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NEED HELP with Cal 2 math

Fri 1 Feb, 2008 02:59 pm
can anyone tell me the formula or how to calculate the length (arc) of a function. i know how to calculate the area under a curve using both the antiderivatives way and using the concept of infinite sums...I am pretty sure the arc of a curve is found by using infinite sums and/or antiderivates but still need help.
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raprap

1
Fri 1 Feb, 2008 03:18 pm
consider what the differential of a function is--in the x, y plane it's the slope (dx/dy) .

Now consider the right triangle describing this slope, the hypotenuse is a combination of the two sides---call this hypotenuse ds. The foot is dx and the height dy. Use Pythagoras to figure out what ds is in terms of dx and dy. Or ds^2=dx^2+dy^2 and isolate the variable ds

ds=(dx^2+dx^2)^1/2
ds=dx((dx/dx)^2 +(dy/dx)^2)^1/2
ds=(1+(dy/dx)^2)^1/2dx

Now the problem becomes, how do you determine the length of s, and the answer is integrate it between the two end points x1 and x2.

let int{ be an intragal sign, and f be some function in x,
so f' is dy/dx=df/dx=f'

so the length of the curve is s and

s = int{ [1+f'^2]^(1/2)dx between x1 and x2

sorry about the unusual symbolism but html doesn't work on this end of the forum.

Rap
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mars90000000

1
Fri 1 Feb, 2008 03:25 pm
its ok with the symbols, i got the picture, i can figure out the rest, i just needed and idea, ill try and check with your answer to see if its the same

THX
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raprap

1
Fri 1 Feb, 2008 03:30 pm
As for the second question--look at the fundamental theorem of integration--it should be in your textbook, it breaks the area under the curve (f) between the two endpoints into an infinite number of rectangles with heights equal to the value of f at a particular x1 and x2. Then all of these areas are summed. As x1 and x2 converge the number of rectangles increase, and the accuracy of the sum's of the areas approach a limit.

To me this is amazing because an extension of the theorem of integration is used by numerical methods in determining areas under curves.

BTW integral is usually synomonius with anti-derivative.

Rap
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mars90000000

1
Fri 1 Feb, 2008 04:05 pm
i know that, im taking cal2 course right now

the definition of area under the curve is the limit as n goes to infinity of the sum of all f(x)d(x) in [a, b] where f(x) is continous throughout the interal and where n is the number of times you cut the interval....i was just wondering if the arc would be the same as the limit as n goes to infinity of sqrt{f(x)^2+d(x)^2} where f(x) is continous throughout the interal and where n is the number of times you cut the interval. since sqrt{f(x)^2+d(x)^2} is the length of the hypotenuse of a triangle as you mentioned
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raprap

1
Fri 1 Feb, 2008 04:37 pm
It will be if the derivative of f(x) is continuous throughout the interval x1, x2. If not the break the interval up into segments such that the interval x1 and x2 is covered (equivalent) and f'(x) is continuous. Then add them up. Note, this procedure can be cumbersome, particularly for elliptical functions.

Rap
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raprap

1
Fri 1 Feb, 2008 04:40 pm
Oh mea culpa, mea maxima culpa. I somehow made the supposition that you were starting calculus. That what happens when you assume, you tend to make an Ass out of U and me

Rap
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mars90000000

1
Fri 1 Feb, 2008 09:24 pm
LMAO!!! Ive taken Cal1 and just started cal2....im a fast learner o its not a big deal.....cal2 is waaayyyy more easier then cal1 anyways
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