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Current time:0:00Total duration:6:23

Video transcript

what I want to do in this video is find the arc length of one petal I guess we could call it of the graph of R is equal to four sine of two theta so I want to find the length of this portion of the curve that is in red right over here and we'll do this in two and I guess two phases first of all I want to set up the definite integral for finding in that arc length and then we'll evaluate it and I'll actually use a calculator to evaluate it because it'll be a little bit more straightforward at least for this integral so let's just start with that I encourage you to pause the video and knowing what we know about the formula for arc length when we have it in polar form see if you can apply it to figure out this arc length right over here so I'm assuming you've had a go at it so let's just remind ourselves that the arc length is going to be the integral from our starting angle to our ending angles will call it from alpha to beta of the square root of of the derivative of our of our function with respect to theta squared plus our function squared plus our function squared D theta so let's figure out what our prime of theta and what R of theta actually are so maybe I'll color code it so our prime of theta our prime actually well we know what R of theta is let me just rewrite it so we know R of theta is equal to 4 sine of 2 theta 2 theta so our prime of theta our prime of theta is just going to be derivative of 2 theta with respect to theta is 2 times 4 is 8 derivative of sine is cosine cosine of 2 theta and so if we wanted to write the the definite integral for arc length it's going to be equal to so it's going to be equal to the definite integral now what's our starting bound well for the petal that we care about we're going to start at an angle of 0 radians when we have an angle of zero radians R is zero that's this point right over here so we're going to start at zero radians and we're going to go all the way up to PI over two when sine of two times pi over two is sine of Pi which is zero so we're get back to this point right over here so we're going to go from zero to PI over two to PI over two radians it's going to be the square root of give myself some real estate here R prime of theta squared it's going to be this thing squared so it's going to be 60 weird isn't that same blue color so it's going to be 64 cosine squared of two theta and then plus this squared which is going to be 16 sine squared of two theta and then of course we have our D theta D theta and you can attempt to solve this or evaluate this analytically it's not too straightforward so I'm going to use a calculator and this gives us practice with knowing that hey are there tools out there that can help us do these types of things so I'll go to the definite integrals here I go to the second calculus on my t 85 I pick this choice which is for the definite integral and so let me express what I'm taking the definite integral of so it's going to be the square root of 64 times cosine of 2 theta squared so cosine and I'm going to use X instead of theta just because it's an easier variable to use on my calculator so cosine of 2 cosine actually let me make sure my parentheses are good so I'm going to say cosine of 2 X cosine of 2 X all right now close that and now I want to square that all right I did that first part plus 16 plus 16 times 16 times sine sine of 2x close that close that squared and actually I already see let make it sure I'm not sure if the calculator knows to interpret that as multiplication so let me insert a times right over here so 64 times cosine of 2x that thing squared plus 16 times sine of 2x that thing squared and now let me go to the end and then that's and then let me close my radical so that's going to close that right over there or that closes what I'm taking the square root of and then comma the variable that I'm integrating with respect to is X in this case everywhere I saw theta here I'm just replacing with an X there and I want to do it from instead of theta equals 0 to theta equals PI over 2 we'll say X is equal to 0 to X is equal to PI over 2 PI over 2 and hopefully I have not made a mistake when I've typed this in and I get it's chugging on it a little bit and we get will it actually come up well finally alright so if we were to round it let's say it around it to the nearest thousandth it'll be nine point six eight eight so approximately nine point six eight eight let's just see does that does that make intuitive sense let's see it goes about it goes as far as four away so if you just went out to four and then back down that would be about eight but of course we have we've actually gone out some so it actually makes intuitive sense that this arc length would be nine point six eight eight anyway hopefully you enjoyed that