 I better not follow these steps. In this video we're going to just do a couple of problems actually only five and we're going to look at the arc length of a parametric curve. Now we've looked at I think when you did for single variable calculus you looked at the length of a curve on some interval and we're going to generalize that and it becomes very easy when we just deal with or change them at least to parametric equations. So any five problems just to get you just to get the idea the first one's going to be how do we derive an equation for the the arc length, the length of a section of a parametric curve and then we're just going to do four problems. These are all pretty much the same and what you might find in your homework or what you have to do is just to dig into single variable calculus trying to remember how to do these problems in single variable calculus. Of course so we'll just do a couple of them just as I said to get the ball rolling. So we get to the arc length of a parametric curve and with these problems we really are just going to well some of them would be easy because we just do derivatives and integrals of polynomials some are going to be more difficult and they might be forcing you just to remember your single variable calculus and how to do derivatives and integrals and all the different techniques. So if you are required to remember all of those these arc length problems are going to be you know very good for you if you don't have to do them you can use the warframe language for instance it's easy to get these derivatives or they might just be given to you and just a sheet you know with all these with all the derivatives of transcendental functions etc. But first of all we need to understand what this what this arc length of a parametric curve is all about and the way to do that is just to derive an equation for a parametric curve. So in the natural sea we have to we have to be clear about a few things so we want this length of the arc over this the length of this parametric curve going from a starting point to an end position on this curve and if so so that means we have this parameter t and we're going from a to b on this parameter but then the first derivative of x with respect to t and the first derivative of y with respect to t they've got to be continuous on this interval for this arc and this arc can also not intersect itself from the starting to the end position of t except perhaps at a finite number of points it can't happen at an infinite number of points it's got to be finite number of points and that must be so because of the way that we're going to construct this derivation. So we've got this arc length that's part of this parametric curve and let's call it c and it's an equation of x as an equation of t and y as an equation of t or function of t and we're going to go from we have this idea that we're going to go from a less than equal to t less than equal to b that's our starting end position on this curve. So we've got to have as I said x and y both the functions of t and the first derivatives with respect to t they must be continuous on this closed interval and then as I said the curve cannot intersect itself other than perhaps at a finite number of points because what we're going to do is we're going to take this closed interval of us a to b and we're going to divide that up divide and conquer divide and and we're going to divide it into equal length sections and we're going to have divide up into end of these sections end of these sections so if we divided that up into these equal lengths little bits so if we have our curve there we're just going to divide it up in these equal little and remember they're all going to have a specific t value so what we're going to have here we're going to start at a and that's a t sub zero for instance and that's going to be less than t1 and less than t2 and then all the way less than tn at the end which is b so the fact that we can write this delta t is this b minus a divided by n that's how long each one of these you know how long each one of these are and what really happens there is or this delta t at least delta t remember that is just if we go from we subtract t sub i wherever we are now and we subtract from that i minus one the previous point where we were because they all have equal length and that means you know when we at let's just look at these these these two points where are we at these two points so when we start so let's just write here when we start what what values of x and y do we have well the value for x and y what we have at the start that's going to be x of t and that's going to be at i minus one and we're going to be at y of t i minus one that's where we start and we're going to end our x and y where we end well that's going to be x of t at this i position and y of t at this i position that's where we start and that's where we end now if you think about it if we zoom in very closely that's this curve running there but we have you know we have an x delta x and a delta y for that and because these are very small we're sort of approximating that by a straight line there we're approximating that by a straight line and you know because we're making these very very small and if we do that we can make use of just the Pythagorean theorem because it's this length squared plus this length squared and we just take the square root of that and we get the length of the hypotenuse so we can really say as let's just take one of these little