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Question 259683: A right-circular cone of base radius 2m and height 3m is filled with water to a depth of x m.
a) Find a formula for the volume of water contained, in terms of x.
b)How deep is the water, if its volume is 1 m^3?
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! A right-circular cone of base radius 2m and height 3m is filled with water to a depth of x m.
a) Find a formula for the volume of water contained, in terms of x.
b)How deep is the water, if its volume is 1 m^3?
R=BC=2
r=DE=r
H=AB=DB+AD=h1+h2=x+h2=3
h1=x
h2=3-x
from similar triangles which ABC and ADE are:
AD/DE=AB/BC (ratios of similar sides are equal)
AD/DE=AB/BC = h2/r=(h1+h2)/R = (3-x)/r=3/2
Rh2=r(h1+h2) = 2(3-x)=r(3) = 6-2x=3r
6-2x=3r
(6-2x)/3=r
-2x=3r-6
x=(-3/2)r+3
Rh2/r=h1+h2 = 2(3-x)/r=3 = (6-2x)/r=3
Rh2=rh1+rh2 = 2(3-x)=r(x)+r(3-x) = 6-2x=rx+r(3-x)=3r
Rh2-rh2=rh1 = 2(3-x)-r(3-x)=rx
h2(R-r)=rh1 = (3-x)(2-r)=rx
h2=rh1/(R-r) = 3-x=rx/(2-r)
(h2(R-r))/r=h1 = ((3-x)(2-r))/r=x
R=(r(h1+h2))/R = 2=3r/2
4=3r --> 4/3=r
Rh2/(h1+h2)=r = 2(3-x)/3=4/3
2(3-x)=4
3-x=2
x=1
6-2x=3r
6-2(1)=3r
6-2=3r
4=3r
4/3=r
h1=x=1
h2=3-x=3-1=2
H=h1+h2=1+2=3
volume V of cone = 1/3*pi*R^2*H = 1/3*pi*(BC^2)*AB
V=1/3*pi*2^2*3=4pi
volume of top cone = 1/3*pi*r^2*h2 = 1/3*pi*(DE^2)*AD
Vtop = 1/3*pi*(4/3)^2*2
Vtop = 1/3*pi*16/9*2
Vtop = 2/3*pi*16/9
Vtop = (32/27)pi
V-Vtop=Vbottom=Vbot
Vbot=4pi - (32/27)pi
Vbot=(108/27)pi - (32/27)pi (27*4=108)
Vbot=(76/27)pi
Vbot=V-Vtop=(pi/3)(R^2(h1+h2)-r^2*h2)
Vbot=V-Vtop=(pi/3)((R^2)(Rh2/r))-r^2*h2)
Vbot=(pi/3)((R^2)( (R(rh1/(R-r))/r) - r^2(rh1/(R-r)) ))
Vbot=(pi/3)( (R^2*1/r*Rrh1/(R-r) - r^3h1/(R-r)) )
Vbot=(pi/3)( R^3h1/(R-r) - r^3h1/(R-r))
Vbot=(pi/3)*h1*((R^3-r^3)/(R-r))
Vbot=(pi/3)*h1*(R^2+Rr+r^2)
(R-r)*(R^2+Rr+r^2(
R^3+R^2*r+R*r^2-R^2*r-R*r^2-r^3=R^3-r^3
Vbot=(pi/3)x(R^2+Rr+r^2)
Vbot=((pi*x)/3)*(2^2+2r+r^2)
Vbot=((pi*x)/3)*(4+2r+r^2)
(6-2x)/3=r
2-(2/3)x=r
4-(4/3)x=2r
(2-(2/3)x)(2-(2/3)x=r^2
((-2/3)x+2)((-2/3)x+2)=r^2
(4/9)x^2+2*2*(-2/3)x+4=r^2
(4/9)x^2-(8/3)x+4=r^2
Vbot=((pi*x)/3)*(4+4-(4/3)x+(4/9)x^2-(8/3)x+4)
Vbot=((pi*x)/3)*(12-(12/3)x+(4/9)x^2)
Vbot=((pi*x)/3)*(12-4x+(4/9)x^2)
Vbot=((pi*x)/3)*((4/9)x^2-4x+12)
Vbot=(pi*x)*((4/27)x^2-(4/3)x+4)
Vbot=(pi*x)*((4/27)x^3-(36/27)x+108/27) (4*27=108)
Vbot=((4/27)x^4-(36/27)x^2+(108/27)x)*pi (formula for the volume of water contained, in terms of x)
Vbot=(4/27-36/27+108/27)*pi
Vbot=(76/27)pi
if Vbot=1, then what is x (x is depth of the water)
Vbot=V-Vtop=(pi*R^2*H)/3 - (pi*r^2*h2)/3)
1=4pi-Vtop
1 = 4pi - (pi*r^2*h2)/3)
1 - 4pi = - (pi*r^2*h2)/3)
4pi - 1 = (pi*r^2*(3-x))/3
12pi - 3 = pi*r^2*(3-x)
(12pi - 3)/(pi*r^2) = 3-x
( (12pi - 3) - (3*pi*r^2) )/(pi*r^2) = -x
( (3*pi*r^2) - (12pi - 3) )/(pi*r^2) = x
(6-2x)/3=r from above
(2-(2/3)x)=r
(2-(2/3)x)(2-(2/3)x=r^2
((-2/3)x+2)((-2/3)x+2)=r^2
(4/9)x^2+2*2*(-2/3)x+4=r^2
(4/9)x^2-(8/3)x+4=r^2
(4/9)x^2-(8/3)x+4=r^2 --> plug in
( (3*pi*((4/9)x^2-(8/3)x+4)) - (12pi - 3) )/(pi*((4/9)x^2-(8/3)x+4)) = x
(3 * pi * 4/9 * x^2 - 3 * pi * 8/3 * x + 12 * pi)/(4/9 * pi * x^2 - 8/3 * pi * x + 4*pi) = x
(12/9 * pi * x^2 - 8 * pi * x + 12 * pi)/(4/9 * pi * x^2 - 8/3 * pi * x + 4 * pi) = x
