Balloon V=4/3pir^3 andA=4pir^2. How fast is surface area changing as a function of r?
929521
V=(4/3)pir^3\ndV/dt=4pir^2*dr/dt\n10=4pir^2*dr/dt\nA=4pir^2\ndA/dt=8pir*dr/dt\n=4pir^2*dr/dt*/r\n=10/r\nwhen r=8 dA/dt\n=10/8cm^2/sec
1047892
OK, let p be pi, OK?\n\nThe volume of the sphere is 4*p*r^3/3...\n\nNOW, the first derivative of that, with respect to the radius, is:\n\n4*p*r^2 dr\n\nSO, you can approximate the answer by:\n\n(4*p*3^2)*0.01\n\nThat'd give you 0.36*pi cubic inches, which would be b
1063841
W = PV = RTn, where P = pressure, V = volume, R = gas constant = 8.31 J/mol.K, T = temperature, n = moles of gas involved.\n\n1) Change in P = 4 x 10^5 Pa, then:\n\nChange in W = 4 x 10^5 Pa x 0.250 m^3 = 10^5 Pa.m^3 = 1 x 10^5 Joules.\n\n2) Change in V = 17.00 x 10^(-2), then:\n\nChange in W = 6 x 10^5 Pa x 17.00 x 10^(-2) m^3 = 102.0 x 10^3 Pa.m^3 = 1 x 10^5 Joules (note: there is only one significant number in result)
1064629
find the surface area first (thats easy). Then given that the emissivity of th sphere is .851, and that the rate of change will be the change of temp/time. Since there is no time involved directly, we need to have the Surface area x emissivity rate / the temperature difference (in relation to the time, which is given within the emissivity rate). This should give u the net rate, depending on the emissivity/time relationship. You do the working out, otherwise you won't learn it.
218464
If by area you mean "Surface area" then\nSA(sphere) = 4(pi)(r)^2\nand\nV(sphere) = 4(pi)(r)^3/3\n\nSince SA = 196 cm^2, then\n\n196cm^n = 4(pi)r^2\nr^2 = 49/pi\nr = 7/sqrt(pi) or approx. 3.949 cm\n\nfor the volume,\nV = 4(pi)(r)^3/3\nV = 4(pi)(7/pi^0.5)^3/3\nV = (4pi/3)(343/pi^1.5)\nV = (1372/(3pi^.5)) or approx. 258.02 cm^3\n\n^_^ Hope this helps ^_^
135768
Area = Pi times the radius squared (A=pr2)\nIf you need to know how to find the radius, go here:\n\nhttp://math.about.com/library/blcircle.htm
739747
p1v1=np2v2\nv2 = 4/3 pi 0.15^3 = 0.014137\nn----> no. of balloons\nputting values we get n = 679.03 so approx 679 balloons.\nhope this helps
1035334
The volume of a sphere with radius r is (4/3)pi * r^3. Your radius is 7yds, so the volume is (4/3) pi (7yd)^3 = (4/3)(343)(pi) yd^3 = 457 1/3 (pi) yd^3
1013776
V of sphere = (4/3) * pi * r^3\n\n= (4/3) * pi * (2 sqrt 3) ^ 3\n\n= (4/3) * pi * (2^3) * (sqrt 3)^3\n\n=(4/3) * pi * (8) * 3 * sqrt 3\n\n=(32 sqrt 3) * pi   cubic units
822086
The volume is V = pi*r^2*h/3 = 27.  So h = 81/(pi*r^2).\n\nThe amount of paper is the surface area without the base of the cone which is A = pi*r*sqrt(r^2+h^2)  Substitute for h:\nA = pi*r*sqrt(r^2+81^2/(pi^2*r^4))\n= sqrt(pi^2*r^4 + 81^2/r^2)\nSince the square root is an increasing function you can minimize sqrt(f) by minimizing f.  So minimize\npi^2*r^4 + 81^2/r^2\nThis differentiates to\n4pi^2*r^3 - 2*81^2/r^3, which must be equal to zero.\n4pi^2*r^3 = 2*81^2/r^3\n2pi^2*r^6 = 81^2\nr=(81^2/2pi^2)^(1/6).  If my calculations are correct you can figure out h from this...