segments and we call its arc length as s i that's the one we're dealing with so that's going to be approximately equal to depending on how small this little segment is that's the square root of the one side squared and the other side squared so when we go from x we will go from there to there in other words we're going from there from this one to this one so it's just the difference of those two squared so let's put it in this order so that's going to be the x of t sub i minus x of t sub i minus one and we square that and we're going to do exactly the same for for the y so the delta y in other words is y of t sub i minus y of t sub i minus one and we square all of that so we just make use there of the Pythagorean theorem it's going from this in position to the start position and we just use the linear this approximation here because it's very small and that piece of the curve there is very close and that's why we also want this finite number of intersections and as much as we can divide this curve up into bits so that we incorporate these intersections so what we have to remember here and that's the sort of the big jump here is just the the uh this mean value theorem and that says look if i've got this curve going here and i connect the start and the end position that gives me a slope somewhere along the line there's lots of slopes here along the line somewhere along the line we've got to have this equality between this slope here connecting the two and the slope right there somewhere along the line at some point we're going to call that point c sub i so we're going to have this fact that dx dt dx dt at this point ci at this point ci that is going to be this delta x divided by delta t at that point and that's you know going from start to end so that's going to be x of t sub i minus x of t sub i minus one the previous one and we divide that by delta t just this little gap that we have and that means we can sort of rewrite this as the x of t sub i minus the x of t sub i minus one and that is just going to equal the x dt at c sub at the c sub i so let's just write that as x prime because that becomes very difficult x prime at that very little point delta t so that was just going to be some just some algebra so i'm just taking this delta t over to the other side and what we'll see is as far as dy dt is concerned if we do dy dt we're going to get exactly the same thing but instead of it being at one point c sub i it's going to be at some other point so let's just make it d sub i so we're going to get this exact same thing that y of t sub t sub i minus y of t sub i minus one that's going to be y prime of let's make a di delta t for that same sort of the same argument but now if delta t goes to zero if it approaches zero these become so small that ci and di they're going to coincide they're going to be they're going to coincide so i can just rewrite this i can rewrite all of this as this si that is going to equal and instead of using this one squared and this one squared as we used it there we're going to just use this we're going to say x prime at c sub i delta t and remember that's all squared and then we're going to have y prime and instead of c sub i i can put d sub i we can put c sub i and delta t and this is all going to be squared so just squaring all of those and then it's easy to see that we can take a delta t out as a common factor delta t squared and the square root of delta t that's this delta t so what we're going to have here is s sub i is going to be approximately equal to so we're going to have this x prime at c sub i and that's all squared plus y prime at c sub i and that's all squared and that's the square root of that oh that's so ugly let's do this so it's all of that delta t because we had a delta t squared there delta t squared we took that out so common factor it is the square root of a square and that just leaves us with that so what we all we have to do now is just sum up all of these to get all of these little segments we just sum them up and if we sum them up s it's still going to be approximately equal to but now we're just summing all of the all of these so we're going to have i equals one to n over all of them square root of x prime c sub i squared plus c prime at c sub i squared delta t and we can easily go from there we can easily go from there to the integral because we're summing over these small things so we just need this limit as n goes to infinity so the arc length then is this going to be this limit as n goes to infinity we just have infinite number of these little segments of this summation here so of this summation the sum of i equals one to n of the square root of x sub i at c sub i remember all of that is squared and y prime c sub i all of that is squared delta t and the limit as n goes to infinity of all of this remember that becomes the integral and from our starting position to our ending position and now the c sub i at that this is now so small as the as n goes to infinity on this delta t goes to zero it becomes so small that that point c sub c sub i well that's just at every point t so we can just have the square root of now x prime let's just write that out as dx dt squared plus dy dt squared dt that's just in the limit and as I say we go from the c to the t there because these little line segments are so short we're taking the we're taking the derivative there with respect to t at every one of those little points they're just on top of each other basically and that is it that