How do I find the are under a parabolic arc?
596087
This is very straightforward. Which step are you having trouble with? The same goes for your other question.
525579
A parabola has a focal point. At this focal point "on" a Parabolic Reflector Microphone is, well, a microphone. The "dish" gathers sound waves (lots and lots) and directs them toward the focal point where the microphone collects/converts the sounds waves into electrical signals. It's similar to cupping your hand behind your ear to hear someone. Your hand reflects more sound into your ear than just your outer ear alone.\n\nFor the mathematics behind how it all works, see any analytic geometry text and search for focus, parabola. It's actually quite simple yet very effective.
353664
find the area of the cross section.... then times the area by the length... that is of course if it's a prism
788027
The graph is a V.  From the 0,0 draw a line with slope = 1 to the right of the y axis and draw a line with slope = -1 to the right of the y axis.  For y<|x|.  All you have to do is shade in the area below the V.  If y>|X|, all you have to do is shade in the area above the V.  Less than, shade below; greater than, shade above.  That should answer all the questions you've submitted.\n\nAnd no, absolute values do not form parabolas.  Some do, not all.  Like y = sqrt[|x-1|].  This forms 2 graphs each going away from each other.\nNice going NerdyKid.  You're not a nerd, you're a dork.
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
688944
View source of the that page.\nYou will see the formulas.
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
450833
join the 2 pointz frm da arc vit da radius.measure out da value of da angle forned here.then substitute da values 4 this formula:  2*pie*radius/da angle u measured\n\nhappy solvin!
95368
No problem!  A parabola is a parabola.  The distance from the parabola to the focus is the same no matter the wavelength.  Your 3 dB beam width and antenna gain will, however, vary with wavelength.  The antenna gain (EIRP) is given as\n\nGmax = nu(pi D f/c)^2\n\nwhere nu is your antenna efficiency (0.5-0.8 is typical), D is the diameter of the antenna, f is your frequency, and c is the speed of light, 3*10^8 m/s\n\nYour 3 dB width (an approximation) is given by:\n\nTheta = 70(lambda/D) = 70(c/fD)\n\nlambda = wavelength, D = diameter of your dish, c = speed of light, f=frequency\n\nIn order to make this work, you will need multiple feeds, typically appropriate length dipoles tuned for each band, at the focus of the parabola.\n\nMake sense?
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
65888
a parabola facing up!\n\nwith center (0,1)
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
232033
you would shade inside the parabola because it is >. you use a dotted line because it is > and not >=.
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
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
358834
KNOW YOUR RELATED RATES FOR CALC!! thats one thing that killed me and know your 2-D rotation around a line intergrals.
638682
Domain ~ x values\nRange ~ y values\n\nYou have to work the function and figure them out.