976222
Area A = area of two ends  +  Curved surfaces.\n  A  =  2π r²  +  2π r h\n\nVolume V = Area of end  x  Heigh.\n  V  = π r²  *  h\n  V  = π r² h\n  h  =  V / π r²\n Now substitute for h into the area formula.\n\n  A  =  2π r²  +  2π r h\n  A  =  2π r²  +  2π r (V / π r²)\n  A  =  2π r²  +  2V / r\nThe volume of the can is 1000cm^3, so substitue in.\n  A  =  2π r²  +  2πV / r\n  A  =  2π r²  +  2000 / r\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\n  A  =  2π r²  +  2000 / r\n  δA/δr  =  4π r  -  2000 / r²\n  4π r  -  2000 / r² = 0\n  4π r  =  2000 / r²\n  4π r^3  =  2000\n  r^3  =  2000/ 4π\n  r  =  (2000/ 4π)^1/3    (that is, cubed root).\n  r  =  5∙419 260 701\n  r  ≈  5∙42 cm.\n\nLet's use the r value to calculate h.\n  V  = π r²  *  h\n  1000  = π (5∙419 260 701)²  *  h\n  h  =  1000 /π (5∙419 260 701)²\n  h  =  10∙838 5214\n  h  ≈  10 51 cm.\n\nFor ratio:\n h/r = 10∙838 5214 / 5∙419 260 701\n h/r = 2/1   That is, a two to one ratio.\n\nFor queation two, follow the same proceedure after you set up your formulas.\nLike wise for question three.\n\nWith question two it's not clear what value you are taken for the length of the side of the square edge.  Because you are still using r, I'm assuming you are taken the largest size of a square that fits into the circle or radius r.  Based on this assumsion, the calculations are are follows:\n\nTo identify the length x of one side of the square of radius r:\nDraw two diameters of the circle at 90º to each other.  Where these diameters make contact with the circumference of the circle will be the corner points of the square.  The area of this square now consists of four triangles.  These triangles will have two sides of length r with the enclosed angle of 90º.  So the length of x will be:\n\n x² = r² + r²\n x² = 2 r²\n x = √(2 r²)   That is, length of one side of the square.\n\nArea A = area of two square ends + parameter of square * heigh.\nA = (2r² * 2) + [(√2)r * 4 * h]\nA =  4r²  +  4√(2)r h\n\nVolume V = square area x heigh.\nV =  2 r² * h\nV =  2 r² h\nh = V / 2 r²\nNow substitute for h into the area formula.\n\nA =  4r²  +  4√(2)r h\nA =  4r²  +  4√(2)r V / 2 r²\nA =  4r²  +  2√(2) V/r\nA =  4r²  +  2000√(2) r^-1     (Volume is 1000cm^3).\n\nTake first derivide and let it equat to zero (zero for max. or min. pt.).\nA =  4r²  +  2000√(2) r^-1\nδA/δr = 8r  -  2000√(2) r^-2\n 8r  -  2000√(2) r^-2  =  0\n 8r  =  2000√(2) r^-2\n 8r  =  [2000√(2)] /r²\n 8r^3  =  2000√(2)\n r^3  =  [2000√(2)]/ 8\n r  =  ([2000√(2)]/ 8)^1/3             (that is, cubed root).\n r  =  (353∙55 33 906...)^1/3\n r  =  7∙071 0678 12\n r  ≈  7∙07 cm.\n\nLet's use the r value to calculate h.\nV = 2r² * h\n1000 = 2r² * h\n1000 = 2 (50) * h\n1000 = 100h\n h = 10 cm.\n\nFor ratio:\nh/r = 10 / 7∙071 0678 12\nh/r = 1∙845 270 149 / 1\n h : r  ≈ 1∙85 : 1\nThese ratios do not take into account the scrap metal from the circle when the square is cut out of it.
801554
Assume spherical approximation for balloon:\n\nVolume of weather balloon at time, t is:\n\nV(t) = 4/3 pi r(t)^3\n\n(1).....dV(t)/dt = 4 pi r(t)^2 x dr(t)/dt  (chain rule)\n\nlet t be in minutes\n\nr(t)= .04t m,  that is (500/1000)*0.08*t m\n\nd r(t)/dt = .04m/min\n\nSo when r(t) = 90cm = .9m, we have from (1)\n\nd V(t)/dt = 4 pi (0.9)^2 x .04 m^3/min\n\n=0.4072 m^3/min
637876
v=4/3 pi r^3 (four thirds x pi x (r cubed))
928305
The volume V=abc\nSurfce area S=2(ab+bc+ca)\nTherefore,2/S(1/a+1/b+1/c)\n=2/2(ab+bc+ca )](ab+bc+ca)/abc\n=1/abc=1/V proved
219209
Surface Area of a Cylinder = 2 pi r 2 + 2 pi rh. (h is the height of the cylinder, r is the radius of the top).\n\nSurface Area = 2(pi r 2) + (2 pi r)* h \n\nSurface Area = Areas of top and bottom +Area of the side
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!
786356
at 3 sec r = 3*8 = 24m\n\na = pi*r^2\n\nda/dt = d(pi*r^2)/dt \nda/dt = 2*pi*r*(dr/dt)\nda/dt = 2*pi*24*8\nda/dt = 384*pi\nda/dt = 1206.4 meters^2 per sec