13.1.24 Find parametric equations for the line that is tangent to the given curve at the given parameter value r(t) = (4P) i+(t+1)j+ () k t=6=3 What is the standard...

Find a vector equation of the line tangent to the graph of $\mathbf{r}(t)$ at the point $P_{0}$ on the curve. $$ \mathbf{r}(t)=t^{2} \mathbf{i}-\frac{1}{t+1} \mathbf{j}+\left(4-t^{2}\right) \mathbf{k} ; \quad P_{0}(4,1,0) $$

Here. We're gonna find the unit tangent vector and we're also going to find a set of parametric equations for the tangent line um of our given space curve. So notice our point is 000. So that is a X, Y Z position. So really you have to find the value of your parametric at this place. But we could say, well T equals zero, T squared equals zero T equals. So pretty much we know here that are actual parameter T also equals zero. Now to the right, our unit tangent vector will be found by first taking our derivative of R. T. And then dividing by the magnitude of our prime of tape. So are derivative will be I plus two T in the J direction, get that J in there and then plus one in that K direction. So our prime of And again I said that R. T would have to be zero because of placing our components equal to the corresponding point. And solving for tea. And so our prime at zero is going to be I plus K. So our prime at zero will be the square root of one squared plus one square. So it's going to be the square root of two. So now we can write our tangent vector at zero as being one over the square root of two times. Um that I plus K. That allows you to set your variables A B and C as that +10 in one. And this is what we'll use to create our parametric equations. So it's always your start value plus a times T. For that X piece. So we end up that X equals T. Ry will be its initial value of zero plus zero times T. So why just ends up being zero and then Z will be its initial value of zero plus the C, which is one times T. So that is also the parametric equations equals T.

Okay, we are going to find the unit tangent vector and then we're also going to find a set of parametric equations for the tangent line to our space curve here at our point given. So I've gone ahead and written the um equation to find a unit tangent vector here to the right now, the other thing that we have to do before we start is we have to figure out at what time we are at. So we're given our X Y z coordinates. Um but we have to consider that T square would equal one or T equals one. Obviously if t squared equals one that could be plus or minus one. But looking at are y parameter, we know that um T is going to equal one. Okay, so now that we know that let's go ahead and find our first derivative and that will be two T in the I direction plus one in the J direction and then zero in the K direction. So for our magnitude we can um first, well, first of all let's evaluate that at one and will replace the one in we get to in the I direction plus one in the J direction. Okay, for our magnitude we will square each of those components added together and take the square root. So that is gonna be the square root of four plus one over the square root of five. So now we can write our tan, our unit tangent vector at T equals one and that will be one over the square root of five. Of our two, I plus J. So what we can do now is we can define our A, B and C as the coefficients and we have a two, A one and a zero. Then when we write our para metrics it's always going to be our an initial value. So the initial value in the X direction was one plus two times because that's our a value T. For our Y direction, we have a start value of one and then we have plus one teas, will just write one plus T. And then in our Z direction we will have 4/3 but we really add no tease to that. So really easy will just be 4/3.

Parametric equations of the tangent line to this curve X. Y. Z. At the 0.21. So for sonny dX DT which will be one over T. An dy DT, which will be two times one half T. To the minus one half Which is one over the square root of T. And then DZ DT equals two teeth. Okay? So at the 0 to 1 um we have t squared equals one. So T. Is plus or minus juan. But why is two squirts of tea? So T. Must be positive one. So these slopes become one one and what's two? So x equals zero plus 1. T. Y. equals two plus 1. T. Z equals I can't see what that oh two plus one cheek. Wait X's wine. Okay. See is one uh just got all confused there. One because that the uh in the point the z coordinate is one and then plus two T. Because that's the slope. So one T. Two plus one T. One plus two T.

So we have are is equal to to co sign pi t times I plus to sign pi t times J plus three t Okay. And we're looking at one t not. It's equal to a one third. Let's go ahead and calculates the derivative of this guy so Khost deserve it of a co Sinus negative sign. Um and we're going to get the pi to come out chain rule. Get that, Get it too high. Co sign high Tea J. And then we have a three k Let's calculate are one third. So that's two times one half I was to square three, divided by two. Knowing our trig, they'll use three times one third k. So this is equal to one square 31 Let's calculate our prime at one third. So this is two times pi square with three divided by two. Too high one half J. Let's through K. Okay, let me rewrite this not simplified way. So this is negative. Square three. Hi. Hi. Three. Okay, so now we have we have the derivative vector when 10 t is equal to one third and with the position vector on t equals one third were good to go to write down a parametric equation for the line. So when x for X, we're going to have negative square three pie. So this thing Times t plus one getting the one from there now for why we have pi t getting pie there. Plus square three. Scary three from there. Okay. And for Z, we have three t plus one. This is the Parametric equation for the long.