radius R of the cone is 10 cm and that the original height H of the cone is 30 cm. Originally, the cone was full of liquid. But, the volume of the liquid has ben decreasing at a rate of 12 cm³/sec, or dV/dt = -12 cm³/sec. At a later instant, the height of the liquid in the cone is h < H and the radius of the liquid in the cone is r < R. At that time, the volume of liquid is

 

V = (1/3) πr²h

 

We have a problem. We have too many variables. We have r and h. We would like to use a relationship to reduce to one variable. We can do this by looking at similar triangles. We know that the ratio of the height and radius of the liquid is the same at any time. So,

 

h/r = H/R

 

We want to eventually find the rate of height, So, let's eliminate the radius variable r. Solve for r.

 

r = (R/H)h

 

Substitute this expression for r into the equation for V above. We get

 

V = (1/3) π(R/H)²h²h = (1/3) π(R/H)h³

 

Remember R and H are constants, R = 10 cm, H = 30 cm

 

Now, we can the time derivative of each side of this last equation.

 

dV/dt = (R/H)h² πdh/dt, using the chain rule and also noting that the 3 from the derivative of h³ canceled out the 1/3.

 

We now want to solve for dh/dt.

 

dh/dt = (H/Rπ)h^-2 dV/dt

 

dV/dt = -12 cm³/sec.

h = 10 cm, since it is 20 cm deep at the time of interest.

 

dh/dt = (30 cm / 10 cm) (10 cm)^-2 (1/π) (-12 cm³/sec) = -0.114 cm/sec