is the equation for the arc length very simple the arc length of a parametric curve going from start to an end positions the definite integral going from the start to the finish of the square root and we just square the first derivative of x with respect to t and we square the first derivative y with respect to t you know we just square them add them take the square root of all of that dt and it's as simple as that now that we know what this equation is and let's just write it down the more you write it down the more that you remember that so that's s that's our arc length that's going to be the square root of dx dt all squared plus dy dt all squared and that's the integral remember going from a to b and there's my ruler dt so it's as simple as that so the first thing we need to do let's just do some side work so we need to know what dx dt is and this first problem dx dt and we can see that that is the equation of a circle if you pop this parametric equation into the Wolfram language you're going to see that on on our parameters theta in this instance but we have this interval for theta being from 0 to 2 pi so let's just do that that's going to be minus r times the sine of theta and if we do dy dt that is going to be r times the cosine of theta and if we square these dx dt should make my axis a bit better that's going to be r squared sine squared of theta and if we do dy dt and we square that that's going to be r squared and we have cosine squared of theta great so let's do let's tackle the problem so the arc length is then going to be the definite integral and going from 0 to 2 pi of the square root and we have r squared sine squared of theta plus r squared cosine squared of theta d theta d theta what am i writing dt for our problem is the moment is d theta theta here is our parameter for this parametric equation so s is going to be the definite integral going from 0 to 2 pi and i see i think you can see let's just do another color again we're going to take under this square sign we're going to take our squared arc as a common factor and then we're going to have cosine squared theta plus sine squared theta and we know that from the trigonometric identities that's just equal to 1 and we have the square root of r squared and that's just r good and you can see where this is going to go so what we have at the moment here is just r d theta we just have r d theta and now what we have to do is we can take the r out because it's not part of the integral it is just a constant d theta and that is just equal to r and then theta but theta going from 0 to 2 pi and that is r of 2 pi minus 0 and that is just 2 pi r so the length of a circle in other words the circumference of a circle is 2 pi r and that's where the equation comes from from for the circumference of a circle let's calculate the length of this parametric curve half t squared that's our x and then for y we have 1 over 9 60 plus 9 to the power 3 over 2 and then t goes from 0 to 4 plot that in the Wolfram language and you can see what this parametric curve looks like first of all let's just remember what our equation is for the arc length of a parametric curve so that's the definite integral going from a to b of the square root of dx dt all squared plus dy dt all squared dt and we just have to close down our square root symbol there so once again let's have a look at this what is dx dt well dx dt as far as x concerned 2 comes to the front so that is just going to equal t that's just going to equal t and if we square that that is going to be t squared and dy dt is a bit more involved we have 3 over 2 which we now have to bring to the to the front so we're going to have 1 over 9 we're going to multiply that by 3 over 2 and we're going to have 60 plus 9 and 3 over 2 minus 2 over 2 that's 1 over 2 and by the chain rule we still have to have a 6 there because we still have to integrate the inside expression so dy dt so it's a bit more involved we can cancel out the 3 there and this becomes a 3 and then we have 6 divided by 6 because 3 times 2 3 times 2 is 6 times that's 6 or 6 times the 6 that's just 1 so we have here in effect the square root of 60 plus 9 because it's to the power half and if we square that dy dt squared that's going to be 60 plus 9 so we can eventually start doing our problem here so we're going to say the arc length is going to be this definite integral in going from and going from 0 to 4 and what we have for x squared plus y squared so all of this is still square root so that's going to be t squared plus 6t plus 9 there's my ruler and then dt well if we look at it what is t squared plus 6t plus 9 well that's just going to be t plus 3 squared and the square root of t plus 3 squared well that's just t plus 3 so when it comes to our s here our arc length that is going to be 0 to 4 t plus 3 dt problem becomes becomes really simple so if we do that that's going to be a half t squared plus 3t and we're going to go from 0 to 4 we can forget the 0 because we've got two t's there so that's going to be 0 plus 0 so arc length is going to be a half 4 squared plus 3 times 4 and 4 squared is 16 divided by 2 is 8 and 3 times 4 is 12 so our arc length is 20 so this is just a nice exercise in a bit of simple algebra and just remembering what the equation is for the arc length problem number four calculate the arc length of the parametric equation x comma y equals the cosine of t plus t times the sine of t and then the sine of t minus t times the cosine of t and we've got this