# Pure Maths

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A container is in the shape of a right circular cone (inverted) with both height and diameter 2 meters. It is being filled with water at a rate of (pi)m^3 per minute. Fine the rate of change of height h of water when the container is 1/8th full(by volume).
(Volume of a right circular cone of radius r and height h is 1/3(pi)r^3h)

Thanks and God bless :)

• Pure Maths -

for any height h, volume is 1/3 PI r^2 h
but for r for any height h is r=h/2. Think on that.

so volume=1/3 PI *(h/2)^2*h

V=1/12 * PI * h^3
and h= cuberoot (12V/PI) solve for h when V=1/8*1/12*PI*2 (max volume, r=1,h=2)
Now for the calculus work.
dV/dh=3/12 PI h^2
but dV/dh*dh/dt=dV/dt

or dh/dt=dV/dt / dV/dh
you area given dV/dt as PI m^3/minute
and you found dV/dh=3/12 PI h^2

so figure dh/dt

• Pure Maths -

I understand the rest but not the 1/8th of the volume part. Can you give me a further explanation on that?

Thanks :)

• Pure Maths -

The volume when it's 1/8th full is Pi/12
I don't understand how they got that....

• Pure Maths -

make a sketch to see that by ratios,
r = h/2 , like bobpursely noted

The sneaky part of the question is that when the cone is 1/8 full , the water is NOT 1/8 of the way up

Full volume = (1/3) π (1^2)(2) = 2π/3 m^3

we want r and h when volume = (1/8)(2π/3) or π/12
(Again, see bobpursely above)

Volume = V = (1/3)π(r^2)(h)
= (1/3)π(h^2/4)(h) = (π/12) h^3

so when cone is 1/8 full,
(π/12) h^3 = π/12
h^3 = 1
h = 1 , and r = 1/2

Now back to actual Calculus,

V = (π/12 h^3
dV/dh = (π/4) h^2 dh/dt

plug in the given dV/dt = π, and h = 1
π = (π/4) (1^2) dh/dt
dh/dt = 1/4

So when the cone is 1/8 full, the height is changing at
1/4 m/minute

We could have done the 1/8 part in our heads by realizing that ..
The volume of two similar solids is proportional to the cube of their sides, and since
(1/2)^3 = 1/8 ......
the height must have been 1/2 of the 2 m of the cone, or 1 m

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