Visualizing a Cone with Constant Altitude and Variable Radius

[regnar]

Hi, I've tried this too many ways and i can't seem to figure it out. the question is:
As sand leaks out of a hole in a container, it forms a conical pile whose altitude is always the same as its radius. If the height of the pile is increasing at a rate of 6in/min, find the rate at which the sand is leaking out when the altitude is ten inches.

It would be great help, if someone could help me. Thank you.

 

[dacruick]

if its height is the same as its radius its not conical is it? isn't it a hemisphere? once you do that, you will find that the radius is increasing by 6 inches a minute. but i feel like there is something missing in your question.

 

[regnar]

It can't be hemispherical; it's saying that the height from the tip to the base is the same length as the radius. what i did was implicitly differentiated the volume formula for a cone and got dV/dt = 1/3*pi(2rh*dr/dt + r²*dh/dt)

 

[PhaseShifter]

Conical with base radius equal to the height...
$$r=h$$

$$V={{1} \over {3}} h A= {1 \over 3} h \pi h^2={\pi h^3 \over 3}$$

So now the question is, what is dV/dt given dh/dt?

 

[regnar]

Thank you. I got the same answer as i did before but in a different way so I know it's right.

 

[dacruick]

if the height is the same as the radius...that sounds pretty spherical to me.

 

[PhaseShifter]

You seem to be having difficulty with visualization.
A sphere's height is twice its radius...but this isn't a sphere, or a hemisphere, it's a cone.

What does a cone look like?
Try drawing cones with various radius to altitude ratios. Post an image here if any of them looks like a sphere or hemisphere.