document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=259683>Question 259683</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A right-circular cone of base radius 2m and height 3m is filled with water to a depth of x m.\r<BR>\n" );
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document.write( "a) Find a formula for the volume of water contained, in terms of x.\r<BR>\n" );
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document.write( "b)How deep is the water, if its volume is 1 m^3? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #191414 by </B><A HREF=/tutors/aboutme.mpl?userid=CharlesG2><B>CharlesG2(834)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CharlesG2><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=191414\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A right-circular cone of base radius 2m and height 3m is filled with water to a depth of x m.<BR>\n" );
document.write( "a) Find a formula for the volume of water contained, in terms of x.<BR>\n" );
document.write( "b)How deep is the water, if its volume is 1 m^3?\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+drawing%28+300%2C+200%2C+-5%2C+10%2C+-10%2C+5%2C%0D%0A++triangle%28+0%2C+0%2C+0%2C+-8%2C+6%2C+-8+%29%2C%0D%0A++triangle%28+0%2C+0%2C+0%2C+-4%2C+3%2C+-4+%29%2C%0D%0A++locate%28+-0.5%2C+0.5%2C+A+%29%2C+locate%28+-0.5%2C+-8%2C+B+%29%2C+locate%28+5.5%2C+-8%2C+C+%29%2C%0D%0A++locate%28+-0.5%2C+-3.5%2C+D+%29%2C+locate%28+2.5%2C+-4%2C+E+%29%0D%0A++%29%0D%0A ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "R=BC=2<BR>\n" );
document.write( "r=DE=r<BR>\n" );
document.write( "H=AB=DB+AD=h1+h2=x+h2=3<BR>\n" );
document.write( "h1=x<BR>\n" );
document.write( "h2=3-x<BR>\n" );
document.write( "from similar triangles which ABC and ADE are:<BR>\n" );
document.write( "AD/DE=AB/BC (ratios of similar sides are equal)<BR>\n" );
document.write( "AD/DE=AB/BC = h2/r=(h1+h2)/R = (3-x)/r=3/2<BR>\n" );
document.write( "Rh2=r(h1+h2) = 2(3-x)=r(3) = 6-2x=3r<BR>\n" );
document.write( "6-2x=3r<BR>\n" );
document.write( "(6-2x)/3=r<BR>\n" );
document.write( "-2x=3r-6<BR>\n" );
document.write( "x=(-3/2)r+3<BR>\n" );
document.write( "Rh2/r=h1+h2 = 2(3-x)/r=3 = (6-2x)/r=3<BR>\n" );
document.write( "Rh2=rh1+rh2 = 2(3-x)=r(x)+r(3-x) = 6-2x=rx+r(3-x)=3r<BR>\n" );
document.write( "Rh2-rh2=rh1 = 2(3-x)-r(3-x)=rx<BR>\n" );
document.write( "h2(R-r)=rh1 = (3-x)(2-r)=rx<BR>\n" );
document.write( "h2=rh1/(R-r) = 3-x=rx/(2-r)<BR>\n" );
document.write( "(h2(R-r))/r=h1 = ((3-x)(2-r))/r=x<BR>\n" );
document.write( "R=(r(h1+h2))/R = 2=3r/2<BR>\n" );
document.write( "4=3r --> 4/3=r<BR>\n" );
document.write( "Rh2/(h1+h2)=r = 2(3-x)/3=4/3<BR>\n" );
document.write( "2(3-x)=4<BR>\n" );
document.write( "3-x=2<BR>\n" );
document.write( "x=1<BR>\n" );
document.write( "6-2x=3r<BR>\n" );
document.write( "6-2(1)=3r<BR>\n" );
document.write( "6-2=3r<BR>\n" );
document.write( "4=3r<BR>\n" );
document.write( "4/3=r<BR>\n" );
document.write( "h1=x=1<BR>\n" );
document.write( "h2=3-x=3-1=2<BR>\n" );
document.write( "H=h1+h2=1+2=3<BR>\n" );
document.write( "volume V of cone = 1/3*pi*R^2*H = 1/3*pi*(BC^2)*AB<BR>\n" );
document.write( "V=1/3*pi*2^2*3=4pi<BR>\n" );
document.write( "volume of top cone = 1/3*pi*r^2*h2 = 1/3*pi*(DE^2)*AD<BR>\n" );
