Infinite series of the sum of areas of spheres of decreasing radius within a cone problem

O

Ognjen

New member
Joined
Jan 4, 2022
Messages
46
Here's the text of the problem:

In a right cone of radius of the base R and the angle at the top of its axial section 2A, a sphere is inscribed. Then, a sphere which touches the first one as well as the wrapper of the cone is also inscribed etc. in infinity. Calculate the sum of the areas of this infinite progression of spheres.

I understood it as if spheres are being piled on top of one another, heading towards the tip of the cone, in infinity. So, this is how I tried to solve it ( and failed apparently ).

[imath]ctgA = h/R\\ h = R * ctgA\\ ctgA = (h - R_1)/R_1 = (R*ctgA - R_1)/R_1\\ R_1 = R*ctgA/(ctgA + 1)\\ ctgA = (h - (2R_1 + R_2))/R_2\\ R_2 = (R*(ctgA)^2 - R*ctgA)/(ctg + 1)^2 = R*ctgA(ctgA - 1)/(ctgA + 1)^2\\ R_2 / R_1 = (ctgA - 1)/(ctgA + 1)\\ R_2^2 / R_1^2 = (ctgA - 1)^2/(ctgA + 1)^2 = q\\ S = 4\pi*R^2*(ctgA )^2/(ctgA + 1)^2*(1 + (ctgA - 1)^2/(ctgA + 1)^2 +...)\\ S = 4\pi*R^2*(ctgA )^2/(ctgA + 1)^2*(1 / (1 - q))\\ S = 4\pi*R^2*(ctgA )^2/(ctgA + 1)^2 * (1/4ctgA)\\ S = \pi*R^2*ctgA*(1/(ctgA + 1)^2)\\[/imath]

However, the final solution is supposed to be:
[math]\pi * R^2 * cosA * ctgA[/math]
This basically means that, if I calculated all the steps properly, my 1/(ctgA + 1)^2 is supposed to equal cosA, which it doesn't.

The ''official solution'' ( presented in the workbook ) it says that the progression goes as follows:
[math]R*cosA/(1 + sinA), R*cosA/(1 + sinA), ((1 - sinA)/(1 + sinA))^2,...[/math]
And it also says: "In solving this problem we use the equation:
[math]tg(\pi/4 - A/2) = cosA/(1 + sinA)[/math]"

Could someone help me out on this one ?

Thank you immensely !
 
Ognjen said:
Why is it wrong ? I'm inspecting a triangle whose one side is R2, and the other, longer side, h (the height of the right cone) minus the radius of the first sphere.
Click to expand...
It's not a [imath]\ctg[/imath].
 
Ognjen said:
ctg = adjacent/opposite

The adjacent side of this triangle is (h - R1) and the opposite is R1. How is this wrong ?
Click to expand...
The angle between those two sides is not right. I.e., the side with length (h-R1) is a hypotenuse.
 
blamocur said:
The angle between those two sides is not right. I.e., the side with length (h-R1) is a hypotenuse.
Click to expand...
How is it not right ? The axial section of cone is a triangle, and in this case, an isosceles one at that ( cause the cone is right and not oblique ). If you split such a triangle into two across its height, you get 2 right triangles, and this is exactly what this is.
 
The ratio of the hypotenuse to the opposite side is [imath]\frac{1}{\sin \alpha}[/imath], not [imath]\frac{1}{\tan\alpha}[/imath]
Do you want to make a drawing?
 
blamocur said:
The ratio of the hypotenuse to the opposite side is [imath]\frac{1}{\sin \alpha}[/imath], not [imath]\frac{1}{\tan\alpha}[/imath]
Do you want to make a drawing?
Click to expand...
 

Attachments

  • 272906230_336589431689632_5267115107389644427_n.jpg
    272906230_336589431689632_5267115107389644427_n.jpg
    39.6 KB · Views: 3
This drawing is incorrect: the radius [imath]R_1[/imath] drawn from the center of the sphere to the side of the cone is not horizontal, but forms angle [imath]A[/imath] with the horizontal plane.
 
blamocur said:
This drawing is incorrect: the radius [imath]R_1[/imath] drawn from the center of the sphere to the side of the cone is not horizontal, but forms angle [imath]A[/imath] with the horizontal plane.
Click to expand...
I really fail to make what you're talking about sensible, sorry. I just can't grasp the picture you're having in your mind, at all. Could you sketch it ?
 
I've attached a drawing of the vertical section going through the center of the cone and thes sphere.
 

Attachments

  • sphere-in-cone.pdf
    49.7 KB · Views: 7
blamocur said:
I've attached a drawing of the vertical section going through the center of the cone and thes sphere.
Click to expand...
But the radius of the sphere is the same on every ''side'' of it: both along its horizontal and vertical axis, as well as anywhere in between. What's the issue here ?

EDIT: Oh I see it now, the horizontal radius doesn't quite reach the wrapper, so it doesn't make a triangle. Thank you so much !
 
Last edited:
You must log in or register to reply here.
Top