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Vi hkch of the following vectors WlicH tangent the curve e(t) (52 40, 7tan t, tat}(0,0.0)(h) (-4,0,0)(c) (-4,7,0)(-4,0.1)(-4,7.1)...

Question

Vi hkch of the following vectors WlicH tangent the curve e(t) (52 40, 7tan t, tat}(0,0.0)(h) (-4,0,0)(c) (-4,7,0)(-4,0.1)(-4,7.1)

Vi hkch of the following vectors WlicH tangent the curve e(t) (52 40, 7tan t, tat} (0,0.0) (h) (-4,0,0) (c) (-4,7,0) (-4,0.1) (-4,7.1)



Answers

Find a tangent vector at the given value of $t$ for the following parameterized curves. $$\mathbf{r}(t)=\left\langle e^{t}, e^{3 t}, e^{5 t}\right\rangle, t=0$$

In order Find the tangent vector here. So first we need to find the derivative. So our prime of tea is going to be equal to now we take the derivative. The first component, so derivative of eat of the tea is just Nativity comma. And then now, to find delivery was either the three t That's gonna be three years of the three T and then, lastly, that are evidence of the each of the five t is going to be able to five e to the five T. Lastly, we're going to plug in our value of equal cereal to find the time director at that point. So our prime at zero this is equal to and then now eat of the zero. Well, even zero is equal to one. So this is just going to be one here and then here again, e to the zero is going to be one. So you're just gonna have three left over. So this is going to be three. The men comma, and then here, lastly, we're gonna have just five. So this is going to be our attention Director. At T equals zero

Okay we are given a vector function here and we're going to find the tangent vector at the time T. So we're given T. Equals negative one. So first of all to find the derivative. I'll take the derivative of each of my components separately. So when I take the derivative of T. I get one so I can just write I and then I take the derivative of negative T. To third. And I get negative three T squared in the J. Direction. Now I'm gonna be placing in a negative one and I get I minus three because remember negative one times negative one is still one. And so that negative three is in the J. Direction. Okay this is my answer to A and now I'm going to graph and there's two different ways to graph. Um One way is to kind of remove your parameter. The other way would be to use a T. Chart and think of sometimes and find your corresponding exes and wise. But here if X equal T. Then my. Why that J. Component would just be a negative X to the third. So I really can draw my negative X. To the third. And then consider that my T. Values are going to go from that left to right. And the last thing to do is I am going to add the tangent vector at my point. So if I placed a negative one in um um as kind of my T. Value then I will get that value of one and then I can draw my um tangent vector and for that I would go to the right one and down negative three. Um Just like how it showed that um I would get that tangent vector from part A.

Okay we are giving the vector function R. Of T equals T. In the I direction minus T squared in the J. Direction. And we are going to first find the tangent vector at T equals one. So first we're going to do the derivative. Now when you do your derivative you take a derivative of each component. So our I components derivative would be one. And then R. J components derivative will be a negative to T. And notice when we had the one in the I direction. We don't have to write the one if you just write I it assumed one. Okay so now we're gonna be putting T equals one in. So there's really nowhere to put it for the eyepiece. It just is I but then we can do that negative two times one and get the two. So that is going to be our answer to part A. So the next thing is we are going to graph the curb. So there's several ways to graph one of them is to consider what removing the parameter T. Would look like. The other way to graph would be to actually come up with some values of T into chart our um function. So I'm gonna say if X equals T then y equals negative X squared. Because the T squared will just replaced with X. Square. So I know what the negative X. Square, that's going to be a problem shaped downward. That goes through the 0.0. Now we're going to add the graph of the tangent vector um at T equals one. So at T equals one. My Y. Value is actually a negative one. And if I draw that vector it should go over one and down negative to um just like um the answer shows us it should in part a.

Okay we are given the vector valued function and we are going to find the tangent to the vector at T. Equals zero. So we can go ahead and find our derivative of R. And we're going to do the derivative each of our components. So when you take the derivative of E. To the two T. You can rewrite it. But then you're gonna be multiplying by two. The derivative of E. To the negative T. Is going to be eating the negative T. But then you multiply by a negative. Now if we place our zero in we will have eat zeros all equal once you're going to have to in the I. Direction -1 in the J. Direction. So let's go ahead now and do some elimination of our variable. So if X equals E. To the to T. Then the square root of X will just equal eat the T. So why ends up equaling one over the square root of X. So this is gonna look a lot like um one over X. Except for being the square root, it just kind of flattens that curve out. Um But you can still see that at T. Equals zero. We would have been at the 00.11. And then um that's going to go negative 22. Yeah To to the right and then one down


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