MHB Rate of change when filling container

[Milly]

The ans is 4. Could someone please help..thanks in advance. :)

 

[Prove It]

Have you tried anything at all?

Similar triangles will need to be utilised to turn the problem into a one-variable problem.

 

[I like Serena]

The ans is 4. Could someone please help..thanks in advance. :)

Hi Milly! :)

Can you set up a formula for $V(h)$?

What is the volume when the container is filled for 1/8?
And what is the corresponding height?

 

[Milly]

I have tried the volume i got is $\pi$/12 and the height i got is 1/4 but i have no idea where to fit them in.

 

[MarkFL]

As suggested, you need to get the volume as a function of height...what is the formula for the volume of a right circular cone? And then how are the height and radius related?

 

[Milly]

I have found dr/dh and dv/dr and substituted h=1/4 in and use them to find dh/dv. Then i multiplied dv/dh with dv/dt but i still cannot get the ans. :(

 

[I like Serena]

I have tried the volume i got is $\pi$/12 and the height i got is 1/4 but i have no idea where to fit them in.

Suppose the cone is filled up to some height $h$.
What is then the radius of the cone as far as it is filled?

Can you substitute that radius in the formula $V=\frac 1 3 \pi r^2 h$?
That will give you the volume $V(h)$ as a function of $h$.

I have found dr/dh and dv/dr and substituted h=1/4 in and use them to find dh/dv. Then i multiplied dv/dh with dv/dt but i still cannot get the ans. :(

Which formulas did you find for each of those? (Wondering)

 

[Milly]

I substituted h=2 and r=1 into V= $\frac{1}{3}$$\pi$${r}^{2}$h and get the original volume which is $\frac{2\pi}{3}$ and using the ans i found to find the height when the volume is one-eighth which i got is 1/4.

 

[MarkFL]

...and substituted h=1/4 in

This would only work for a container that has a constant cross section (like a cylinder)...a cone oriented as in the diagram has more volume up top than at the bottom...

 

[I like Serena]

I substituted h=2 and r=1 into V= $\frac{1}{3}$$\pi$${r}^{2}$h and get the original volume which is $\frac{2\pi}{3}$ and using the ans i found to find the height when the volume is one-eighth which i got is 1/4.

What is the radius when the container is filled up to height $h$?
Hint: it is not $r=1$.

 

[Milly]

What is the radius when the container is filled up to height $h$?
Hint: it is not $r=1$.

$\frac{1}{4{r}^{2}}$ ? :/

 

[I like Serena]

$\frac{1}{4{r}^{2}}$ ? :/

When the container is full, the height is 2 and the radius is 1, which is half of the height.
If it is filled up to some height $h$, the radius will be half of the height.
So:
$$r = \frac 1 2 h$$

What do you get if you substitute that in the volume formula for a cone? (Wondering)

 

[Milly]

When the container is full, the height is 2 and the radius is 1, which is half of the height.
If it is filled up to some height $h$, the radius will be half of the height.
So:
$$r = \frac 1 2 h$$

What do you get if you substitute that in the volume formula for a cone? (Wondering)

$\frac{{h}^{3}\pi}{12}$...

 

[I like Serena]

$\frac{{h}^{3}\pi}{12}$...

Good!

Can you find $\d V h$ now?

And how might you get to $\d h t$, which is what the problem asks for?

 

[Milly]

Ohhh i got it$\frac{{h}^{3}}{12}$$\pi$ Is equal to 1/12 $\pi$ So h is 1.
Thank you so muchh!