Find the length of the loop of the curve 9y^2=x(x-3)^2?
553747
Topacestriker, I have a doubt in your equation: is the square on the LHS only for (x-3), or is it for the whole product x(x-3)?\n\nAssuming the square is only for (x-3), then follow exactly what 'ringm' said. \n\nThe length is equal to 4*sqrt(3), which is twice what you got - this is because the integration you performed gave you the result only for the part of the curve above x-axis. But you know that it is a loop that also goes below x-axis. To account for this, you will have to perform a separate integration - but you can use a shortcut to avoid all this hassle - by judging that the curve is symmetric about the x-axis, so just double the result you got on the first integration, i.e. twice of 2*sqrt(3) = 4*sqrt(3)
797189
x - 3y = 9\n\nTo solve for "y" you have to isolate it on one side of the equation. First subtract x from both sides to get: \n\n-3y = 9 - x\n\nthen divide both sides by -3 to get:\n\ny = (9 - x) / -3  \n\nWhich is y = -3 + 1/3 x or in slope-intercept form: \n\ny = 1/3 x -3 \n\n\nNow graph the y-intercept at -3 and use the slope (go 1 up + 3 across) to get the next point and so forth. \n\nHope this helps...Good luck! =)
928715
Say the length in centimeters is l\nThen the bredth(b) is 3 less than that, \nb = l - 3\n\nNow the bredth, length and diagonal(d) form a right triangle, \nso we can say d^2 = b^2 + l^2 by pythagorus.\nd^2 = (l-3)^2 + l^2\n15^2 = (l-3)(l-3) + l^2\n225 = l^2 - 6l + 9 + l^2\n216 = 2l^2 - 6l          div by 2\n108 = l^2 - 3l            move over the 108\n0 = l^2 - 3l - 108\nYou could use the quadratic formula, or factor.\n0 = (l-12)(l+9)\nl = 12\n\nb = 12 - 3 = 9\nThe bredth is 9cm and the length is 12cm.\n\nSince the other diagonal is just the diagonal flipped along the center, it is the same length: 15 cm.
97125
use the distance formula\nsqrt(3^2+(-sqr3)^2)=?
861574
first find the centripital acceleration\na_c = v^2 / r\na_c = 140^2/1000\na_c = 19.6\n\nthe total acceleration at the bottom of the loop is:\na = a_c + g\na = 19.6 + 9.8\na = 29.4\n\nF = m*a\nF = 70*29.4\nF = 2059 Newtons\n\n29.4/9.8 = 3 so the pilot is pulling 3 Gs at the bottom of the loop
1092235
Use FOIL\n(3 - 2sqrt3)^2\n(3)(3) + (3)(-2sqrt3)  + (-3sqrt3)(3) + (-2sqrt3)(-2sqrt3)\n= 9 - 6sqrt3 - 6sqrt3 + 4*3\n= 21 - 12sqrt(3)
745065
(3m/n)(n^2/9) \n=(3mn^2)/(9n)\n=(3mn)/(9)\n=(mn)/3
157955
9*9\n=(3^2)(3^2)\n=3^(2+2)\n=3^(4)\n=3*3*3*3\n=(3*3)*(3*3)\n=(9)(9)\n=1(9*9)\n=1(9+9+9+9+9+9+9+9+9)\n=1(18+18+18+18+9)\n=1(36+36+9)\n=1(72+9)\n=1(81)\n=1*81\n=81\nFINALLY, I GOT IT!\nThe answer to this complex rational algebraic problem is 81\n\nDuh!  ;-p
256079
x(x^2 -9)= x (x-3)(x+3)
316497
If a curve in n dimensions is given parametrically as:\n\n( f1(s), f2(s), ..., fn(s) ) for a <= s <= b\n\nthen you can approximate the arc length by dividing the domain [a,b] into many small pieces a = s0, s1, s2, ..., sk = b and summing the lengths of the chords (secants? straight line segments. whatever) between s_{i} and s_{i+1}.\n\nThe length of the curve is defined as the least upper bound of all such sums (if that set has an upper bound).