Find the Length of the Curve?
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The arc length of the curve y = f(x) from x = a to x = b is\n\nL = ∫[a to b]√{ 1 + [f '(x)]² } dx\n\nRewrite the curve as f(x) = 3 x^(4/3) - (3/16) x^(2/3) from x = -216 to 216. [I believe you made a typo with 3/12, else the following will get messy!]\n\nf '(x) = 4 x^(1/3) - (1/8) x^(-1/3)\n\nAfter simplifying, and using symmetry\n\nL = 2 ∫[0 to 216]√[ 16 x^(2/3) + (1/64) x^(-2/3) ] dx\n\nLet t = x^(2/3) and get\n\nL = 3 ∫[0 to 36]√( 16 t² + 1/64 ) dt\n\nLet u = 4t to get\n\nL = ¾ ∫[0 to 144]√( u² + 1/64 ) du\n\nThis evaluates to\n\n¾ [0 to 144 of \n½ u √( u² + 1/64 ) + ½ (1/64) ln | u + √( u² + 1/64 ) | ]\n\nwhich simplifies to\n\n54 √( 144² + 1/64 ) + \n( 3 / 512 ) { ln [ 144 + √( 144² + 1/64 ) ] - ln ( 1/8 )\n\n= (27/4) √1327105 + ( 3 / 512 ) ln ( 1152 + √1327105 )\n\n= 7776.048295\n\nVerify using WIMS Function calculator
1057756
lets say length= L\nwidth=W\nperimeter=P=22=2L+2W\nL=W+7\n\n22=2(W+7)+2W\n22=4W+14\n8=4W\n2=W\n9=L
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Example: Find the length of a helix which winds three times around a cylinder of height 12" and radius 3"\n\nAnswer: divide the cylinder into three sections. The helix goes once around a cylinder of height 4" and radius 3"\n\nThe surface of the cylinder can be "peeled" into a rectangle 4" by 2 * pi * 3" = 18.850". The helix is a diagonal of that rectangle. So its length is sqt(4^2 + 18.850^2) = 19.27"\n\nIt goes around the cylinder three times and its total length is therefore 57.81"
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If L = Length\nW = width\nPerimeter = P = L + L + W + W\nP = 2L + 2W ... (1)\n\nBut in this question \nL - 5 = 3W\nL = 3W + 5 ... (2)\n\nSubstitute this value of L in (1)\nP = 2(3W + 5) + 2W\n\nP = 46\n46 = 2(3W + 5) + 2W\n46 = 6W + 10 + 2W\n46 - 10 = 6W + 2W\n36 = 8W\nW = 36/8\nW = 4.5 cm\n\nSubstitute this value of W in (2)\nL = 3W + 5\nL = 3(4.5) + 5\nL = 13.5 + 5\nL = 18.5cm\n\nCheck ...1\nP = 2L + 2W\n46 = 2(18.5) + 2(4.5)\n46 = 37 + 9\n46 = 46!\n\nCheck ... 2\nL - 5 = 3W\n18.5 - 5 = 3(4.5)\n13.5 = 13.5!
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this question will require calcs using logarithm.\nhence will take time
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Let's call the entire track length x, measured in meters.\n\nTo arrive at the time, we have to divide the distance by the speed:\n\nx / 8.5 + x/8.0 = 99\n\nmultiply all sides by 8.5 and by 8.0\n\n8.0 X + 8.5 X = 8.0 * 8.5 *99\n\n16,5 X = 8*8.5*99\n\nX = 8*8.5*99/16.5\n\nTime for the first lap: \nT1 = X / 8.5 s\nT1 = 8*99/16.5 s\n\nT2 = X / 8.0 S\nT2 = 8.5*99/16.5 s\n\n\ncalculator time....\n\nSolution D
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Formula for Arc length: (2pir) x (arc length/360)\n\n(2)(pi)(10) x (5/360)=\n62.8 x 0.01388=\n\n.87 cm
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How long is a piece of string ? Twice the distance from the centre to one end !
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a) Assume the following:\n\nAngles a= 10 deg, b= 80 Deg, c=90 deg \nHypotenuse length = 2.5 m\nheight = 2.5 * cos 80 Deg= .434 m\nlength = 2.5 * sin 80 Deg= 2.462 m\nacceleration = "g" * Sin 10 Degree
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An approximation for the length is the circumference of cylinder for each turn (as discussed in previous answer).\nThe exact helix angle is inverse tangent of elevation through one turn divided by the approximate length you just calculated.\nThen the exact length is the approximate length divided by cosine of angle you just calculated.