document.write( "Vtop = 1/3*pi*(4/3)^2*2<BR>\n" );
document.write( "Vtop = 1/3*pi*16/9*2<BR>\n" );
document.write( "Vtop = 2/3*pi*16/9<BR>\n" );
document.write( "Vtop = (32/27)pi<BR>\n" );
document.write( "V-Vtop=Vbottom=Vbot<BR>\n" );
document.write( "Vbot=4pi - (32/27)pi<BR>\n" );
document.write( "Vbot=(108/27)pi - (32/27)pi  (27*4=108)<BR>\n" );
document.write( "Vbot=(76/27)pi\r<BR>\n" );
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document.write( "Vbot=V-Vtop=(pi/3)(R^2(h1+h2)-r^2*h2)<BR>\n" );
document.write( "Vbot=V-Vtop=(pi/3)((R^2)(Rh2/r))-r^2*h2)<BR>\n" );
document.write( "Vbot=(pi/3)((R^2)( (R(rh1/(R-r))/r) - r^2(rh1/(R-r)) ))<BR>\n" );
document.write( "Vbot=(pi/3)( (R^2*1/r*Rrh1/(R-r) - r^3h1/(R-r)) )<BR>\n" );
document.write( "Vbot=(pi/3)( R^3h1/(R-r) - r^3h1/(R-r))<BR>\n" );
document.write( "Vbot=(pi/3)*h1*((R^3-r^3)/(R-r))<BR>\n" );
document.write( "Vbot=(pi/3)*h1*(R^2+Rr+r^2)\r<BR>\n" );
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document.write( "(R-r)*(R^2+Rr+r^2(<BR>\n" );
document.write( "R^3+R^2*r+R*r^2-R^2*r-R*r^2-r^3=R^3-r^3\r<BR>\n" );
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document.write( "Vbot=(pi/3)x(R^2+Rr+r^2)<BR>\n" );
document.write( "Vbot=((pi*x)/3)*(2^2+2r+r^2)<BR>\n" );
document.write( "Vbot=((pi*x)/3)*(4+2r+r^2)\r<BR>\n" );
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document.write( "(6-2x)/3=r<BR>\n" );
document.write( "2-(2/3)x=r<BR>\n" );
document.write( "4-(4/3)x=2r<BR>\n" );
document.write( "(2-(2/3)x)(2-(2/3)x=r^2<BR>\n" );
document.write( "((-2/3)x+2)((-2/3)x+2)=r^2<BR>\n" );
document.write( "(4/9)x^2+2*2*(-2/3)x+4=r^2<BR>\n" );
document.write( "(4/9)x^2-(8/3)x+4=r^2\r<BR>\n" );
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document.write( "Vbot=((pi*x)/3)*(4+4-(4/3)x+(4/9)x^2-(8/3)x+4)<BR>\n" );
document.write( "Vbot=((pi*x)/3)*(12-(12/3)x+(4/9)x^2)<BR>\n" );
document.write( "Vbot=((pi*x)/3)*(12-4x+(4/9)x^2)<BR>\n" );
document.write( "Vbot=((pi*x)/3)*((4/9)x^2-4x+12)<BR>\n" );
document.write( "Vbot=(pi*x)*((4/27)x^2-(4/3)x+4)<BR>\n" );
document.write( "Vbot=(pi*x)*((4/27)x^3-(36/27)x+108/27)   (4*27=108)\r<BR>\n" );
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document.write( "Vbot=((4/27)x^4-(36/27)x^2+(108/27)x)*pi (formula for the volume of water contained, in terms of x)\r<BR>\n" );
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document.write( "Vbot=(4/27-36/27+108/27)*pi<BR>\n" );
document.write( "Vbot=(76/27)pi\r<BR>\n" );
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document.write( "if Vbot=1, then what is x (x is depth of the water)\r<BR>\n" );
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document.write( "Vbot=V-Vtop=(pi*R^2*H)/3 - (pi*r^2*h2)/3)<BR>\n" );
document.write( "1=4pi-Vtop<BR>\n" );
document.write( "1 = 4pi - (pi*r^2*h2)/3)<BR>\n" );
document.write( "1 - 4pi = - (pi*r^2*h2)/3)<BR>\n" );
document.write( "4pi - 1 = (pi*r^2*(3-x))/3<BR>\n" );
document.write( "12pi - 3 = pi*r^2*(3-x)<BR>\n" );
document.write( "(12pi - 3)/(pi*r^2) = 3-x<BR>\n" );
document.write( "( (12pi - 3) - (3*pi*r^2) )/(pi*r^2) = -x<BR>\n" );
document.write( "( (3*pi*r^2) - (12pi - 3) )/(pi*r^2) = x<BR>\n" );
document.write( "(6-2x)/3=r from above<BR>\n" );
