Applications of Derivatives

[Cyto]

A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate 10cm^3/min. When the depth of water in the cone is 8cm, the depth is decreasing at 2cm/min. What is the ratio of the height of the cone to its radius...

I'm stuck with this one.. can anyone help?


BTW.. the answer is 8(squareroot5pi) over 5

 

[PrudensOptimus]

[tex]

\frac{dV}{dt} = \frac{10cm^3}{1min} = S(h)[/tex]

[tex]\frac{dV(8cm)}{dt} = \frac{2cm^3}{1min} = S(8cm)[/tex]

[tex]S = \pi * r(r + \sqrt{r + h})[/tex]

[tex]\frac{2cm^3}{1min} = \pi * r(r + \sqrt{r + 8}) = S(8cm)

[/tex]

Solve the last equation for r. h is given, 8 cm.

 

[Hurkyl]

Let give names to things so we can talk about them precisely.

[tex]\mbox{V(t)}[/tex] = the volume at time t.
[tex]\mbox{h(t)}[/tex] = the height of the water at time t.
[tex]\mbox{r(t)}[/tex] = the radius of the (surface of the) water at time t.
[tex]\mbox{p}[/tex] = the height divided by the radius.

Let's look over what you are told:

A coffee filter has the shape of an inverted cone.

So what do we know about cones? Well, we know that when the water fills the cone, it will take the shape of the cone, and we know the volume of this cone:

[tex]
V(t) = \frac{1}{3} \pi r(t)^2 h(t)
[/tex]

We also know that the cone of water will have the same ratio of height to radius as the big cone. Thus

[tex]
p = \frac{h(t)}{r(t)}
[/tex]

Water drains out of the filter at a rate 10cm^3/min.

So we have:
[tex]
V'(t) = 10 \frac{\mathrm{cm}^3}{\mathrm{min}}
[/tex]

When the depth of water in the cone is 8cm, the depth is decreasing at 2cm/min.

Which is translated as:

[tex]
h(t) = 8 \mathrm{cm} \rightarrow h'(t) = 2 \frac{\mathrm{cm}}{\mathrm{min}}
[/tex]


Do you understand how I translated the problem into equations? See if you can now solve for p; if you get stuck, show us everything you tried then we'll give you more pointers.

 

[PrudensOptimus]

Hurkl, I think he is deeply confused. You should give him a equation to solve R.

 

[Hurkyl]

Hurkl, I think he is deeply confused. You should give him a equation to solve R.

If I do that, he'll still be confused.

 

[HallsofIvy]

Part of the confusion (my confusion, anyway!) is the original problem:

A coffee filter has the shape of an inverted cone. Water drains out of the filter at a rate 10cm^3/min. When the depth of water in the cone is 8cm, the depth is decreasing at 2cm/min. What is the ratio of the height of the cone to its radius...

The question is "What is the ratio of the height of the cone to its radius?" which is the opposite of the usual "related rates" problem where we are given the shape and asked for the rate of change of height.

Since, as Hurkyl said, V= (1/3)πr²h. The question asked for the ratio of "height of the cone to its radius" so let x be that ratio: x= h/r so h= xr (x is a constant) and dh/dt= x dr/dt,
dr/dt= (1/x)dh/dt.

Now, dV/dt= (π/3) (2rh dr/dt+ r²dh/dt)
= (π/3) (2rh/x+ r²)dh/dt.

We are given that, when h= 8, dV/dt= -10 and dh/dt= -2. Of course, since 8= xr, r= 8/x.

Putting all that into the above equation,
dV/dt = (π/3) (2rh/x+ r²)dh/dt,
-10= (π/3)(128/x²+ 64/x²)-2 so
(π/3) 192/x²= π (64/x²= 5

x²= (64/5)π Taking square roots,

x= (8/√(5))√(pi) which we can write (rationalize the denominator) as x= (8/5)√(4π)

 

[vapatel1]

Let give names to things so we can talk about them precisely.

[tex]\mbox{V(t)}[/tex] = the volume at time t.
[tex]\mbox{h(t)}[/tex] = the height of the water at time t.
[tex]\mbox{r(t)}[/tex] = the radius of the (surface of the) water at time t.
[tex]\mbox{p}[/tex] = the height divided by the radius.

Let's look over what you are told:



So what do we know about cones? Well, we know that when the water fills the cone, it will take the shape of the cone, and we know the volume of this cone:

[tex]
V(t) = \frac{1}{3} \pi r(t)^2 h(t)
[/tex]

We also know that the cone of water will have the same ratio of height to radius as the big cone. Thus

[tex]
p = \frac{h(t)}{r(t)}
[/tex]



So we have:
[tex]
V'(t) = 10 \frac{\mathrm{cm}^3}{\mathrm{min}}
[/tex]



Which is translated as:

[tex]
h(t) = 8 \mathrm{cm} \rightarrow h'(t) = 2 \frac{\mathrm{cm}}{\mathrm{min}}
[/tex]


Do you understand how I translated the problem into equations? See if you can now solve for p; if you get stuck, show us everything you tried then we'll give you more pointers.

ok i understand how you set it up. But how do you actually solve it? i don't get it. it's not clear in your response. can u please help me?