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3. Given r(t) = (sin 2t , cos 2t cos? 2t finda) the unit tangent vector T() at t=T 2b) the curvature at t= T 2...

Question

3. Given r(t) = (sin 2t , cos 2t cos? 2t finda) the unit tangent vector T() at t=T 2b) the curvature at t= T 2

3. Given r(t) = (sin 2t , cos 2t cos? 2t find a) the unit tangent vector T() at t=T 2 b) the curvature at t= T 2



Answers

\begin{equation}
\begin{array}{l}{3-8} \\ {\text { (a) Sketch the plane curve with the given vector equation. }} \\ {\text { (b) Find } \mathbf{r}^{\prime}(t) .} \\ {\text { (c) Sketch the position vector } \mathbf{r}(t) \text { and the tangent vector } \mathbf{r}^{\prime}(t) \text { for }} \\ {\text { the given value of } t \text { . }}\end{array}
\end{equation}
$$\mathbf{r}(t)=4 \sin t \mathbf{i}-2 \cos t \mathbf{j}, \quad t=3 \pi / 4$$

You miss problem? We have a curve R t defined by a three course 90 till scientist. So the irritated off this curfew. Belief minus three slightly in the tool. CropScience When he caused this year old 10 jobs actor are crying. Zero he crossed the zero to So the union taken back at this phone you teach. Is there one the key close to I have the detention vector. Our prime off high half equals two minus 30 old. So the unique endure vector p off high half. We have B minus 10 and what he closed to minus pi. Half our prime off miners have equals 23 0th So the unit had your Magda tee off high at minus. I have It goes to one here.

Here we are asked to sketch the plane curve for the vector valued function R f t equals t squared comets. He cute. Okay, so let's go ahead and sketch that here. All right, so we need to make some substitution to express this vector valued function as a function of X and Y. So we can write out the equations X equals t squared from the X component. And why equals T. Cute From the Y component. And from this we can just plug in Y equals X two the three halves. And you can see you get that by writing Y equals t cubed which is T to the three halves or rather t squared to the three halves. And so when we substitute in X equals C squared, we get y equals X to the three halves, notably. We should also make sure that we write plus or minus, X equals or plus or minus X to the three halves. Because when we take the square root from the denominator that will give us a plus or minus and then when you cube that plus or minus, that uh positive or negative sign will be retained when you cubit. And so each value of X is going to give us two different values of why on the right hand side. So now for instance if we try applauding X equals zero equals 0. If we try x equals one and we will get y equals plus or -1. If we try X equals four, which is the next number, that kind of square roots nicely. And we will get two cubed or eight 345678. So about the top of the graph up there. 234567. About down here. So here is what this plane curve looks like, It's curving slightly. Mhm. And part B is asking us to find the derivative of this function, which we do by just taking the derivative of each individual component, which will be to T three T squared. And then part C. Asks us to sketch the position vector and the tangent vector for the given value of T. Which is one. So our of one Is going to give us 1:01. And you see if we had plugged in T equals negative one. Here, we still would have gotten one and the y component will have the negative one giving us the bottom half of the curve. But here this sits in the top half so we can go ahead and write Are one right there. Yeah. And then Our prime of one Is given by 2:03. So this is saying that the rate of change at this point Is equivalent to going over two and up three. Yeah. That is the answer to this question.

