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# Worked example: Parametric arc length

AP.CALC:
CHA‑6 (EU)
,
CHA‑6.B (LO)
,
CHA‑6.B.1 (EK)

## Video transcript

let's say that X is a function of the parameter T and it's equal to cosine of T and Y is also defined as a function of T and it's equal to sine of T and we want to find the arc length of the curve traced out so length of curve from T is equal to zero to T is equal to PI over two so pause this video and see if you can work that out based on formulas that we seen in other videos all right so first I'm going to look at the formula and then we're going to visualize it and appreciate why what we got from the formula actually makes sense so the formula tells us that arc length of a parametric curve arc length is equal to the integral from our starting point of our parameter T equals a to our ending point of our parameter T equals B of the square root of the derivative of X with respect to T squared plus the derivative of Y with respect to T squared DT DT this could also be re-written as this is equal to the integral from A to B of the square root of DX DT squared plus dy DT squared DT but either way we can now apply it in this context what is DX DT so DX DT is equal to the derivative of cosine of T is equal to negative sine of T negative sine of T and what is dy DT the derivative of Y with respect to T derivative of sine of T is cosine of T cosine of T so our arc length up here is going to be equal to the integral from T is equal to zero to PI over two that's what we care about our parameter is going from zero to PI over two of the square root of the derivative of X with respect to T that's a negative sign of t squared well if you square it the negative is going to go away you're going to negative sign times negative sign is positive sine squared so I could write this as sine squared of T and then dy DT squared that's just cosine squared T plus cosine squared T and then we have our DT out here now lucky for us sine squared plus cosine squared of some variable is always going to be equal to one so that just that's one of our most basic trig identities comes straight out of the unit circle definition of sine and cosine and so we have the square root of one the principal root of one which is just going to be one so all of this thing everything here is just simplified to digital from zero to PI over two DT well this is going to be equal to so you could view it this is a one here the antiderivative of 1 with respect to T is just going to be T we're going to evaluate that from PI over 2 we're going to evaluate that at PI over 2 and then subtract it evaluated at 0 so this is going to be equal to PI over 2 minus 0 that's going to be equal to PI over 2 now let's think about why that actually does make sense let's plot this curve so that is my y-axis this is my x-axis right over here when T is equal to 0 you have X of 0 cosine of 0 X is equal to 1 X is equal to 1 and Y sine of 0 is 0 so y is equal to 0 so at this point right over here at t is equal to 0 and then as T increases up to PI over 2 we trace out the top right corner of the unit circle and we end up right over here when T is equal to PI over 2 you could view T in this case as some type of an angle in radians and so the arc length is really just the length of quarter of a quarter of a unit circle well we know what the circumference of a circle is it is 2 pi R in the unit circle case the radius is 1 so the circumference of the entire circle is 2 pi 1/4 of that is going to be pi over 2 so it's nice when we this fancy thing that we feel good about in calculus is consistent with what we first learned in basic geometry
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