At a sand plant, sand is falling off a conveyor and onto a conical pile at the rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of the pile changing when it is 15 feet high? Answer: .006 ft/min (8/405pi)

Question

At a sand plant, sand is falling off a conveyor and onto a conical pile at the rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate is the height of  the pile changing when it is 15   feet high? Answer: .006 ft/min (8/405pi)

anonymous · Answer

@phi

phi · Answer

Here are a lot of examples of this type problem http://tutorial.math.lamar.edu/Classes/CalcI/RelatedRates.aspx

anonymous · Answer

oh ok thx but can u show me how to do this one particular problem? so far i put
d=3h
2r=3h
$$r=\frac{ 3 }{ 2 }h$$

anonymous · Answer

my internet is kinda laggy :x

phi · Answer

Start with the volume of a cone
$$ V= \frac{1}{3} \pi r^2 h$$
they say the diameter is 3 times the height

phi · Answer

they give dV  change in volume
they want dh  change in height.  so we want to replace r with h in the formula

anonymous · Answer

so we take the derivative of $$V=\frac{ 1 }{ 3}\pi r^2h$$ in terms of h? :)

phi · Answer

first plug in r= 3/2 h, then take the derivative

anonymous · Answer

after i plug in r i take the derivative in terms of "h"?

phi · Answer

$$ V=\frac{ 1 }{ 3}\pi r^2h $$ 
with  r= 3/2 h  you get
$$ V=\frac{ 3 }{ 4}\pi h^3 $$ 
now take the derivative
$$ dV=\frac{ 3 }{ 4}\pi d\left(h^3\right) $$

phi · Answer

better, take the derivative with respect to time
$$ \frac{dV}{dt}=\frac{ 3 }{ 4}\pi \frac{d}{dt}\left(h^3\right)$$

anonymous · Answer

ok then what

phi · Answer

what did you get?

anonymous · Answer

how did u get "d" in $$dV = \frac{ 3 }{ 4 }\pi  d{}(h^3)$$

anonymous · Answer

where did the random d come from

phi · Answer

I meant to say derivative wrt time

anonymous · Answer

how do i get h by itself now

phi · Answer

did you take the derivative?  what did you get?

anonymous · Answer

i take the derivative of $$\frac{ 3 }{ 4 }\pi h^3$$ rite?

phi · Answer

take the derivative of both sides of the equation with respect to time

anonymous · Answer

how do i do that

phi · Answer

$$ \frac{dV}{dt}=\frac{ 3 }{ 4}\pi \frac{d}{dt}\left(h^3\right) $$
d/dt   of V  is dV/dt
d/dt (h^3)  is 3 h^2 dh/dt

anonymous · Answer

like this? $$\frac{ dV }{ dt } = \frac{ 3 }{ 4 }\pi 3h^2$$

anonymous · Answer

@phi

phi · Answer

with these problems you cannot leave out the dh/dt part

if you take the derivative of x^3 with respect to x  you do
3 x^2  dx/dx   and the dx/dx  is 1.  you get  3x^2
but if you take the derivative with respect to t  you do
3 x^2 dx/dt    and the dx/dt stays

anonymous · Answer

so how do i get h by itself

phi · Answer

you don't.  you should now have
$$ \frac{dV}{dt}=\frac{ 9 }{ 4}\pi h^2 \frac{dh}{dt}$$

now you plug in what you know and solve for dh/dt, the rate that the height is changing

anonymous · Answer

answer was .006ft/min

phi · Answer

dv/dt = 10 cu ft/min
h= 15 ft
dh/dt will be in ft/min

phi · Answer

$$ \frac{dh}{dt}= 10 \frac{4}{\pi \cdot 9 \cdot 15^2} $$

phi · Answer

it's algebra at this point.

anonymous · Answer

how did u get $$\frac{ dh }{ dt }=10\frac{ 4 }{ \pi  \times 9 \times 15^2 }$$

phi · Answer

look at my post where we have the derivative  dV/dt =  ...
we know dv/dt,  and h.  sub in their values, and solve for dh/dt

anonymous · Answer

ohh okay ty :)

anonymous · Answer

can u help me with another problem? ^^

phi · Answer

ok, please post it.