Related rates problem

  • Thread starter regnar
  • Start date
  • #1
regnar
24
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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.
 

Answers and Replies

  • #2
dacruick
1,039
1
if its height is the same as its radius its not conical is it? isnt 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.
 
  • #3
regnar
24
0
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)
 
  • #4
PhaseShifter
276
1
Conical with base radius equal to the height...
[tex]r=h[/tex]

[tex]V={{1} \over {3}} h A= {1 \over 3} h \pi h^2={\pi h^3 \over 3}[/tex]

So now the question is, what is dV/dt given dh/dt?
 
  • #5
regnar
24
0
Thank you. I got the same answer as i did before but in a different way so I know it's right.
 
  • #6
dacruick
1,039
1
if the height is the same as the radius...that sounds pretty spherical to me.
 
  • #7
PhaseShifter
276
1
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.
 
Last edited:

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