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Is/pearl/Downloads/maths"o2OdiffpdfQuestion 3. (Total matks: 16 Find tle: solution_ ineluding domain of Ihe initial value problemwith V = "bahMake suTe&#x...

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Is/pearl/Downloads/maths"o2OdiffpdfQuestion 3. (Total matks: 16 Find tle: solution_ ineluding domain of Ihe initial value problemwith V = "bahMake suTe' explain the se]k of JUIu sulutiola clueck Hol reqquirexi but_ Iot full Matky WIsI juclul cuough words Aldl symlals make VT solutiou cle;tActnvate . 06pafcn

Is/pearl/Downloads/maths"o2Odiffpdf Question 3. (Total matks: 16 Find tle: solution_ ineluding domain of Ihe initial value problem with V = "bah Make suTe' explain the se]k of JUIu sulutiola clueck Hol reqquirexi but_ Iot full Matky WIsI juclul cuough words Aldl symlals make VT solutiou cle;t Actnvate . 0 6pafcn



Answers

In Problems $11-16,$ solve each vector differential equation with the given condition. $\mathbf{r}^{\prime}(t)=e^{t} \mathbf{i}-\ln t \mathbf{j}+2 t \mathbf{k}, \quad \mathbf{r}(1)=\mathbf{j}+\mathbf{k}$

Okay, so in order to find the domain of a vector function, we actually look at the domain of all of our components. So if you consider CO. Sign well, any real number. So all real numbers. Sign again. Anything can go into that, and then to um doesn't even have a variable. So there's nothing that cannot go into this. So our domain is all real numbers. I'll write that as negative infinity to positive infinity.

So in this problem were given the initial value problem. GRD t is equal to negative. T i minus t j minus t k And our zero is I plus two Jabo streaking. So ah, if we integrate grd t e t people get negative piece wearing over to I s t squared over 20 k J and s Peace bird over to you. Okay, listen. Constant in our three. So next. So we know that our zero is equal to I plus two j plus three k. And this is equal Teoh this evaluated at zero, which means zero plus c. So that's e equals I plus two j plus three k. That means our tea is going to be so we gotta we gotta add C to this and then combine like terms so we'll have negative t squared over two plus one. I and they were looking at J component, so we have negative. He squared over two plus two today, and now we're looking at the K component. So negative cheese squared over two plus three. Okay. And that's our

So in this problem we want to find the solution to the initial value problem. D are equal. Teoh de Cubes plus 40 I plus t j plus jue Qi Square K d t and r of zero is I A plus j So if we integrate both sides will get una role of tea cubed plus 40 I plus t j plus two t squared K gt. So these are owners. Give me power rules so we don't have to do any funny stuff. So that's gonna be, um, Peter the fourth over four plus t smart over two. That's time I plus t squared up. Shoot. Oh, your horses. So we should have ah four times t squared over juice. That'll be two key square. So times I plus he squared over two j plus huge. He cubed over three K plus some constant. Okay. And we know that our zero is equal to I plus J This is gonna be equal to so anything was t zero. So we'll have zero i zero j plus this year. Okay, close a constant. That concept is going to be I plus J. So then our tea will be equal Teoh. Um, so he gets Tito fourth over four plus two t squared, plus one times I plus So we have the jail term. So he squared over two plus one j plus ju t cubed over three.

All right. In this problem we wish to find DZ DT derivative of Z. With respect to T. Using chain will one where Z is E. To the X and Y. To find parametric lee with X equals T. Y equals pi T, his wife and his child. Your understanding of multi very different aviation. To solve what we want to do is utilize the chain rule for function of several variables. The general one safe spaces for one independent variable. We have DZ DT equals easy. Dx dx DT plus DZ Dy Dy DT. Thus the salt. We must find the four derivatives given in our formula for D C. T. First differentiate EMC. We have D C D X equals X and Y and the Z D Y equals E. Coast wide differential. The X and Y. Inspector T X equals one over to routine. Dy DT is high. Thus plugging into DZ DT gifts the formula here, plugging in X and Y simplifies to get to the root tea time. Sign pretty over to routine plus pie. Co sign pipe T. As it is shown


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