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Find the direction of the line tangent to the curve r(t) = (sin(t) ~ t,cos(t), e2t 1 atthe point (~T, -1,e2)(~T,-1, e2r(0, 0,2)(-2,0, 2e2t =(0,1,1)...

Question

Find the direction of the line tangent to the curve r(t) = (sin(t) ~ t,cos(t), e2t 1 atthe point (~T, -1,e2)(~T,-1, e2r(0, 0,2)(-2,0, 2e2t =(0,1,1)

Find the direction of the line tangent to the curve r(t) = (sin(t) ~ t,cos(t), e2t 1 atthe point (~T, -1,e2) (~T,-1, e2r (0, 0,2) (-2,0, 2e2t = (0,1,1)



Answers

Find the unit tangent vector to the curve at the indicated points. $\mathbf{r}(t)=\left\langle 4 t, 2 t, t^{2}\right\rangle, t=-1, t=0, t=1$

In this question, we have a curve are key defined by three key and the piece worth its curative with respect to pee will be three to t. So when P causes zero, um, the tinges vector up, Frankie a friend zero will be 30 old. So the union tend your record. P zero is just 10 And when he closed to minus one, the tension vector our pride minus one Because doe three minus two and the units tend your record, he minus y was through three overweight, off 30 minus two over route off 13. And that's when he closed toe one. Um, the tension vector up from one because the three to the unit tender vector t one because the three over eat off 30 to overeat off 30.

In order to find the tangent vector. First we need to find the derivatives. So that's our prime of tea. This is going to be able to. And then now the derivative of the first component here is going to be eight t cubes and then comma. Now here, um, and power. Also three halves times six. That's gonna be called to nine First out front and then T to the three high was minus, um, one is 1/2 and then comma. Now, remember, this is 10 t to the negative one, so the negative one comes outfront. So it becomes a negative 10 and then this becomes t to the negative one minus one. So that's gonna be negative too. So it's divided by a T squared Now, finding the time Inspector at T equals one. Now, this is going to be cool, too. Well, one cube is just one. So that's gonna be eight and then comma here. Nine. I'll t to the will want to. The 1/2 is also one. So we're gonna have here just nine and then comma. And then here one squared his negative 10. So this is going to be our counter factor here

Find the unit tangent vector um at T equals zero. So in order to do that, we're going to be taking the derivative of R. O. T. And we're going to be putting that over the magnitude of our prime of T. Okay, so far derivative. Notice in our I component, we are going to have to use product role. We have a keep and then the derivative of cosine of T. Is a negative sign of T. And then plus derivative. And of course the derivative either T. Is either the T. And so we have that multiplied by Cassina T. That's all the I. Direction. And then in the J direction we will just have the the tea. Now if we place zero in each of the zero powers one, however, a sign of zero is zero But the co sign of zero is 1. So we end up with just I plus J. So to find that magnitude we're just taking the square root of one squared plus one square. So that's just gonna be a square root of two. So now we can write our tangent vector at zero by placing each of our values of one over the square root of two in both the I and J direction.


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