interesting interval for our parameter goes from pi over 6 to pi over 4 so the reason why we're doing all these problems to write down these equations until they become second nature so the arc length is going to be the definite integral and going from a to b square root of dx dt all squared plus dy dt all squared and we close our square root symbol there all of that with respect to t so let's do some side work let's just have a look at what dx dt is going to look like so dx dt we have cosine of t there that's first derivative with respect to t is negative the sine of t and then we add to that now we have to make use of the product rule so let's keep t first and then take the derivative of sine t with respect to t that's this cosine of t and then we're going to keep the sine of t but we take the first derivative of t with respect to t that's just one so we left with sine of t and then very nicely for us let's take another color just to show negative sine of t positive sine of t so that's very nice and if we square dx dt that just means it's t squared plus cosine squared of t let's have a look at dy dt so dy dt we have the sine of t first derivative of that with respect to t is the cosine of t then we have a minus sign so let's just keep that separate so let's keep the t and then differentiate cosine of t so that's negative sine of t so that becomes negative t sine of t and then the other way around we're going to keep cosine of t take the first derivative of t with respect to t that just leaves us with a cosine of t and so if we write this out that's going to be the cosine of t minus times minus is positive t sine of t and we have negative the cosine of t and again again that's very nice because we have cosine of t negative cosine of t that's the way this problem was designed so if we square this we're just going to be left with t squared plus sine squared of t so let's get some work done so we're going to have the arc length that's going to be the definite integral going from pi over 6 to pi over 4 again draw that curve in the Wolfram language you can see what it looks like and that's going to be the square root of t squared cosine squared of t plus t squared sine squared of t and I think you can immediately see what's going to happen there let's just close our square root symbol there this is do it on the side so if we have the square root we're going to take t squared out as a common factor and we left with cosine squared of t plus sine squared of t and that's a trigonometric identity remember which is just one so we left with the square root of t squared and that's just t so that makes life very simple for us so we're going to have arc length that's going to be this definite integral and going from pi over 6 to pi over 4 and that's just going to be t all of that just becomes t dt and if we take the integral of that that's going to be a half t squared and we have to go from pi over 6 to pi over 4 so that's going to be a half and what do we have we have pi squared over 4 squared oh what am I writing now pi over 4 pi squared over 4 squared and we subtract from that pi squared over 6 squared so that's still going to be this half that we have and we have pi squared over 16 minus pi squared over 36 and so we have to have a common factor there and I think I did it before that's going to be 144 times the half that we have to have there and that leaves us with check just check that it's correct 5 pi squared divided by 288 there we go so again this is just a just a friendly exercise and doing these very simple integrals and I'm going to do one more problem and what you're going to see is with your problems doing this arc length you're just going to get more and more complicated single variable calculus problems you're going to have single variable integration and it's going to become more and more complex and you might have to do that or you might you know just use the orphan language to do those those integrals for you or you'll just have to remember all the techniques that we do have to get these definite integrals calculate the arc length of the parametric curve x comma y that's half times the natural log of 1 plus t squared and then the inverse tangent or arc tangent function of t over the interval for this parameter t going from 0 to 1 as always just do it every time you do these exercises arc length is the definite integral and going from the a to b and that is a there remember let's be there of the square root of dx dt all squared plus dy dt and that's all squared and we close our square root symbol there dt and now we just have to do some side work let's have a look at what the x dt is the x dt and that's why I said these problems are now going to become that you just have to remember what your first variable calculus single variable calculus was was all about and how do we take the the derivative first derivative of the natural log of something with respect to to your variable and if we just have a look at at at doing this dx dt and you'll remember that we have the half out in the front that's just a constant and as far as the natural log of something is going to constants so for instance what is the natural log of x so natural log of x first derivative of that with respect to x well that's just going to be one over x so this is going to be one over and the x here was this t squared plus one or one plus t squared and with a chain rule we just have to remember that we still have to do we still have to do the inside