\n\nUnder suitable hypotheses you can compute this as the definite integral of ``ds'' as s varies from a to b, using the Euclidean rule\n\n(ds)^2 = (df1(s)/ds)^2 + ... + (dfn(s)/ds)^2\n\ni.e., your length is the definite integral from a to b of the square root of the sum of the squares of the derivative of each coordinate function.\n\nFor example for a circle of radius one at the origin parameterized as ( cos(t), sin(t) ) as t varies from a = 0 to b = 2 pi you get ... 2 pi ... which is what you would expect.  :-)  :-)\n\nOr for the arc length of ( (s^3)/3 - s, s^2 ) from s = 5 to s = 8 the length is the integral from 5 to 8 of\n\nsqrt( (deriv (s^3)/3 - s w.r.t. s)^2 + (deriv s^2 w.r.t. s)^2 ) ds\n\n= sqrt( (s^2 - 1)^2 + (2s)^2 ) ds\n= sqrt( s^4 - 2s^2 + 1 + 4s^2 ) ds\n= sqrt( s^4 + 2s^2 + 1 ) ds\n= (s^2 + 1) ds\n\ni.e., integral from s=5 to s=8 of (s^2 + 1) ds.
436114
well, you could be an architect and be using a 30-60-90 triangle and you needed to know a particular side of it. you can use x times the square root of 3, x being the shorter leg, to find the length of the longer leg.
481212
the area under the curve f(x)!
910754
the length of any line passing through the origin is sqrt(x^2 + y^2) The equation of the line is y = ax, so the length L = x*sqrt(1+a^2), and x = L/(sqrt(1+a^2).
305313
4x^3y^2-16y^3-9y=\ny(4x^3y-16y^2-9)
4959
=8(y^3)z + 12(y^2)(z^2) - 20y^2
1033128
Rewrite the equation as y=9/x. If we take one point on that curve and rotate it around the axis, the circle generated has a circumference of 18π/x and an area of  81π/x² (2πr and πr², respectively). Each of these circles is as an infinitesimally thin slice of the entire solid of revolution, and so we can find the surface area and volume of it by integrating the circumference and area of these circles, respectively. Thus the surface area of the entire curve is (n→∞)lim (1, n)∫18π/x dx (that should be read as "the limit as in approaches infinity of the definite integral from 1 to n of 18 pi over x with respect to x"). The antiderivative of 18π/x is 18π ln (x) and thus the by the fundamental theorem of calculus, (n→∞)lim (1, n)∫18π/x dx= (n→∞)lim (18π ln (n) - 18π ln (1)). Since the limit of ln n as n→∞ is ∞, the surface area is infinite.\n\nRepeating for the volume: the volume is (n→∞)lim (1, n)∫81π/x² dx . Our antiderivative is then -81π/x, and so the final volume is (n→∞)lim -81π/n - (-81π/1). Since the limit of 1/n as n→∞ is zero, the first term in this expression is zero, thus the final answer is - (-81π/1) or simply 81π.
498166
Draw a right triangle and let one of the angles be the arccosine of 1/2. That means the length of the adjacent leg divided by the length of the hypotenuse is 1/2, so you may write the length of the adjacent leg as 1*x and the length of the hypotenuse as 2*x, where x is some unknown scaling factor. \nThe sine of this angle is the length of the opposite side divided by the length of the hypotenuse, 2x. We get the length of the opposite side from the Pythagorean theorem: Call that length s. Then s^2 + x^2 = (2x)^2, so s^2 = 3x^2, so s = Sqrt[3]x, the positive square root since we're dealing with lengths. Now we have the sine ratio as opposite/hypotenuse = Sqrt[3]x/(2x) = Sqrt[3]/2.