Function geometry problem

[Guest]

A right circular cone is inscribed in a sphere of radius 16 inches.

Express the volume of the cone as a function of it's altitude.

Please show full working out to show thought process of solving problem.


Thanks in advance

[Guest]

Is that the answer ???

V=[tex]\frac{\pi(32h^{2}-h^{3})}{3}[/tex]

[Guest]

Please can you show your working out as this is MUCH MORE VALUABLE than the answer it's self!!!!

[Guest]

The computer will translate,heres the solution:
Cone with peak-C,radius of the base r=AH,height-CH=h.Let AB be the diameter of the circle that is the base of the cone.Then point H is AB medium.
(Volume of straight circular cone) V=[tex]\pi[/tex][tex]\frac{r^{2}h}{3}[/tex] (1)
We hawe to eхpress r by h.Let the rights CH cross the sphere at point D.
Sections (chordi) AB,CD are intersecting at point H [tex]\Rightarrow[/tex] AH.BH=CH.DH

r.r=CH(CD-CH) [tex]\Leftrightarrow[/tex] [tex]r^{2}[/tex]=h(32-h) [tex]\Leftrightarrow[/tex] [tex]r^{2}[/tex]=32h-[tex]h^{2}[/tex] (2)

Replace (2) in (1) V=[tex]\pi[/tex][tex]\frac{(32h-h^{2})h}{3}[/tex]

V=[tex]\frac{\pi(32h^{2}-h^{3})}{3}[/tex]

[Guest]

The above solution is understood.

What I don't understand is why my solution fails.

AGREED! Radius(r) rewritten in terms of (h) becomes [tex]r = ( 32-h )[/tex]

The solution reads [tex]r^{2} = ( 32-h ) * h[/tex] since [tex]h = r[/tex] ( in this example )

But why can't [tex]r^{2}[/tex] be rewritten as [tex]r^{2}[/tex] = [tex](32-h)^{2}[/tex] and replaced for [tex]r^2[/tex] in the cone volume formula...

i.e

Cone_Vol [tex]= (1/3)\pi * (32 - h )^2 * h[/tex]

My formula works for [tex]h = 16[/tex] but falls apart for all other values.

[Guest]

How do you know that r=32-h
Did you prove it ?
I think you have a mistake here.

[Guest]

How do you know that r=32-h
Did you prove it ?
I think you have a mistake here.

The above proof is not my work and so I don't take credit for it, but I think it's correct!

My understanding of the proof is not 100 percent but here's what I've learned.

Remember the question is...

can the volume of a right circular cone inscribed in a sphere of radius 16 be written in terms of it's height?

Cone Volume = [tex]\frac{1}{3}\pi r^2 h[/tex]

i.e rewrite [tex]r^2[/tex] in terms of ( h )

We take the case of a right circular cone that shares the radius of the sphere it's inscribed in.

If a circle representing a sphere is diagrammed having a VERTICAL diameter of length 32 from point(C) to point(D) on the circle

and a HORIZONTAL diameter of length 32 from point(A) to point(B) on the circle with...

point(H) at the circle centre i.e point(H) is the median of diameters AB and CD

From the diagram [tex]CH = h[/tex] ( representing the cone height )

[tex]AH = BH =[/tex] Circle radius( r ) and so...

[tex]AH * BH = r * r = r^2[/tex]

and

[tex]r^2 = CH * DH[/tex]

[tex]DH = CD - CH[/tex]

Therefore [tex]r^2 = CH * ( CD - CH )[/tex] or...

[tex]r^2 = h * ( 32 - h )[/tex]

replacing [tex]r^2 = h * ( 32 - h )[/tex] in cone volume formula equals [tex]\frac{1}{3}\pi [ h ( 32 - h ) ] * h[/tex]

i.e

[tex]\frac{1}{3}\pi ( 32h^2 - h^3 )[/tex]


That ends the proof.

I didn't have the smarts to come up with this proof despite spending hours on it - to my everlasting shame.

[Guest]

Thanks to the last guest ! Through it i understood the reason-ing and the decision.
I think the solution is wrong ,because you are look private case .
The condition does not say that the radius of the cone is equal to the radius of the sphere.
If r(cone)=R(sphere)=h,then your solution will be true.
I will write the formula as follows:

V=[tex]\pi[/tex][tex]\frac{r^{2}h}{3}[/tex]=[tex]\pi[/tex][tex]\frac{16^{2}h}{3}[/tex]=[tex]\pi[/tex][tex]\frac{256h}{3}[/tex]

The formula you get is valid only for this particular case,when r=h=16 .

[Guest]

Thanks to the last guest ! Through it i understood the reason-ing and the decision.
I think the solution is wrong ,because you are look private case .
The condition does not say that the radius of the cone is equal to the radius of the sphere.
If r(cone)=R(sphere)=h,then your solution will be true.
I will write the formula as follows:

V=[tex]\pi[/tex][tex]\frac{r^{2}h}{3}[/tex]=[tex]\pi[/tex][tex]\frac{16^{2}h}{3}[/tex]=[tex]\pi[/tex][tex]\frac{256h}{3}[/tex]

The formula you get is valid only for this particular case,when r=h=16 .

First of all I want to thank you for your reply.

I know the question did not ask the reader to assume the cone radius to equal the radius of sphere.

It was my understanding to write the volume of ANY right circular cone in terms of it's height inscribed in a sphere of radius 16.

Because of the information given, the case was taken of the cone that shared the radius of the sphere it was inscribed in.

Finding the above formula to that case would also be the general formula.

If the answer is not [tex]V = \frac{1}{3}\pi(32h^2 - h^3)[/tex]

and the answer is also not

[tex]V = \frac{1}{3}\pi * 256h[/tex]

then please could you show your reasoned steps to the correct answer.

[Guest]

I think the solution,which was sent on August 25.2018.at 12:58 is true.(She writes everything there.)