function and that's just going to give us a two t so if we put all of that together the twos are going to go so I'm left with t over t squared plus one or one plus t squared and remember we have to square that so we have to square all of that let's look at dy dt so the first derivative of the inverse tangent of t with respect to t well that's very simple if you remember your arc tangent derivatives that's just going to be one over one plus t squared or t squared plus one and if we square that we have to square that so that's why I say these problems are just going to be either they're going to give you this in the exam or you just have to remember sort of these single variable calculus derivatives and how to do the derivatives in the integrals so that's going to go from we have zero to one and we have the square root of that squared plus that squared so that's just going to be t over t squared plus one all of that squared and then we're going to have one over t squared plus one all of that squared dt so we have this common denominator here so let's just do that on the side as well so we've got this t squared plus one all squared is the common denominator and so that's just going to be on the top we have a t squared plus one and now we can cancel one of them out so that will go and that will go and that will just be one so that's one over t squared plus one and I think you remember those kind of substitutions so that goes from zero to one so we still we are still left with the square root symbol here so that's one over t squared plus one and close our square root symbol there oh why am I writing d theta I don't know where that d theta came from dt of course well I suppose it comes from the fact that we now have to do some substitution so let's do that so let's say t equals the tangent of theta remember those substitutions when we have one over x squared plus one so dt d theta so we're going to have dt d theta remember what that's going to be what is the first derivative of the tangent of a function well that's the secant squared of that function so we have the fact that dt is now the secant squared of theta d theta so just bringing that d theta over to the other side so we can replace that or substitute that dt and all we have to remember now what you can do otherwise you can bring it back is just to say what is going to be when is the tangent of theta going to equal zero because we're going from zero to one remember so we just want to replace those well that happens when theta equals zero and when is the tangent of theta equal to one well that's when theta equals pi over four remember so we can replace those so we can now say the arc length is going from zero to pi over four got a slightly different blue color here and now we have this one over and we have this secant or remember t is now the tangent of t so that's going to be the tangent squared plus one remember this this t here we're just replacing that and then dt we're replacing this is closed there dt we're replacing with secant squared secant squared d theta so now it's this whole problem in theta so we are fortunate here because we remember that the tangent squared of an angle the tangent squared of an angle plus one well that's just the secant squared of that angle so we have one over the secant squared but the square root of all of that oops that would be totally the wrong color let's just go back so s is going to be this definite integral and going from zero to pi over four so it's one over secant squared but the square root of all of that so that's just one over secant but we've got secant squared of theta there so let me just write it out secant squared of theta forgive me for leaving out these parameters so there should be a theta there as well over secant of theta d theta but that is just the different integral going from zero to pi over four of the secant of the secant without the square the secant of theta d theta and now you'll just have to remember what that is and if you do remember what that is that's going to be the absolute the natural log of the absolute value of the secant i can't write today secant of theta plus the tangent of theta okay and all of this has got to go from zero to pi over four so that leaves us with the natural log of the absolute value of so what is the secant of pi over four that's square root of two that's going to be square root of two and then plus a tangent of that that's one and we're going to subtract from that the natural log of the absolute value of if we put zero in there we just left with one and because this is a negative remember we can also just divide those so s is this going to be this natural log of the absolute value or then the positive square root of two plus one you just write that as natural log of the square root of two and it's just the positive one that we're taking plus one and there you go so this all your other problems you're basically going to see that it all boils down to can you remember your single variable calculus so as I mentioned it's either going to be given to you or you'll just have to remember these and if you have to remember your single variable calculus how to do derivatives and integrals of that these problems are good and just to do your exercises to remember all those techniques of getting the derivative and the integral so that's it for these five problems and arc length quite a bit of fun and of course it has a lot of real-world uses to calculate the arc length of a parametric curve