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ffuh205

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Mar 17, 2010
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Ok. I am really lost with this one. I have the question "Sand falls from a hopper at a rate of 0.1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle of ?/6 radians with the vertical, find the rate at which the height of the cone increases. At what rate does the radius of the base increase? Give both answers in terms of h, the height of the pile in meters."
What I think I need to do is substitute for either the radius or the height in terms of ?/6 radians and plug that into the Volume of a cone = 1/3?r2h but I am not sure where to go from here.
 
"Sand falls from a hopper at a rate of 0.1 cubic meters per hour and forms a conical pile beneath. If the side of the cone makes an angle of ?/6 radians with the vertical, find the rate at which the height of the cone increases. At what rate does the radius of the base increase? Give both answers in terms of h, the height of the pile in meters."
What I think I need to do is substitute for either the radius or the height in terms of ?/6 radians and plug that into the Volume of a cone = 1/3?r2h but I am not sure where to go from here.
Click to expand...

You are on the right track. Find the relation between r and h.

Replace r in the volume equation so that the only variable on the right side is h.

Take the derivative of the volume equation with respect to time. You will now have an equation that looks like
dV/dt = (something)(dh/dt)

You know that dV/dt = .1 m^3/hr. Plug that in and solve for dh/dt (the rate of height increase).

Now, you've already found the relationship between r and h, so you can also find the relationship between dh/dt and dr/dt. Make sense?
 
Makes sense but "Find the relation between r and h." is what I'm stuck on.
 
Ok. I see that h = sin pi/6 * r = .5r; plugging .5r back into 1/3?r2h we get 1/6 ?r3.
 
Ok. I see that h = sin pi/6 * r = .5r
Click to expand...

Neither h nor r are the hypotenuse of a triangle; you will not be using sine. They are the legs of the triangle; you will be using either tan or cot. Which one? (Hint: always draw a picture and label everything you know on it.)
 
We then substitute .1 cm^3*t = .5774/3 ?r3; .5196 cm^3 = ?r3; . 5196/? = r3;
 
the side of the cone makes an angle of ?/6 radians with the vertical
Click to expand...

Draw the cone. Place a vertical line from the top of the cone down to the base/ground. The angle formed at the top, between the vertical line and the side of the cone is the one referred to. That angle is pi/6.

tangent theta = opp./adj
tan(pi/6) = r/h
r = (h)tan(pi/6)

Substitute r into the volume equation.
 
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