document.write( "(2-(2/3)x)=r<BR>\n" );
document.write( "(2-(2/3)x)(2-(2/3)x=r^2<BR>\n" );
document.write( "((-2/3)x+2)((-2/3)x+2)=r^2<BR>\n" );
document.write( "(4/9)x^2+2*2*(-2/3)x+4=r^2<BR>\n" );
document.write( "(4/9)x^2-(8/3)x+4=r^2<BR>\n" );
document.write( "(4/9)x^2-(8/3)x+4=r^2 --> plug in<BR>\n" );
document.write( "( (3*pi*((4/9)x^2-(8/3)x+4)) - (12pi - 3) )/(pi*((4/9)x^2-(8/3)x+4)) = x<BR>\n" );
document.write( "(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<BR>\n" );
document.write( "(12/9 * pi * x^2 - 8 * pi * x + 12 * pi)/(4/9 * pi * x^2 - 8/3 * pi * x + 4 * pi) = x<BR>\n" );
document.write( "(12/9 * x^2 - 8 * x + 12)/(4/9 * x^2 - 8/3 * x + 4) = x<BR>\n" );
document.write( "12/9 * x^2 - 8 * x + 12 = 4/9 * x^3 - 8/3 * x^2 + 4x<BR>\n" );
document.write( "-4/9 * x^3 + 12/9 * x^2 + 8/3 * x^2 - 8 * x - 4 * x = -12<BR>\n" );
document.write( "-4/9 * x^3 + 36/9 * x^2 - 12 * x = -12<BR>\n" );
document.write( "-4/9 * x^3 + 4 * x^2 - 12 * x = -12<BR>\n" );
document.write( "4/9 * x^3 - 4 * x^2 + 12 * x = 12<BR>\n" );
document.write( "1/9 * x^3 - x^2 + 3x = 3<BR>\n" );
document.write( "x^3 - 9 * x^2 + 27 * x - 27 = 0<BR>\n" );
document.write( "(x-3)(x-3)(x-3)=0<BR>\n" );
document.write( "(x-3)(x^2-6x+9)=0<BR>\n" );
document.write( "x^3-6x^2+9x-3x^2+18x-27=0<BR>\n" );
document.write( "x^3-6x^2-3x^2+9x+18x-27=0<BR>\n" );
document.write( "x^3-9x^2+27x-27=0\r<BR>\n" );
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document.write( "x=3 which is how deep is the water, if its volume is 1 m^3\r<BR>\n" );
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document.write( "x=3 is the same as the height of the whole cone\r<BR>\n" );
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document.write( "the cone would be all the way full then\r<BR>\n" );
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document.write( "hope this was not all too confusing\r<BR>\n" );
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document.write( "am hoping you can figure out what I did\r<BR>\n" );
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document.write( "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.\r<BR>\n" );
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document.write( "!!!!  REVISITING PROBLEM !!!!  (sorry for caps just meant to draw attention)\r<BR>\n" );
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document.write( "I got this: -->   Comment from student: The answers should be : a ) V(x) = (4pix^3)/2 b ) 1.290 m deep Thanks for trying.\r<BR>\n" );
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document.write( "Well that set me to wondering well what if the problem was not talking about an upright right-circular cone but an inverted one? \r<BR>\n" );
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document.write( "One with the point at the bottom instead of the top like I was thinking?\r<BR>\n" );
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document.write( "Well then the total depth (H) would be 3 m,  and the radius R at the top would be 2 m.\r<BR>\n" );
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document.write( "The depth of the water would still be x, and the radius of the water would be r.\r<BR>\n" );
