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Answers

Solve the initial value problems in Exercises $11-16$ for $\mathbf{r}$ as a vector function of $t .$
$$\begin{array}{ll}{\text { Differential cquation: }} & {\frac{d \mathbf{r}}{d t}=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}\end{array}$$

Okay we are going to integrate our prime of T to get our of tea and then we'll be solving for our constant of integration given um the condition at our of one. So first of all to integrate E F T becomes E F T in the I direction. Are integral of Ellen. We need to do integration by parts. So we'll make our U R L N F T and R D V R T. So that means do you will be one over T and v will be T. So what we'll get here is a T. Ellen of tea. And then when we leave our table we have a negative D. T. To still integrate. So that will become a minus T. So that allows us to now put in R. J. Component for our empty and then as far as archaic component we are going to go up a power and divide by two so that two divided by two just becomes a one. Um then we have the plus C. And remember we had constants for R. I. J. And K. So when we're considering that when we place in a value of one for T we need to get out a value of zero in the I. Direction, one in the J. And one in the K. So here I've started to solve for my eyepiece, I'm gonna say zero equals E. To the first plus C. One. So C. One actually equals negative E. Now my next one is a value of one for the J. And then I'm also going to be putting in a value of one. So I have that negative one Ellen of one minus one. Now the Ellen of one is zero. So I really have a negative of a negative one there. And so when I subtract one from both sides I see that see to dis equals zero and then my third one. Um when I place a one in there, I also get a constant of zero. So the main thing is I need to have that minus E. In my eye component and then I can rewrite the rest of my R of T. And the vector differential has been assault.

So this problem we have the initial value problem. TRT t is equal to three halfs route t plus one I plus e to the negative t j plus 1/80 plus one K. And we know that our zero is equally Kate. So we want to find our our teeth. Okay, so if we integrate this are looking at two So three halfs group t plus one I plus e to the negative TJ lost one over to t plus one k t. And so let's split up things with T plus one on things that we can just integrate directly. So we're gonna have Negro three halfs through key posts. One I plus 1/30 plus one case DT plus in a role of each of the negative T j. Okay, so if you was equal that t plus one do you people? It's DT. So have three halfs route. You I plus one over you jook. This is the same. Okay, so now we get any rate. So you're gonna get you thirds you to the three halfs times three halfs. That's one. And as the coefficient at times I plus Ellen of U K minus E to the negative. T j close in constant. So we have you to the three. So what is your you wa steeples one? It will have t plus one to the three halfs I plus Ln US three Absolutely absolute value of T plus one. Okay, fine ass either. Negative T J put some constant. Okay, so let's figure out with Constant. So we know that our zero equals K which is going to equal. So we have one eye. Plus, So Elena one is zero. So that's easy. And then so, actually, an me minus either the zero. What's eating the chair? That once somebody, it's one J plus zero. Okay, let's see. So c is going to be negative. One I plus one j uh, plus one. Okay. All right. So now I know our teeth. There's going to be They have a TV plus one to the three halfs minus one. I less Ln Absolutely your t plus one close. Absolute value plus one j and then plus one minus e to the negative. T. OK,

