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Suppose the motion of a particle is given by parametric curve r R2 by r(t) cos? (t)i + Asin" (t)j Find the set A of t values for which has CUSp_ (b) Find the v...

Question

Suppose the motion of a particle is given by parametric curve r R2 by r(t) cos? (t)i + Asin" (t)j Find the set A of t values for which has CUSp_ (b) Find the values of r at each of the values in A, There are four distinct values. What is the speed of a parametric curve at cusp? d) Find the turning points in the speed function and decide for each of them whether the speed is at a local maximum Or minimum Your answers to and should be useful.

Suppose the motion of a particle is given by parametric curve r R2 by r(t) cos? (t)i + Asin" (t)j Find the set A of t values for which has CUSp_ (b) Find the values of r at each of the values in A, There are four distinct values. What is the speed of a parametric curve at cusp? d) Find the turning points in the speed function and decide for each of them whether the speed is at a local maximum Or minimum Your answers to and should be useful.



Answers

Suppose that the position function of a particle moving in 3 -space is $\mathbf{r}=3 \cos 2 t \mathbf{i}+\sin 2 t \mathbf{j}+4 t \mathbf{k} .$ (a) Use a graphing utility to graph the speed of the particle versus time from $t=0$ to $t=\pi$ (b) Use the graph to estimate the maximum and minimum speeds of the particle. (c) Use the graph to estimate the time at which the maximum speed first occurs. (d) Find the exact values of the maximum and minimum speeds and the exact time at which the maximum speed first occurs.

Okay notice the position vector that we have been given. We are first going to take the velocity of that by taking the derivative of each of the components. So V of t equals four, I minus three T squared J. Now, in order to get the acceleration will go to the next derivative. So the I. Component will go away but we have a negative six T. In the J. Direction. Okay, so um in order to graft so first we're going to find speed. So speed is the magnitude of velocity. So we will do the square root of four squared plus negative nine t squared squared. And so we get that square root of 16 plus 92 the fourth. Okay so now we're going to graph it. So in order to do that we come up with some teas and it's usually good to have both positive and negative teas and then we put the tea into the I. Component and that will give us our X value. So negative two times four will be a negative eight negative 40. And then we'll go back to four and then eight um are Y. Component is put into the negative T. To the third. So t to the third of negative two would be a negative eight but then it goes positive And then it's the next value would actually be a -8. So let's go ahead and sketch some of those numbers out. So kind of using my bearings on my X. Axis. I'm able to kind of quickly figure out what those values are and then knowing that I'm going to um 8 -8 is useful. Okay so now I'm finding V. Of zero so I go put zero in for time and I get four. I plus zero J. So my velocity at zero. That instantaneous velocity is all to the right and it's a magnitude of four. Now my acceleration when I put a. T. Value in I get zero. Now does that make sense? Well remember your second derivative, which acceleration is the second derivative of that position function? Your second derivative has to do with con cavity. So we go from being concave up to convict cave down. So right there are curvature actually is zero. And so that's why we get that value.

Okay first we're going to find the velocity acceleration and speed of our given motion for our particle. So in order to find our velocity, we take our first derivative. So our Vot is just going to equal one eye so we can just write I and then minus three T squared in the J direction. Our acceleration will be the next derivative, no I. Component. But we have a negative six T in the J direction. So now our speed is the magnitude of velocity. So we can go ahead and express that as the square root of one square plus that negative three T squared. So it's a one plus 92 the fourth. So now we're gonna go ahead and graph this so notice that were linear in the X. Direction but we're a degree three in that Y direction. And so with that you know, we could obviously put some points in but what we end up with is a function that this would be a positive X. to the three. But because that is a negative T. To the third, it ends up going this direction. Now when we go to actually um find our values for Visa zero, if you put a zero in notice you get I minus zero J. So because of that I we know our velocity is one unit to the right now. Our acceleration when we put a zero and we just get zero and what you might notice is we go from a positive curvature to a negative curvature for that second derivative. So you can kind of see how it changed. So it's not surprising that our second derivative ends up being zero at that point.

Problem number 15. Um, you are. He is equal to sign. Three tee off time minus two. Poor signed. Three tee off Jane as the mounts between zero and 25 over to buy over three s o b off T is equal to ah three. Full sign three team, We're fine. A plus seats flying three t o j. It's were off. The normal b is equal to square root nine hole science squared 80 both 36 sine squared three Teen which is the speed James off time. So, um, in order to get the maximum and minimum speed, we need to differentiate the over bt off the speed a square 15 which is equal to ah, the over Bt off nine plus 27 sign. It's queer three t is equal to zero so three times to sign three D at times co sign three t is equal to zero, so sign 60 is equal to zero. Eso 60 is equal to zero by to buy three point and so on. So t is equal to zero point over six boy over three by over through and so on for maximum speeds. Ah, we differentiate the over DT off sign A 16. Just a quick six whole sign. 16. It's morning Banville. So full sign 60 in smaller than here's OT is able to buy over overseas. Who is the medicine is?

Okay the first thing we're going to do is notice our position vector and we are going to take its derivative to find velocity. So the velocity is going to be three T squared in the I. Direction and then three in the J. Direction. Okay so now we are going to find acceleration and we're going to do that by taking the derivative of velocity. So that will give us a six T. In the I. Direction and then there is no J direction left. So our speed is going to be the magnitude of velocity. So we'll do the square root of that three T squared plus three squared. And that gets us that square root of nine T. To the fourth plus nine. We could factor out of nine. Take a square root, put a three in front um we'll just leave it like that. Okay so um I'm going to create a chart so they can graph this so I'm picking a range of T. Values that are both negative and positive. And then I'm placing um t into my eye component to get my ex cordant and in my J component to get my why coordinate. And that's of course in my original position function or vector. Okay, so let's go ahead and graph this. Get some important places down. Okay. And then we'll join those um dots and kind of put some orientation arrows on there. Okay, so now where we're going to actually find values that V. Of zero and a. Of zero. So if we put a zero into our velocity vector notice we're going to get three J. So we can go ahead at 00 and sketch that 3J. For our velocity vector. Now for acceleration when you put a zero in you get a acceleration of zero and notice how your actual curvature change changes in your function. You know, we're a lot more used to thinking of just first derivatives and second derivatives. Well, acceleration is a second derivative, so you can see it's common cavity changes at that point.


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