Here we are asked to plot the plane curve given by the vector valued function R. Of T equals co sign T plus one. I had Plus sign T -1. J. had. Yeah. So let us cross axes And then note that in parametric form x equals co scientific plus one. Why equal sign T -1. And so now we have to figure out a way to get X as a function of Y or vice versa. So if you are familiar with your trig identities, you know that you can combine co sign and sign to form a constant one. If you do cosine squared sine squared. And so let's manipulate these equations a bit to where we can do that. So this can become X -1 equals co sign T. This will become my plus one. Equal scientific. And then we can get the equation x minus one squared plus y minus y plus one squared equals cosine squared T plus sine squared T. She holds one or in other words x minus one squared plus y plus one squared equals one. And so this is a circle of radius one that is simply offset Uh 1 to the right From the X -1 and one down from the Y Plus one. So if we draw That circle here and remember Radius one And then let's also figure out where what point is that at T equals zero in which direction is turning in. So when you plug in T equals zero, you get co sign T is one plus one is two. So the X coordinate is too in other words, that only gives us one starting point here At T. zero. And we see that as T increases co sign T will shrink from one towards zero. Um sign T -1 will grow from -1-0. And so we are going mhm counter clockwise along the circle. Sure. Part B asks us to find the derivative our prime of T. Which we get by taking the derivatives of individual components of our ft. So derivative of cosine T plus one is simply negative sci fi times I had And then derivative of Sine T -1 is just KK 70. And part C asks us to use this information to sketch a position and tangent vector at the point T equals negative pi over three. So first given this point, let's find what are our and our prime vectors are So our of negative pi over three. Co sign of negative pi over three is going to be positive one half. So this will be 3/2. I had sign of negative pi over three is going to be Negative Route 3/2. Yeah, so this will be minus Route 3/2 plus one J hat. Yeah. And then our prime of negative pi over three. Again, a sign of negative pi over three is negative. Route 3/2 times negative gives us positive route 3/2 I hat and co sign of negative pi over three is positive one half, 1/2 a half. Now we are asked to sketch these on the curve. So first of all, intuitively you go around a full rotation over two pi radiance. And so if we are looking for the point negative pi over three, we would expect it to maybe be about 1/6 of the way around the circle going in the opposite direction of T of increasing T. Um Specifically if we look here, we need three halves I hat uh minus Route 3/2 plus one J hat. And you can estimate that that will be about here, which looks to be Uh about negative 1/6 of the way around the circle. Yeah. And so that vector here is our prime or rather our of negative pi over three, remove that up here and then our prime of negative pi over three is showing a increase in X and an increase in why? However, the increase is larger and X. And so we'll draw a vector that looks something like this And it should be about a 3060 90 triangle if you want to be exact. Mhm. And this shows the position and the tangent vectors. This tangent vector is good because it matches the direction that we expect the circle to move in as we increase T.

Yeah. So here we are asked to find a few things related to the specter valued function harv t equals t minus two squared plus one. First we are asked to sketchy playing curb with the given vector equation. We can do this by writing out two equations. X equals to my t minus two from the X. Component and why equals t squared plus one from the Y component. And from here we're going to solve the system of equations. So we can change this, X equals two plus two to be T equals X plus two. Yeah. And then we can substitute in that too. The second one. So we have Y equals X plus two squared Plus one. And this gives us a relationship between X&Y for this parametric curve. And so we can go ahead and just write down what that curve looks like. It is a parabola along the Y axis. Okay because there's a plus two inside the parentheses we shift left to negative two and because there's a plus one we shift up to one so the apex of the problem is right there. Then we go up by one on each side and continue curving up like that. Okay. Mhm. Okay. Yeah. Yes. Mhm. There we go. Heart be is asking us to find the derivative of this vector valued function which is as simple as taking the derivative of each component one at a time. So the T derivative of the X. Component here is just one meaty derivative of the Y component is to t. Playing power rule to the T. Square. And then finally part C asks us to sketch the position vector R. T. And the tangent vector R. Prime of T. For the given value of T. Which is T equals negative one. So when t equals negative one are negative, one is going to be negative three comma the one squared plus one is two. Okay. And our prime of -1. Only one comma negative tooth. We can go ahead and sketch this on the graph here. This first one are of negative one is giving us the vector too the point on the curve. And then This our prime of negative one here is telling us what the rate of change is at that point. So we are told that it is one common negative too. And so we will draw that here and that is all that the question is asking for. Yeah.


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