(12/9 * x^2 - 8 * x + 12)/(4/9 * x^2 - 8/3 * x + 4) = x
12/9 * x^2 - 8 * x + 12 = 4/9 * x^3 - 8/3 * x^2 + 4x
-4/9 * x^3 + 12/9 * x^2 + 8/3 * x^2 - 8 * x - 4 * x = -12
-4/9 * x^3 + 36/9 * x^2 - 12 * x = -12
-4/9 * x^3 + 4 * x^2 - 12 * x = -12
4/9 * x^3 - 4 * x^2 + 12 * x = 12
1/9 * x^3 - x^2 + 3x = 3
x^3 - 9 * x^2 + 27 * x - 27 = 0
(x-3)(x-3)(x-3)=0
(x-3)(x^2-6x+9)=0
x^3-6x^2+9x-3x^2+18x-27=0
x^3-6x^2-3x^2+9x+18x-27=0
x^3-9x^2+27x-27=0
x=3 which is how deep is the water, if its volume is 1 m^3
x=3 is the same as the height of the whole cone
the cone would be all the way full then
hope this was not all too confusing
am hoping you can figure out what I did
Am not entirely certain I am right in this, If anyone ever figures out if I am or not and can explain why, I spent a ton of time on this one, too much for me, am throwing my hands up in the air on this one. So anyway if anyone can explain why I might be in error please let me know. Sorry.
!!!! REVISITING PROBLEM !!!! (sorry for caps just meant to draw attention)
I got this: --> Comment from student: The answers should be : a ) V(x) = (4pix^3)/2 b ) 1.290 m deep Thanks for trying.
Well that set me to wondering well what if the problem was not talking about an upright right-circular cone but an inverted one?
One with the point at the bottom instead of the top like I was thinking?
Well then the total depth (H) would be 3 m, and the radius R at the top would be 2 m.
The depth of the water would still be x, and the radius of the water would be r.
The remaining height would be 3-x, so 3-x would be height not containing water.
The total volume would still be (pi*R^2*H)/3 or
(pi * 2^2 * 3)/3 or 4pi.
The water volume would be (pi*r^2*x)/3.
From ratios of similar triangles x/r = H/R = 3/2
solving for x and r we get:
2x=3r
x=(3r)/2 and r=(2x)/3
and r^2=(4x^2)/9
plugging the solved r^2 into water volume and we get:
(pi * (4x^2)/9 * x)/3
(4*pi*x^3)/9 * 1/3
(4*pi*x^3)/27 which should be the volume of the water in terms of x
(which solves a) Find a formula for the volume of water contained, in terms of x. )
Now we want to solve: b)How deep is the water, if its volume is 1 m^3?
1 = (4*pi*x^3)/27 (solve for x which would be how deep the water is)
27 = 4*pi*x^3
27/(4pi) = x^3
3/(cubed root of 4pi) = x
1.290381 = (approximately) 1.290 m deep = x , which is answer the person said it was, but wait they said the formula was V(x) = (4pix^3)/2 , well that can just not be possible (and if you can prove me wrong please do so), I will show why
(4pix^3)/2 = 1 (trying their answer out and solving for x to show you)
4*pi*x^3 = 2
2*pi*x^3 = 1
pi*x^3 = 1/2 = 0.5
x^3 = 1/2 / pi = 1/2 * 1/pi = 1/(2pi) = 0.159155
x=0.541926 m deep which is way off 1.290 m deep
27/(4pi) / 1/(2pi) = 27/4pi * 2pi/1 = (54pi)/(4pi) = 13.5 = 27/2 which is factor the formula they had is off by
cubed root of 13.5 is 2.381102 which is factor their x was off by
0.541926 * 2.381102 = 1.290381082452 which is the proper answer
There am finally done with this problem.
It would of been a heck of a lot easier if known it was an inverted cone to begin with.
Hope any of this helps someone.
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