Find the length of the arc from x = 0 to x = a of the catenary y = 1/2 * a [ e^(x/a) + e^( -x/a) ]?
483089
First write your equation using hyperbolic function:\n\ny = a cosh(x/a)\n\nDifferentiate, dy/dx = sinh(x/a)\n1 + (dy/dx)^2 = 1 + sinh^2(x/a) = cosh^2(x/a)\nSQRT (1 + (dy/dx)^2 ) = cosh(x/a)\n\nIntegrate from x = 0 to x = a:\n\nAnswer = [asinh(x/a)]@x=a  - [asinh(x/a)]@x=0\n       = asinh1 - asinh0\n       = asinh1\n       = a(e - (1/e))/2\n\nRefer to the following website for details on the catenary
829160
y'=2 cos(7x)*(-sin 7x)*7 By the Chain Rule
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
508595
You have two ways:\nOld one: Thomson divide the arc into strips at equal intervals and solve each.\nNew one: Integrate from y=0, first root to y=0 second root
1096282
cos^2(A) / ( 1 - sinA )\n=( 1 - sin^2(A) ) / ( 1 - sinA )\n=( 1 + sinA )( 1 - sinA ) / ( 1 - sinA )\n=1 + sinA
986397
dear u are integrating a derivative. so the answer is\ne^(arctan x) 0 through 1\n= e^(π/4) - e^0\n= e^(π/4) - 1
187232
LET:\nr = radius of pulley\ns = arc length = belt length that moved around the pulley.\nt = time of belt to move that length\nw = angular speed of pulley\nv = linear speed of belt\n\nGIVEN:\nr = 13.34 cm\ns = 52 cm\nt = 16 sec\n\nSOLUTION:\n\na) w = v/r = s/(rt) = 52 cm/[(13.34 cm)(16 sec)] \n = 0.24 rad/sec ANSWER\n\nb) v = s/t = 52 cm/16 sec = 3.25 cm/sec ANSWER\n\nI hope the solution is clear enough...
217051
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.
510334
x = a*sin(ang)\ny = b*cos(ang)\n\nang goes from 0 to 360 degress, a and b determine how oval it is (if a=b it is a circle)
930129
r = √1² + (-1)² = √2\nθ = arctan(1/-1) = -π/4 ± n2π, where n is any integer.
198537
arctan'(x) = 1/(1+x^2), so differentiating using the good old rules, we get the result (using 'sq' for the square root to avoid yahoo cutting the equation short):\n\nsq(a+x^2)-1\n-------------------------------\n(x*sq(a+x^2)(1+(x-sq(a+x^2))^2\n\nMan, that's ugly. :-)
93089
This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
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The IMA of a machine can be found with the following formula:\n\nIMA = DE / DR \nWhich, this is lacking some specific effort and resistance distances from the question so I can't solve it this way.\n\nThis one might give the right answer:\nMA = length of slope ÷ height of slope \n\n15/3=5
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There is no answer to that question.\n\nMillimeters is a measure of length and arc seconds are a measure of angle. It's like asking how many mm in 30 degrees.\n\nYou need to know the distance to the object or thing that you are measuring to be 0.2 arc seconds across. Then you can use the pythogorean theorem to calculate the length in mm. (As long as the object is reasonably far away, if it is very close then you can't make the approximation of a right triangle.)\n\nadded:\nkirchwey is absolutey correct!\n\nach! its been too many years since I did this stuff!
970181
1) This has no elementary solution\n2) This one, however, does have a solution, with the substitution e^x = tanu.  You are at the answer arctan(e^x)\n\nSteve
385689
You want arc length or straight line?\n\n934.59 nm\n\naccording to the reference below.
437123
Looks like you're missing a few things here.  To find an arc length, you need a radius along with the measurements you gave.  If you had a length of one of the sides, say AP, then you could find the arc length by taking a ratio of that length compared to the circumference of a circle:\n\narc length / circumference = angle / 360\n\nIf you solve for the arc length and plug in your known values, you'd have\n\narc length = 2*pi*AB * 25 / 360\n\nFrom this, you can see that angle CPB is irrelevant.\n\nAlso, it's not clear that this is a circle.  You list two angles, and the only thing we can figure is that they both share a vertex at P.  It's not clear if A is between B and C or outside of it.  But again, it doesn't really matter how angle CPB plays into this, since the arc length is really a function of the angle it contains and the radius of the circle.