The topic of this question is directional derivatives and gradient vectors. This question asks us to find the directional derivative of G. At the origin in the direction specified by this vector. A. To start off recall that the directional derivative of a function at a certain point in a certain direction is the gradient vector of that function. Uh at that point in the direction are dotted with the unit vector specifying the direction. So in our case we want to find the gradient of G at this point peanut, the origin and the unit vector in the direction of a. Yeah. So first let's look for this unit vector. Um The unit vector in the same direction as a will be just a vector that is a divided by a certain value, a scaled by a certain constant value. Yeah. And that specific value will just be the length of A. Since a shortened by the length of a has length one. And that's what we want. So to find the length of A. We take the square root of the sum of the squares of the components of a. Oh and in that case we get three. Therefore you is the vector a shortened by a factor of three. These components about it by three. No, let's find the other vector in our dot product the gradient vector of G. At the origin. The gradient director at a generic point X. Y. Z. Is the vector whose first component is the derivative of the with respect to X. Who's second component is the derivative of the with respect to why and who's third component is the derivative of G. With respect to Z. Yeah. To find this derivative with respect to X. Note that this term is constant with respect to X as is three. The only part that we have to differentiate is E. To the power of X. And it is its own derivative. So keep the constants multiplied on to the E. To the power of X. And there is our derivative. It's just the same thing as cheap. Yeah. Now the derivative with respect to why this time this function it's constant and it is just a multiplier onto the derivative that we are taking. Now the derivative, of course, Y. X. With respect to Y. Uh, is a little bit more complicated because why is multiplied by zero inside another function? So we apply the chain rule. First, we have the derivative of coast Y. Are the derivative of coast wise said with respect to Y. Zed. It's variable and that is negative sign. Why is it? And then we also have the term that we multiply onto that. The derivative of wizard with respect to Why here, is that in a bit more neat of a form. And now finally we take the derivative with respect to Z. Which will be the exact same as before when we differentiated with respect to why? Except in the end we take the derivative of Y. Said with respect to zed. And so this will now be a Y. Yeah. And so that gives us this gradient factor. G D L G. Mhm. No. We want to find what that would be at the origin. 000 So all we have to do is substitute X equals zero, Y equals zero and said equals zero. Yeah. Mhm. What what we find is that the first component, Which is one times 1 times three, is just three. As I said, the second component has a zero multiplier and the third also has a zero multiplier since both zed and why are zero. And so the gradient vector is only in the direction of I at the origin. I had what Now we are ready to take the directional derivative here, which is what the whole point was. Yeah. So simply taking the dot product of 300 with two thirds, one third and negative two thirds, We get three times 2/3 Plus zero Plus 0. And so that's just too. And since we have found that the directional derivative is to out to the origin, we have completed our solution. And so that ends this video. Thank you for watching.

The topic of this question is directional derivatives and gradient factors. This question asks us to find the derivative of this function at this point in this direction given by the specter. So to do that, we use the fact that the directional derivative of earth mm at any point peanut in any direction given by the unit vector U. Is the gradient factor at the point peanut dotted with the unit factor you. Yeah. So this is relatively simple to do. All we have to do is find two vectors. So let's start with the gradient vector of F. At peanut first let's find out for a generic point X. Y. Z. And then we'll substitute the values of X. Y. Z. Given by peanut. In this case each of X. Y. Z. one. Each is one. So what is the Great Aunt Director? The gradient vector is the vector whose first component is the derivative of F. With respect to X. Who's second component is the derivative with respect to said. And who's third component is the derivative of F with respect to Oh sorry, the second with respect to Y. And the third is with respect to set. So differentiating this with respect to X. We see that these are constants and the derivative of X squared is two X. Mhm. The derivative with respect to Y is just the derivative of two high squared. Yeah, I'm some of my variables all the time here. And the derivative with respect to said Is -60. Mhm. Mhm. You know now we want to substitute X equals one, Y equals one and that equals one. Uh huh. Right. This is the victor we get now we want to find the other vector in R dot product you. Yeah. Since we want to find the directional derivative in the direction of you. Right. And the directional derivative in the direction of a. You will be the unit factor in the direction of a. Yeah. Oh right. Any vector in the direction of a is just a scalar multiple of a. Yeah. And that's Kellar. And our case will have to be one divided by the length of it. So let's find this length which is the square root of the sum of the squares of the components and then let's express you As one of the route three times a. Yeah. Yeah. Now we can evaluate our directional derivative. It is simply this gradient factor that we found no dotted with fine The vector that is the same as a. Only with the components divided by square root of three. Mhm. Okay. Gives us To over route three plus four over Route three minus six. Over Route three and that is simply zero. Which brings us to the end of our solution. Thank you for watching.


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