Conceptual introduction to the formula for arc length of a parametric curve.
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- Why doesn't the integral give us the area under the curve?(6 votes)
- Normally, you're integrating a quantity like f(x) dx, which is a rectangle of height f(x) and infinitesimal width dx, so the sum of infinitely many rectangles gives you area. However, here you're integrating a quantity ds = sqrt(dx^2 + dy^2), which is a line segment of infinitesimal length; the sum of infinitely many line segments gives you length. Let me know if this helps.(8 votes)
- Hi, About 2 minutes in to the video Sal explains that dx=(dx/dt)*dt and that dy=(dx/dt)*dt. I have a hard time understanding how that works and why that is the case. Can someone help me understand this better?(8 votes)
- In the video, Dx is the rate of change our function X. Our function X is written in terms of t, so the derivative of X(t) will be dx/dt, the derivative of our function X with respect to t, multiplied by dt, the derivative or rate of change of the variable t, which will always be equal to 1 here. It's basically the same thing as taking the derivative of any other function with the variable x in it, but in this case its replaced with the variable t. For example, the derivative of x^2 is equal to 2x(dx) , where d/dx=2x and dx=1. So in the video, dx/dt is like d/dx and dt=dx.(2 votes)
- It doesn't make sense to me conceptually that dx=(dx/dt)*dt. Why are we multiplying times dt, besides the fact that it cancels out with dx/dt to give dx? Is there another reason besides this, or is that the reason? Please explain.(3 votes)
- That is the reason. The thing is that technically you are not supposed to treat the dx and dt like quantities but we do it because it is convenient and easy. Hope this helps!(2 votes)
- Why can't you just find dy/dx (by doing (dy/dt)/(dx/dt) then use the arc length equation: int(sqrt(1+(dy/dx)^2)). You can also find your limits by plugging the starting and end point of the interval in your paramteric equations and solving.(3 votes)
- What is the formula of area under a parametric curve ?(2 votes)
- If x=f(t) and y=g(t) and the parametric curve is traced out exactly once as t increases from a to b, then the area is the integral from a to b of g(t)*f'(t) dt (not sure how to do integral formatting).(1 vote)
- Why didn't we solve it by turning the parametric equation into y(x)= expression with x? Would it still work if we did like that or is it problematic because we lose the information of time related to position?(1 vote)
- This isn't always possible because y may not be a function of x. Consider the curve given by
<x, y>=<tcos(t), tsin(t)>. This is a spiral centered on the origin, so it fails both the vertical line test and the horizontal line test infinitely many times.
We use parametric equations because there are lots of curves that just can't be described by y as a function of x. This gives us a more powerful language, but it also means we can't always convert back to the 'familiar' y=f(x) setting.(3 votes)
- [Instructor] Let's say we're going to trace out a curve where our X coordinate and our Y coordinate that they are each defined by or they're functions of a third parameter T. So, we could say that X is a function of T and we could also say that Y is a function T. If this notion is completely unfamiliar to you, I encourage you to review the videos on parametric equations on Khan Academy. But what we're going to think about and I'm gonna talk about in generalities in this video. In future videos we're going to be dealing with more concrete examples but we're gonna think about what is the path that is traced out from when T is equal to A, so this is where we are when T is equal to A, so in this case this point would be X of A, comma Y of A, that's this point and then as we increase from T equals A to T is equal to B, so our curve might do something like this, so this is when T is equal to B, T is equal to B, so this point right over here is X of B, comma Y of B. Let's think about how do we figure out the length of this actual curve, this actual arc length from T equals A to T equals B? Well, to think about that we're gonna zoom in and think about what happens when we have a very small change in T? So, a very small change in T. Let's say we're starting at this point right over here and we have a very small change in T, so we go from this point to let's say this point over that very small change in T. It actually would be much smaller than this but if I drew it any smaller, you would have trouble seeing it. So, let's say that that is our very small change in our path in our arc that we are traveling and so, we wanna find this length. Well, we could break it down into how far we've moved in the X direction and how far we've moved in the Y direction. So, in the X direction, the X direction right over here, we would have moved a very small change in X and what would that be equal to? Well, that would be the rate of change with which we are changing with respect to T with which X is changing with respect to T times our very small change in T and this is a little hand wavy, I'm using differential notion and I'm conceptually using the idea of a differential as an infinitesimally small change in that variable. And so, this isn't a formal proof but it's to give us the intuition for how we derive arc length when we're dealing with parametric equations. So, this will hopefully make conceptual sense that this is our DX. In fact, we could even write it this way, DX/DT, that's the same thing as X prime of T times DT and then our change in Y is going to be the same idea. Our change in Y, our infinitesimally small change in Y when we have an infinitesimally small change in T, well, you could view that as your rate of change of Y with respect to T times your change in T, your very small change in T which is going to be equal to, we could write that as Y prime of T DT. Now, based on this, what would be the length of our infinitesimally small arc length right over here? Well, that we could just use the Pythagorean theorem. That is going to be the square root of, that's the hypotenuse of this right triangle right over here. So, it's gonna be the square root of this squared plus this squared. So, it is the square root of, I'm gonna give myself a little bit more space here because I think I'm gonna use a lot of it, so the stuff in blue squared, DX squared we could rewrite that as X prime of T DT squared plus this squared which is Y prime of T DT squared and now let's just try to simplify this a little bit. Remember, this is this infinitesimally small arc length right over here. So, we can actually factor out a DT squared, it's a term in both of these and so, we can rewrite this as, let me, so I can rewrite this and then write my big radical sign, so I'm gonna factor out a DT squared here, so we could write this as DT squared times X prime of T squared plus Y prime of T squared and then to be clear this is being multiplied by all of this stuff right over there. Well, now if we have this DT squared under the radical, we can take it out and so, we will have a DT and so, this is all going to be equal to the square root of, so the stuff that's still under the radical is going to be X prime of T squared plus Y prime of T squared and now we took out a DT and now we took out a DT. I could have written it right over here but I'm just writing it on the other side, we're just multiplying the two. So, this is once again just rewriting the expression for this infinitesimally small change in arc length. Well, what's lucky for us is in calculus we have the tools for adding up all of these infinitesimally small changes and that's what the definite integral does for us. So, what we can do if we wanna add up that plus that plus that plus that and remember, these are infinitesimally small changes. I'm just showing them as not infinitesimally just so that you can kind of think about them but if you were to add them all up, then we are essentially taking the integral and we're integrating with respect to T and so, we're starting at T is equal to A, all the way to T is equal to B and just like that we have been able to at least feel good conceptually for the formula of arc length when we're dealing with parametric equations. In the next few videos we'll actually apply it to figure out arc lengths.