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#7. Let be the vector function that satisfies 1"(t) = (~e-t; -4sin 2t) , r(0) = (0,2) , and r' '(0) = (2,2) . Find r(t) (if r(t) is the position vect...

Question

#7. Let be the vector function that satisfies 1"(t) = (~e-t; -4sin 2t) , r(0) = (0,2) , and r' '(0) = (2,2) . Find r(t) (if r(t) is the position vector of an object at time t, then r(0) is its initial position and r (0) is its initial velocity).

#7. Let be the vector function that satisfies 1"(t) = (~e-t; -4sin 2t) , r(0) = (0,2) , and r' '(0) = (2,2) . Find r(t) (if r(t) is the position vector of an object at time t, then r(0) is its initial position and r (0) is its initial velocity).



Answers

In Problems $17-24,$ find the velocity, speed, and position of a particle having the given acceleration, initial velocity, and initial position.
$\mathbf{a}(t)=-32 \mathbf{k}, \quad \mathbf{v}(0)=0, \quad \mathbf{r}(0)=0$

Okay, This question starts us off the acceleration vector and wants us to use the initial position and velocity to find our of tea. So we'll start with the fact that acceleration is just the derivative of velocity. And then we can rearrange this equation to show the V F t is the integral of acceleration. So all we need to do is integrate each component of the acceleration vector to get our velocity vector. So let's go do that. So this gives us a velocity vector of integrating each term t squared plus see what x cause you can't forget your plus c. Then we get negative co sign X or sorry co sign T plus C two c one. Why? And then we get 1/2 sign of to t plus C one z. Remember, you have to use the chain rule to do the anti derivative for that one. Are you so and then we'll use the fact that we know the initial velocity is 100 and plug that in AT T equals zero so we can see that we get C one acts minus one plus C one why, and see one z so This means this means that our integration Constance, are equal to one, one and zero. So we now know that V of tea is just those integration constants at it aren't so. Make it t squared plus one. Then we get negative co sign T plus one and then 1/2 a sign of to t. So that's our velocity. And now we'll use the fact that philosophy is the derivative of position to give us the similar relation that R of T is just the integral of e ft. So now let's integrate this thing again, which will give us r of T being equal to t squared plus T plus C two x that our second component, we get negative sci fi plus t plus C two. Why and then for our third component, we get negative 1/4 co sign of to t cause remember, we have to use that chain roll factor. One a half again. Well, and of course, we have another integration constant. So now we know that our zero is equal to 010 We can use that to solve for integration, Constance. So you get c two x c two. Why and negative 1/4. So this tells us. Sorry, missing that. So this tells us that our integration Constance are equal to 01 and 1/4. So this will give us a final position function R of t of t squared plus t on top for the X component, then negative scientist E plus T plus one for a second component and then 1/4 minus 1/4 co sign of to t for third component. And this is our final function.

So in this video we're given the acceleration, which is one of my hat gloves to J hat and were given the initial velocity and the initial position and were asked to determine the velocity function. Yes, so that the lost city folks should have the position. So first of all, we know that the velocity is the integral of the X celebration. So and were given that the acceleration is I had was to her I must do jihad. So we're going to need to grow that. So the integral of one GT is just t the integral off to this to t do you have? And then we have to add a C because this isn't indefinite. Well, we're given a useful piece of information so that we can determine this. See, we're told that the of zero is one k. So instead of TV, put zero. So zero I have lost to time zero j have plus C needs to equal to one. So this is zero. This is zero. So we're just left with C is equal to one k, huh? All right. Great. So we could take this on pulling you back in. So we pluck that in for See, we get that dft this t I have austerity jihad. So that's our for lost. All right, now the position is just the integral off the velocity. So the integral of tea I hat was to t j a happ plus one k had tea the integral of tea with respect cities just t squared, divided by two integral of two. Tea is just Tea Square and the integral off warden It's just And then, of course, plus, C because this is indefinite were also given that are of zero is one I have so free plug in zero for see we get zero squared divided for plug in zero for tm Sorry, we get zero squared divided by two I have was Nero Square J a happ zero k half see needs to equal to one I had so c is equal to one So now we take this see and we look it back in And if we rearrange, we get he's squared divided by two plus one My hat Dusty square jihad plus TK hat And this is our physician

We condemn the processing function as they go to the in tiny relative the with the D. T. And in this question we've given the velocity v t ego to the to society with the initial value answer zero equal to zero. And now we need to find position function SD So it's an equal to the integral on the choco side d d d Here we know we can bring the constant outside They're gonna chew integral on the cost side d d d and untidy riveted because I would you go to the side, the Prince of constancy So this will be the SD here. And given the initial value here and there are ego to the to size zero plus C equals zero Listen, here we get it on equal to zero plus C equals zero and listen implies that c equals zero and therefore we have the position function as the first have the form to side day

This question asked us to find the position function as a T. What we know is that velocity is the first derivative of position. Therefore, what we know is that the integral of D of us is the integral of in this contact sign of T minus co sign T d t. So what is the integral of sine? The answer's negative. Cosa Negative Co signed t If you take the derivative, you got paws of scientific. What's the integral of co sign? The answer is a sign of tea and then we have plus C. We know the initial displacement is s of +00 Therefore, plug in negative co sign of zero minus sign of zero plus C is equivalent to zero. This gives us see as one, which means you can plug back into the S A T equation you can put I put the one at the front


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