Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**gourish****Member**- Registered: 2013-05-28
- Posts: 153

"A tower has the following shape: a right circular truncated right circular cone (one with radii 2R(the lower base) R(the upper base), and height R bears a right circular cylinder whose radius is R and height is 2R. Finally a semisphere of radius R is mounted on the cylinder. suppose the cross sectional area S of the tower is given by f(x) where x is the distance of the cross section from the lower base of the cone"

define the function.

i know that the function is (pie)*R^2 for the interval R≤ x≤ 3R

but what about the other intervals and yeah the domain of the function is is from 0 to 4R. i am facing problems with relating the area of the cross section with the radii and x of both the cone and semisphere help me please!

"The man was just too bored so he invented maths for fun"

-some wise guy

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,012

hi gourish,

Let's see if I can 'talk' my way up this tower.

(i) At the bottom it has radius 2R and this tapers to a radius of R over a distance of height R

(ii) Then it becomes a cylinder. I agree with your formula for that.

(iii) At the top it is half a sphere, radius R.

So now for the formulas:

I'll do the radius at each stage. You can convert that to the area of a circle (they all have circular cross sections)

(i) This is a linear function as the taper is regular. I'll use y for the radius at any point

At x = 0, y = 2R. At x = R, y = R. So if you graphed this the gradient would be -1 (R across is R down * that's putting the tower on its side so that x is across as usual and y up. I've added a diagram to show this)

y = 2R -x

(iii) Here the x and y coordinates lie on the sphere so you can use Pythagoras to write an equation linking them:

(x-3R)^2 + y^2 = R^2

Re-arrange to make y the subject:

y = √(R^2 - [x-3R]^2)

Hopefully that should do it.

Bob

ps. Looks a bit like a Dalek.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**gourish****Member**- Registered: 2013-05-28
- Posts: 153

that was a great explanation thanks bob. it was really helpful but i have another of similar kinda this time i am asked to find out the volume of triangle inscribed in a sphere of radius R. where x is the height of the triangle inscribed in it. i am having troubling finding the domain of the function as a starter and that's the reason why i am not even able to define this function. hope you can explain it to me like you did for my earlier question. i would really appreciate it

"The man was just too bored so he invented maths for fun"

-some wise guy

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,012

volume of triangle inscribed in a sphere

triangle??

Do you mean cone?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**gourish****Member**- Registered: 2013-05-28
- Posts: 153

well i posted that question on the forum

"The man was just too bored so he invented maths for fun"

-some wise guy

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,012

What has me puzzled is you have 'triangle'. That is a flat shape and has no volume. ???

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**gourish****Member**- Registered: 2013-05-28
- Posts: 153

it was my mistake the actual question was about a cone which i posted already and found the answer

"The man was just too bored so he invented maths for fun"

-some wise guy

Offline

Pages: **1**