8. Let rlt) = 2cos(t), 4t, Zsin(t) vector-valued function: Calculate the derivative r(t) .Find the indefinite integral frlt) dt:Find the arclength of the curve tr...

In Exercises $5-8, \mathbf{r}(t)$ is the position of a particle in the $x y$ -plane at
time $t .$ Find an equation in $x$ and $y$ whose graph is the path of the par-
ticle. Then find the particle's velocity and acceleration vectors at the
given value of $t$ .
$$
\mathbf{r}(t)=e^{t} \mathbf{i}+\frac{2}{9} e^{2 t} \mathbf{j}, \quad t=\ln 3
$$

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.

And this problem we have the premature ization of a problem are tee equal to t I plus t squared plus want quantity times J And we want to find the acceleration and velocity AT T equals negative 10 and one then plotted on a little graph 50 is gonna be the derivative of rt. So we'll have one I plus two TV j And the acceleration is gonna be the truth of the velocity. We'll have zero I plus to J So if we want to make our ah little chart, just make it easier. So have t V and a so negative 101 So at there's no more space here, that's OK. We'll figure it out. We'll do that. Uh, okay, so negative one. So have one eye minus two j and zero have one I plus zero j and at one, we'll have one I plus two j and then for acceleration. We'll have, um so that's your eyes across the board. And it's constant. So we're just gonna get the same value no matter what t we pull again. All right. Blue is gonna be my velocity and green will be my acceleration so plot the blues first. So at negative one, which is about here was on the bottom. So negative 10 and a one. So for blue, we have positive acts native way. So to be like so And for J, we always just have Constance all around is constantly going up. All right, So t equals zero. Yeah, a little, little less velocity. That's much magnitude there and at equal. The one going back up. So there we go.

One Today they are going to solve problem number eight. His Articles x a Plus Lady with Jessica's Cause, Toe I plus three Sign to T D. X equals Cause duty. Why? Because the signed toe pick So bye bye three calls sign toe X squared because cause square too deep y squared by nine because Science Square to keep it's a square plus y squared by nine equals one nine X square plus y squared equals night. We cause our dish, which is because minus sane to be into two in the way plus three cause to teach in two. Includes it vehicles minus two saying toe take a plus six. Cause to be did equals Viewed icicles minus two cause too deep into in do I minus states saying toe deep into into the which is equals a equals minus four cause toe Enjoy minus well sign to t did beyond zero equals minus two. Signed toe into zero I plus six calls. Two in zero, the the off zero equals 60 A of zero equals minus four. Cause to into zero I minus tool signed toe into zero j A. Of zero equals minus four. I 1960 then is this one beats this one. So that's another question

Okay, we're giving a position effect er and the first thing we're going to do is find velocity. So to do that, we're going to take a derivative of all of our components. So think of that I component as a 16 minus t squared. All raised to the one half power be a little bit easier to do chain rule like that. So our Vot Will equal a 1/2 16 minus t square. Then we go down a power to a negative one half and then we multiply by the drift of the inside. That negative to T. That's all of our I component. RJ component will be a two T and RK component will be a one. So I'm going to clean that up the two's cancel out. So I have a negative T in the numerator and then I put that 16 minus t squared in the bottom so I can write it as a positive exponents of a hat and then I just continued to write my other components. Okay, the next thing we're going to do is we are going to find acceleration. So we are going to take the derivative of velocity to do that. Now I'm going to go over here to the side and I am going to write my negative T. And then I'm going to read that 16 minus t square to the negative one half. So I can use product rule. So with product, will you do to keep drip plus trip? Keep so if you write your derivatives right underneath each other, you can kind of do a criss cross um since we've already taken the derivative of that um 16 minus T square. Now we're just going down to a negative three halves power. So you can see again your two's cancel out. You have a negatives cancel out actually we have three negatives so we still have a negative, make sure to get that there. Yes. Let's just check. 1, 2, three. So we do still have a negative on that one and then we have a one over the t minus t squared all to the half power. So that whole piece there is my eye component. You could do common denominators if you would like. Um you can multiply both top and bottom by a half power. And so I'm just showing you how to do that. If that if you do do that, you're going to have two of the t squares minus the 16 and then you'll be over that common denominator of 16 minus t squared all to the three house power. So R. I. P. S was the more complex. R J P s just became that nice to J. And then RK piece um did you know, taking the derivative of one? It went away. Okay, so the last thing to do is for us to find our speed. So for speed, that is the magnitude of velocity. So if we go back up to our velocity vector, we can consider um that we are going to have to square each of our parts, some of them together and take the square root. So we'll have a t squared over a 16 minus t. Square. Now will be to the first power because we squared it plus a four t squared and then plus form.