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"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
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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
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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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
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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
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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well i posted that question on the forum
"The man was just too bored so he invented maths for fun"
-some wise guy
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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
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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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
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