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December 22, 2014

December 22, 2014

Posted by **Ariana** on Saturday, August 16, 2014 at 4:11am.

(Volume of a right circular cone of radius r and height h is 1/3(pi)r^3h)

Step by step working please.

Thanks and God bless :)

- Pure Maths -
**bobpursley**, Saturday, August 16, 2014 at 5:57amfor any height h, volume is 1/3 PI r^2 h

Notice your formula is wrong.

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 -
**Ariana**, Saturday, August 16, 2014 at 6:52amI understand the rest but not the 1/8th of the volume part. Can you give me a further explanation on that?

Thanks :)

- Pure Maths -
**Ariana**, Saturday, August 16, 2014 at 6:58amAnswer is dh/dt= m/min

The volume when it's 1/8th full is Pi/12

I don't understand how they got that....

- Pure Maths -
**Reiny**, Saturday, August 16, 2014 at 8:16ammake 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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