Transcript
Hello, Tom for everystepcalculus.com and everystepphysics.com this is a another nonsense problem by calculus said to me by a student. I would imagine it never appear out of test but I’m going to do it anyways regarding an arc. Arc length and the unit tangent vector okay so, let’s do it index 8 to get to my menu we’re going to scroll down here to arc length in RTC. The RT here that’s a position vector, a vector has magnitude and direction in this case its tangent. So, therefore whatever curve this turns out to be on a graphing which I don’t know how to graph it. But somebody made this function here some mathematician or person made this function and they’re trying to convince us that this solves stuff in life and I say it doesn’t I say it comes up with a number which will show you this number right now and here it comes up with units. As if that was important to anything and it’s also one numbers like taking six times seven coming up with forty-two and saying hey there you see it’s that’s the orbit around Neptune, Neptune or something like that you know some wild statement.
So, anyways let’s do it arc length and RT, we’re going to choose number 3 RT I have program the X and Y values also and I’ve already entered these to save time okay. All these functions but you’re gonna press alpha and put every one of these variables in here okay exactly like you see it, make sure you put time sign between the T etc. But I’ve already added them so here they are two times T times cosine not this right here I guess what the original function this one here okay and – T etc. So, I always show you, you can change it if you want I say it’s okay and we’re gonna do number four arc length okay. You can do all these things with what you’ve entered here. We’re gonna do and here’s the formula you write this stuff down in your paper for your test or homework or whatever. L is length of an arc it’s really a definite integral and that it uses the magnitude of the derivative of the RT function to come up with this stuff here and I show you all this you just keep writing in your paper putting it down you look like a genius. Now X of T is equal to remember I is X, J is Y, K is Z okay those are the parameters that they give us. So, the derivative of X of T this one right here is this right here two times cosine of T I’m not going to go through each one of it because it takes too much time for this video and here’s why T the next one – right here – 2 times T sine of T. Derivative of that is this right here of course you knew that didn’t you, you didn’t pull that stuff out of the air just like anybody can. Here Z of T here’s this right here and that turns out to the derivative is 2 times the square root of 2 times T okay.
Now, we’re going to find the magnitude we have to square all this stuff and here we have the integral here over A and B that range and we’re going to put in the range okay. So, I haven’t program that’s where we’re going to go alpha 0 for the value of T in the upper range is alpha second I. Okay, so I show you this is the integral we’re doing with regard to T I just put that in there in case you made a mistake on the range. So, X prime of T is equal to this and then we’re going to so then X prime of T2 is equal to this right here. Okay, the same thing with YT here’s the square of it and here’s the square of the Z parameter K. So, now we’re gonna do it over the range we got this that all in here we’re still doing the integral of this all stuff okay. You notice we got the square root of this here okay square root of this remember when you do an integral of the square root you change it to half, your exponent half right here. So, here’s what we’re still completing and of course the integral then is 2 plus 1 times the absolute depth T plus 1 okay and at T is 0 the problem as you put zero in for all the T’s and the answer for the integral. We get one and for pi we get PI plus 12 upper – lower always when you’re doing these this with a range a definite integral equals here right here so this is the answer pi squared plus 2pi.
Now, this is the answer that they get here on the right, right they giving you the answer okay so it’s 16.513 units. Now, what does that mean I mean you know it doesn’t mean anything to me doesn’t mean anything to you. It’s completely irrelevant you’re not gonna use this to go into space with let me tell you okay. So, this is the length of this unit tangent mention vector okay now we’re going to do more calculation we’re going to go back here number one and we’re going to scroll down here too unit tangent vector. You know tangible is given I’m against teaching this stuff but I’ll give you a little hint of what you’re talking about here and of course the length is one in a unit tangent vector okay. You can read that in your spare time and here’s the formula unit tangent vector is written just like this and all the calculus books derivative RT is over the magnitude of it and I show you what you’re doing here. Here’s the derivatives of each one of those pi J and K and the magnitude how you come up with that okay and so we’re going to do the derivative of this that equals this is what we found before. I drew the magnitude right here of course this is the numerator and this is the denominator.
We’re trying to and here’s the answer now you can see the answer that they have two times cosine of t minus two T sine of T okay, that’s this one here this one here over here for the J is minus two T cosine T two times sine of T okay. It’s all over four times T plus one squared now they took the square root outside the radical sign but no problem go to my site buy my program there only forty bucks and you’ll be able to do all this stuff forever you’ll notice when I. This is the greatest notebook ever these in the calculator here because my brain tells me why would I ever study something hard and not write it down somewhere where he could refer to it in the future. Well, I programmed it so in the calculator so that I can do actually do the problem no matter what variables they get. So, think about buying my programs and you’ll pass calculus you’ll get six or seven problems right in any test compared to the guy sitting next to you or the girl sitting next to you and because it’s all scored on partial credit in the class curve that’s all you need to pass the class and I never wanted to get a I did get A’s and in calculus but I never wanted to I’ve got a D I don’t care. Just get me out of there, I don’t ever let me touch this stuff again okay. Hey have a good one
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