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To find arc length, you need the angle that the arc subtends and the radius of the circle.  To find the angle, you'd have to do some geometry...\n\nImagine a triangle with sides 10, 10, and 5.  This is pretty much a chord on your circle (5) and the radii from the center to the ends of the chord (10).  If you draw a line from the center perpendicular to the chord, it bisects the chord.  This gives you two right triangles, with hypotenuse 10, side 2.5 (half the chord), and side unknown (the segment from the center to the chord).  To find the angle at the center formed by the radius and the unknown, you can use the sine function:\n\nangle = arcsin(opposite/hypotenuse)\nangle = arcsin(2.5/10)\nangle = 14.48 degrees or 0.25 radians\n\nThe total angle is then 0.5 radians, since the above only found half of that total angle.\n\nThe formula for arc length is\n\narc length = radius * angle\n= 10 cm * 0.5 rad\n= 5 cm\n\nAnd this kinda makes sense - if the total circumference is 20pi, and the arc is 28.96 degrees, then the arc length can also be found by\n\narc length = angle * circumference/360\n= 28.96 * 20pi/360\n= 5 cm
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The length is 10 miles and the width is 8 miles.\n\nOkay... this is what you do:\n\nLet w = width\nLet w + 2 = length, because the length is 2 miles larger than the width.\n\nArea = (Length)(Width)\n80 = (w+2)(w)\n80 = w^2 + 2w\n0 = w^2 + 2w - 80\n0 = (w+10)(w-8)\nw = - 10 miles or 8 miles. Since there is no such thing as a negative distance, it's 8 miles.\nw + 2 = length, so...\n8 + 2 = length\n10 = length
939597
measure the distance between the beginning of the positive transition to the end of the negative transition.\nThis will be the wave length.
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x=r(t-sint) and y= r(1-cost) are the parametric equations of the cycloid where r is the radius of the rolling circle and t is the angle through which the radius of the circle has turned from its starting point. The angle t must be measured in radians.\n\nFor simplicity let r =1 so that we have the unit circle.\nwith t=60 = pi/3 we get\nx = pi/3-sin(pi/3) = pi/3 -(sqrt 3)/2\ny = 1-cos (pi/3) = 1- 1/2 = 1/2\n\ndy/dx= sint/(1-cost)\n=[(sqrt 3)/2]/(1/2) = 3^0.5 (at the point where t= 60= pi/3)\n\nFrom this , you can find the slope at any point define by the parameter t.\n\nb) The tangent is horizontal when t= pi and is perpendicular at the cusp which occurs at 2pi and multiples thereof. The maximum value  is 2r (or 2 for the unit circle). The length of the base of the arch is 2pi *r or just 2pi for the unit circle.
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First figure out whether to use sin or cos, I decided on cos because we look for the max and min.\n\nNext find the amplitude which is 6. a = amplitude/2 so a*cos(x).\n\nThird find the midpoint of the curve which is (max+min)/2 = m. a*cos(x) + m.\n\nThe fourth step is to get the max or min at the x cord. that you want. With cos you subtract max from x. a*cos(x-max) + m.\n\nFinally you find your wave length, which is simply (max-min)/pi = wl for cos. a*cos((x-max)/wl) + m\n\nso you go from (-2,1) to (10,7)\n\na = 3\nm = 4\nmax = 10\nwl = 12/pi to give us\n\n3*cos ((x-10)/(12/pi)) + 4
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Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
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Your question is not well posed, but I'm going to restate it in the way that I'm guessing you're thinking.  \n\nSuppose you have a 20 inch diameter circle.  Take two points on the circle and draw a straight line segment between them that is precisely 8 inches in length (this is termed a "secant", or a "chord").  What is now the arc-length of the segment of circle between those two points?\n\nIf that is your question, then here is how I work the problem:\n\nYou have defined an equilateral triangle, two edges of which are the radii of the circle, 10 inches each, and one which is the chord of 8 inches.  We can now determine the ANGLE between the two 10 inch edges using basic trigonometry.  The easiest way for me to think about this, offhand is to divide the equilateral triangle in half, making two right triangles.  The hypotenuse is 10 inches, the opposite leg is 4 inches, and therefore the sin function of the angle is equal to 4/10.  the angle is therefore (approximately) 23.6 degrees.  Since we "trigged" a right triangle that was half of the original equilateral triangle, we need to double the result, yielding 47.2 degrees.\n\nThe angle between the two radii that subtend the arc who's chord is 8 inches (on a 20 inch circle) is 47.2 degrees.\n\n47.2 degrees represents 47.2/360 of the circle (or 0.13 of the circle).  Therefore the arc length of the segment in question is given by 0.13 x the circumfrence.\n\nThe circumfrence is pi times the diameter or 3.14 x 20 = 62.8 inches, therefore the arc length is given by 0.13 x 62.8 =\n8.2 inches\n\nThe answer here isn't terribly surprising.  The 20 inch diameter circle is quite large compared to the short segment of an 8 inch straight line.  The "gentle curvature" of the circle compared to the line means that the arc length of the circular segment isn't going to be much longer than the straight line.  If  you took it to the extreme and compared the circular segment from a circle who's DIAMETER was about 8 inches, then the arc length would become nearly half the circle, 3.14 x 8 / 2 =12.5 inches.  On the other hand, if the diameter of the circle got really really big (like the diameter of the earth for example) then you'd not be able to tell the difference between the straight 8 inch rod and the curve of the circle at all.  The answer would approach exactly 8 inches.\n\nI hope that answered your question!