A cylinder of length 2x is inscribed in a sphere of radius a. Between one end of this cylinder and the sphere, another cylinder of length 2y is inscribed with one end on an end of the first cylinder so that the axes of the cylinder are collinear.
Show that the sum of the volume of the two cylinder is
2pi(x+y)(a^2 - x^2 -4y^2)
how can i solve this?
1. Make a sketch. (I've made a little mistake: The red line should be labeled as 2y)Originally Posted by MilK
2. Calculate the radius of the greater cylinder:
3. Calculate the volume of the greater cylinder:
4. Calculate the radius of the smaller cylinder:
5. Calculate the volume of the smaller cylinder:
6. Add the volumina, expand the brackets and factor again. You'll get the result given in the problem.
do you mean treat a as a constant when finding partial x and y OR do u mean keep treat a as a constant for the whole equation? it is not stated in the question than a is a constant. if a is not a constant in the eqn, do i need to find partial a?Keepas a constant. Then you'll get x as a function of
do you mean treat a as a constant when finding partial x and y OR do u mean keep treat a as a constant for the whole equation? it is not stated in the question than a is a constant. if a is not a constant in the eqn, do i need to find partial a?is the radius of the sphere. There is absolutely nothing in the question (that I can see) that suggests that this radius can change. It is clearly an implied constant for the problem and obviously should be treated a such.
If it worries you that much, give it a value like a = 1. Then forget about it and get on with solving the problem.
Then do the problem again but this time keep theas
i got these 2 eqn for partial x and y
for partial x = 2pi( -2x(x+y) +a^2 -x^2 -4y^2)
for partial y = 2pi( -8y(x+y) +a^2 -x^2 -4y^2)
subtract these two eqn,
i get 2pi(-6x)(x+y) = 0
therefore x+y = 0 and -12pi(x) =0
solving i get x=0, y=0
the question stated : find the max value of this eqn :
2pi(x+y)(a^2 - x^2 -4y^2)
something is wrong somewhere, after finding fxx=0, fyy=0 and fxy=0, (0,0) is not the max point.
what when wrong?
I haven't checked the details but I think it should be blue sky .....i got these 2 eqn for partial x and y
for partial x = 2pi( -2x(x+y) +a^2 -x^2 -4y^2)
for partial y = 2pi( -8y(x+y) +a^2 -x^2 -4y^2)
subtract these two eqn,
i get 2pi(-6x)(x+y) = 0 Mr F says: This is wrong.
The correct result is..........
Therefore x = -y OR x = 4y. x = 4y is the one you want since x = -y is impossible since x> 0 and y > 0. So sub x = 4y into either of your original equations and solve for y. once you have y, get x. Then test the nature of the solutions.
therefore x+y = 0 and -12pi(x) =0
solving i get x=0, y=0
the question stated : find the max value of this eqn :
2pi(x+y)(a^2 - x^2 -4y^2)
something is wrong somewhere, after finding fxx=0, fyy=0 and fxy=0, (0,0) is not the max point.
what when wrong?
equation 1
y=a/60
a = 60y
equation 2
x = 4a / 60
4a = 60x
a = 15x
so 60y=15x
but that's not what you want to do ....i think?
you could solve the equation for a in terms of y?
2pi(-8y(x+y) + a^2 - x^2 - 4y^2)=0
2pi(-8y((4y)+y) + a^2 - (4y)^2 - 4y^2)=0
2pi(-8y(5y) + a^2 - 16^2 - 4y^2)=0
-60y^2 + a^2 = 0
a = (+)root(60y^2) and (-)root(60y^2)
a = + 2y(root15) and - 2y(root15)
but god knows what the question was ?
AHHHH LOL page 2/2 >.<
You don't! It is a constant!
Why are you so determined to get rid of a in your answer??
[snip]
so my final ans is it terms of a?
sub y = a/60 and x =a/15,
fxx = -26(pi)a/30
fyy = -28(pi)a/15
fxy = 64(pi^2)(a^2) /225
there its max point since fxy^2-fxxfyy is -ve and fxx is -ve
and to find its max value, i sub y=a/60 and x=a/15 into fxy which gives max value as -8(pi)a / 15
correct?
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