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document.write( "The remaining height would be 3-x, so 3-x would be height not containing water.\r<BR>\n" );
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document.write( "The total volume would still be (pi*R^2*H)/3 or<BR>\n" );
document.write( "(pi * 2^2 * 3)/3 or 4pi.\r<BR>\n" );
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document.write( "The water volume would be (pi*r^2*x)/3.\r<BR>\n" );
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document.write( "From ratios of similar triangles x/r = H/R = 3/2<BR>\n" );
document.write( "solving for x and r we get:\r<BR>\n" );
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document.write( "2x=3r<BR>\n" );
document.write( "x=(3r)/2 and r=(2x)/3<BR>\n" );
document.write( "and r^2=(4x^2)/9\r<BR>\n" );
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document.write( "plugging the solved r^2 into water volume and we get:\r<BR>\n" );
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document.write( "(pi * (4x^2)/9 * x)/3<BR>\n" );
document.write( "(4*pi*x^3)/9 * 1/3<BR>\n" );
document.write( "(4*pi*x^3)/27  which should be the volume of the water in terms of x\r<BR>\n" );
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document.write( "(which solves a) Find a formula for the volume of water contained, in terms of x. )\r<BR>\n" );
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document.write( "Now we want to solve: b)How deep is the water, if its volume is 1 m^3?\r<BR>\n" );
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document.write( "1 = (4*pi*x^3)/27 (solve for x which would be how deep the water is)<BR>\n" );
document.write( "27 =  4*pi*x^3<BR>\n" );
document.write( "27/(4pi) = x^3<BR>\n" );
document.write( "3/(cubed root of 4pi) = x<BR>\n" );
document.write( "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\r<BR>\n" );
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document.write( "(4pix^3)/2 = 1 (trying their answer out and solving for x to show you)<BR>\n" );
document.write( "4*pi*x^3 = 2<BR>\n" );
document.write( "2*pi*x^3 = 1<BR>\n" );
document.write( "pi*x^3 = 1/2 = 0.5<BR>\n" );
document.write( "x^3 = 1/2 / pi = 1/2 * 1/pi = 1/(2pi) = 0.159155<BR>\n" );
document.write( "x=0.541926 m deep which is way off 1.290 m deep\r<BR>\n" );
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document.write( "27/(4pi) / 1/(2pi) = 27/4pi * 2pi/1 = (54pi)/(4pi) = 13.5 = 27/2 which is factor the formula they had is off by\r<BR>\n" );
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document.write( "cubed root of 13.5 is 2.381102 which is factor their x was off by<BR>\n" );
document.write( "0.541926 * 2.381102 = 1.290381082452 which is the proper answer\r<BR>\n" );
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document.write( "There am finally done with this problem.\r<BR>\n" );
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document.write( "It would of been a heck of a lot easier if known it was an inverted cone to begin with.\r<BR>\n" );
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document.write( "Hope any of this helps someone.\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );