# Thread: related rates word problem

1. ## related rates word problem

Water is poured into a conical funnel at a rate of 1 cm'3/s. The radius of the top of the funnel is 10cm and the height of the funnel is 20cm. Find the rate at which the water level is rising when it is 5cm from the top of the funnel.

Formula for volume of a cone, V=1/3 pi r'2h

I did a diagram, found out what I have to find and what I am given, and I believe I know the rest of the steps but am not sure. My question is do you sub the 10cm for r and 20cm for h in the volume equation given, take the derivative of that, then use the chain rule to work out the equation?

If so, could you show and explain how do do those steps up to the derivative. (I can work out the rest from there)

If your really feeling generous you could find out the final answer, just so I know I'm on the right track for the rest of the question.

Thx in advance for the help.

2. Originally Posted by dcfi6052
Water is poured into a conical funnel at a rate of 1 cm'3/s. The radius of the top of the funnel is 10cm and the height of the funnel is 20cm. Find the rate at which the water level is rising when it is 5cm from the top of the funnel.

Formula for volume of a cone, V=1/3 pi r'2h

I did a diagram, found out what I have to find and what I am given, and I believe I know the rest of the steps but am not sure. My question is do you sub the 10cm for r and 20cm for h in the volume equation given, take the derivative of that, then use the chain rule to work out the equation?

If so, could you show and explain how do do those steps up to the derivative. (I can work out the rest from there)

If your really feeling generous you could find out the final answer, just so I know I'm on the right track for the rest of the question.

Thx in advance for the help.
Umm, it's Saturday morning, no work, lots of sparetime. An related rates problem. Why not.

We are asked to find the rate at which the water is rising. So that is dy/dt if the height of the water is y. And so we base the y from the bottom of the inverted cone.
Imagine, or draw the figure on paper. Draw a vertical cross-section of an inverted cone that's supposed to be 10cm round at the top and 20 cm tall. So you have an isosceles triangle that is standing on its apex. Draw the altitude from the apex to the midpoint of the 10-cm base at the top. Now you have two congruent rigth triangles, any of which has:
---vertical leg = 20cm
---horizontal leg = 10/2 = 5cm.

Now draw a horizontal line, supposed to be the level of the water rising in the cone, somewhere below the top base. Again, you have another pair of congruent smaller right triangles, each of which has:
---vertical leg = y cm
---horizontal leg = x cm

Volume of the water, V = (pi/3)(x^2)y -------(1)
V is in terms of x and y. For simpler differentiation, we need to express V in terms of one variable only. Since we are interested on dy/dt, we choose V in terms of y only.

So we express x in terms of y.
In the large and small right triangles in the figure, by proportion,
x/y = 5/20
So, x = 5y/20 = y/4
Substitute that into (1),
V = (pi/3)[(y/4)^2]y
V = (pi/48)y^3 ---------(2)

To have dy/dt, we differentiate (2) with respect to time t,
dV/dt = [(pi/16)y^2](dy/dt) ----------(3)

Since dV/dt = 1 cu.cm/sec, and y = (20-5) = 15cm, then dy/dt at that instant is:
Plugging in all known values into (3),
1 = [(pi/16)(15^2)](dy/dt)
1 = (225pi/16)(dy/dt)
dy/dt = 16/(225pi) = 0.022635 cm/sec -------------answer.

================
Ooopppss, sorry, the radius at the top is 10cm. I thought I read 10-cm diameter.
Some changes on the computations above then.

x/y = 10/20
x = y/2

V = (pi/3)[(y/2)^2]y = (pi/12)y^3

dV/dt = [(pi/4)y^2](dy/dt)
1 = (pi/4)(225)(dy/dt)
dy/dt = 4/(225pi) = 0.005659 cm/sec ------------final answer.