Level of water rising in a cone.

[Shawn Garsed]

Homework Statement

A right circular cone with radius r and height h is being filled with water at the rate of 5 cu in./sec. How fast is the level of the water rising when the cone is half full.

Homework Equations

V=r²h∏/3

The Attempt at a Solution

V=5t. The level of the water is determined by h, so I need the find the derivative of h. h=15t/∏r², but I'm pretty sure that's not right because r is not a constant, it depends on h. That's where I'm stuck.

 

[eumyang]

Looks like some info is missing in the problem. For instance, if we knew the relationship between r and h, then we can rewrite r in terms of h and get rid of the r. Can you double check and make sure you have the entire problem?

 

[Shawn Garsed]

That's what I was thinking too. Here's the question in it's entirety:
"A cup in the form of a right circular cone with radius r and height h is being filled with water at the rate of 5 cu in./sec. How fast is the level of the water rising when the volume of the water is equal to one half the volume of the cup?"

 

[azizlwl]

Find r in term of h.

 

[Shawn Garsed]

How can I do that if I don't know the angle between h and the slant height.

 

[Shawn Garsed]

Here's my (attempted) solution given the information I have.
r=htanθ, θ is the angle between h and the slant height.
V=5t=tan²θh³∏/3, therefore h=(15t/∏tan²θ)^1/3
and h'=(5/∏tan²θ)*(∏tan²θ/15t)^2/3.
When the cup is half full t=tan²h³∏/30 and
h'=5*4^1/3/(∏tan²θh²).

 

[ehild]

It is correct if you add the parentheses, but you can replace tanθ=r/h and get the result in the given radius of the cone.

ehild

 

[skiller]

Consider the cone (or at least the cross-section of it) on the x-y plane, such that the point is at the origin and the open end is at x=h, y=-r to +r.

For any x (from 0 to h), you know:

y = ?

V = ?

dV/dt = ? (this will involve a dx/dt term) - and you know this equals 5

So you want dx/dt, where x solves V(x)=(∏r²h/3)/2

I'm not sure if this is the simplest way, but that's how I'd approach it.

(Hope I'm right!)

 

[Curious3141]

Homework Statement

A right circular cone with radius r and height h is being filled with water at the rate of 5 cu in./sec. How fast is the level of the water rising when the cone is half full.

Homework Equations

V=r²h∏/3

The Attempt at a Solution

V=5t. The level of the water is determined by h, so I need the find the derivative of h. h=15t/∏r², but I'm pretty sure that's not right because r is not a constant, it depends on h. That's where I'm stuck.

I would normally prefer to let the constant dimensions of the cone be represented by the big letters, but I'm stuck because the problem used small letters. Never mind.

Let R = R(t) and H = H(t) represent the instantaneous radius of the circular surface of the water and the instantaneous height of the water above the vertex, respectively.

Use similar triangles to determine the relationship between R and H in terms of the constant radius r and height h of the conical container. You don't need trig.

Once that's done, use V = \frac{1}{3}{\pi}R^2H to find an expression for V = V(t) in terms of R and H. That represents the instantaneous volume of water at time t. Use the previously derived relationship between R and H to express V only in terms of H. In other words, you now have V as a function of H, i.e. V = f(H).

Now find \frac{dV}{dH}. By Chain Rule, you know that \frac{dV}{dt} = \frac{dV}{dH}.\frac{dH}{dt}, so \frac{dH}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dH}}. Put the expression for \frac{dV}{dH} into the denominator, so that there will be a H term in the RHS of the equation.

You're already given \frac{dV}{dt} (constant at 5 cu in/sec.), so all you need now is to find the value of H (using V = f(H)) when the cone is half full (i.e. the volume of water is half the volume of the container), put it all into the last equation and compute.

 

[ehild]

Here's my (attempted) solution given the information I have.
r=htanθ, θ is the angle between h and the slant height.
V=5t=tan²θh³∏/3, therefore h=(15t/∏tan²θ)^1/3
and h'=(5/∏tan²θ)*(∏tan²θ/15t)^2/3.
When the cup is half full t=tan²h³∏/30 and
h'=5*4^1/3/∏tan²θh².

It needs some parentheses: h'=5*4^1/3/(∏tan²θh²)

Have you all got different result from that of the OP? I think it is correct.

ehild

 

[Shawn Garsed]

@Curious3141 How would you relate r to h without using trig, cause I see no other way, though I could be missing something.

@ehild You're right I forgot some parentheses. Fixed it.

 

[Curious3141]

@Curious3141 How would you relate r to h without using trig, cause I see no other way, though I could be missing something.

@ehild You're right I forgot some parentheses. Fixed it.

I'm using big R and big H for the "water cone" and small r and small h for the conical container. Of course, the dimensions of the latter are bigger.

Sketch a vertical cross-section through the middle of the cone. The container will be an isosceles triangle made up of two identical adjacent right triangles. The shape of the water will just be a scaled down version of the same, made up of two smaller identical right triangles. Can you see this?

Then, by similar triangles, R/H = r/h. That's it.

 

[Shawn Garsed]

I see what you mean, R then equals Hh/r which eliminates the need for the angle θ. I kept getting confused because there are two values for both the radius and the height, one is a constant and the other is a variable.

 

[Curious3141]

I see what you mean, R then equals Hh/r

That should be R = \frac{rH}{h}.

I kept getting confused because there are two values for both the radius and the height, one is a constant and the other is a variable.

Which is why I prefer to use the small letters for the variables and the big letters for the constants. If we did this, then r \leq R and h \leq H, where the small letters refer to the variables describing the water, and the big ones the constants describing the cone. I think you'll find this far more natural.

Unfortunately, the question used small letters for the cone, so we have to make do with the opposite arrangement.

 

[HallsofIvy]

@Curious3141 How would you relate r to h without using trig, cause I see no other way, though I could be missing something.

Similar triangles. If the cone has radius R and height H, then the radius, r, of the circular top of water that fills the cone to height h satisfies r/h= R/H because they are "corresponding parts of similar triangles".

@ehild You're right I forgot some parentheses. Fixed it.

 

[Shawn Garsed]

@Curious3141 Once again you're right. I really need to get some sleep.

Thanks for the help guys.

 

[skiller]

Have you all got different result from that of the OP? I think it is correct.

Hi ehild,

Yes, I think we're all singing from the same hymn sheet.