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4) 10 points. A student is asked to find the normal component of acceleration for the position vector given by r(t) = 3ti - tj +t2k 21 The student has already deter...

Question

4) 10 points. A student is asked to find the normal component of acceleration for the position vector given by r(t) = 3ti - tj +t2k 21 The student has already determined thatr' (t) = 3i - j+#k and that r"c '(t) 2k. Complete the problem

4) 10 points. A student is asked to find the normal component of acceleration for the position vector given by r(t) = 3ti - tj +t2k 21 The student has already determined thatr' (t) = 3i - j+#k and that r"c '(t) 2k. Complete the problem



Answers

Use the given information to find the normal scalar component of acceleration at time $t=1$ $$ \|\mathbf{a}(1)\|=9 ; a_{T}(1) \mathbf{T}(1)=2 \mathbf{i}-2 \mathbf{j}+\mathbf{k} $$

Okay. So we're trying to find the position vector of a particle that has the given acceleration as F. T equals T E. T. And E. Two negative T. With the The initial value as zero equals Is your 01 and uh video equals zero 11 Okay now we know that the function can be written as FFT- zero plus Integral from T 0 to T. F prime of you. Do you? Okay. The way you can confirm this is by substituting a 50 0 and then notice that the Right hand side is also a 50 because Is the integral from 0 to 0 is just zero. Right? So this expression is what we're going to use to find the velocity vector and the position text from the acceleration vector. Okay, so we have a O. T. Equals key HD E to negativity. Sorry? David, L C. With the even initial value is equal to The F0 plus from zero to T. And then the acceleration. Do you eat a negative view? Do you G cause 001 plus use square Over two and then eat you negative. Each negative view From 0 to T. And I am denoting this by the integration component wise. Soy Eur 01 plus T square over to E. T. And then negative each negativity minus zero one negative one. And then you get T square over to negative one plus E to T. And two minus. Eat negativity. And in a similar manner you can find the position vector L. T. Equals I was your plus The integral from 0 to T. And then I'm gonna substitute the expression of VF T. Here of your view negative one plus eat you two minus eat negative. You do you She calls 011 plus You to the 3rd over six and then negative. You plus E. D. U. And then to you plus eat negative you From 0 to T. She calls 011 plus T. to the 3rd over six negative T plus E. T. And two T. Plus each negativity minus zero 11 Which he calls T. to the 3rd over six negative T plus DT and the last component. Okay? So now that we found the expression of the position vector, we can graph it using a computer. And when you grow up it with the computer you get the following graph.

Okay, This question. Who wants us to find the tangential and normal components for the acceleration of a particle whose position is given by this r of t So we're gonna need to know the magnets, you know, velocity and acceleration. So let's start by calculating those functions. So v A T is the derivative of the position function. So going by components derivative of Ln is one over tea derivative of T squared is to t derivative three teas three. Then we have to over squared of tea for the last year. Now for acceleration, we just take another their evidence. So our top term becomes negative one over t squared. Then we get it too. And then we could a negative t to the negative three halves because we have to bring down a negative 1/2 hour and then subtract. So now let's go about calculating what we needed. So for the magnitude of velocity, we just some the squares of the components, then at the very end, will take a square root. So just plugging in our derivatives. Earlier this works out toe one over T squared plus 40 squared plus 12 t plus nine plus four over tea or getting everything over a common denominator, we get one plus four t to the fourth plus 12 t cubed plus nine T squared plus for tea, all divided by t squared. So the magnitude of the is just the square root of that whole thing itches. 42 the fourth plus 12 t cubed close to 90 squared, plus 40 plus one, all the vital by t squared, for we can pull out a one over tea, and then we still have the square root of four T to the fourth plus 12 t cubed plus 90 squared close 40 plus one. And we'll just leave it like that and we'll mess with it later if we need to turn this into a sort of perfect square. So what we'll do now is calculate z dot eh over magnitude TV because that's the tan, gentle acceleration so V dot eh is equal to one over two to t plus three and two over root tea dotted with negative one over T squared two and negative T to the negative three halves, which gives us negative won over T Q plus 40 plus six minus t the negative three halves times t to the negative 1/2 with the two in front for v dot a equals negative one R t cubed plus 40 plus six minus two over t squared or if we want. This is a common denominator. We can get negative one plus 42 the fourth plus 60 cube minus two tea. So si dot a over magnets you divvy is just this stroppy over t to the thank you. So 40 to the fourth plus 60 cubes. Waas. Sorry, Linus minus two tea minus one, divided by t cubed, divided by one over tea and then our square root from earlier cube plus 90 spirit close for two people. Okay, so now what can we do? Well, we can simplify this a little bit. Or we could just plug in right here, which I think is the quicker option. So feed on a over magnitude of the in general. We just write this in a final little compact form. Here. There's four t to the fourth plus 60 huge minus minus two tea minus one, divided by T squared times the square root of for two to the fourth plus 12 t cubes plus 90 square. It was 40 plus one. So wants us to evaluate this at a specific X, Y and Z So are of tea equals 044 So that man's the X equals zero Michael's for a Z equals for so we can pick whatever one of these coordinates. We want to find the tea value, so exit T equals zero. So this man's the Ellen of T equals zero and Ellen of what zero? Well, Ellen of one equals zero, so t equals and you could get the same answer. Doing all their ways are using extra using wires. He just be careful that you're getting a unique T. If you get multiple options for tea, make sure you're getting the right one by testing the other components. So now that we know T equals on, we can plug that in here. So we get four plus six minus two minus one, divided by one times the square root of move This four plus 12 plus nine plus four plus one. And what does this simplify too? Well, we get seven on the top, divided by square roots of 12 plus eight is 20 20 plus turns 30. So after all that work, we get the tangential acceleration at that 0.7 over square root of 30 no, onto the normal component. And that is just equal to the cross. A two butted by magnitude of the And we already know that the magnitude that the magnitude of V and are given point is just squared of 30 because we plugged into the formal already. But well, still calculate the cross, eh? So we have won over route 30 times V cross A, which is won over tea to t plus three to over rooty crossed with negative won't over t squared to t and negative to you to the three halves so we know what time we're looking at here. So at first were cut. We can plug in t now because t equals one. So this is one over route 30 times looking and one everywhere we get 152 crossed with negative one to a negative one, which is much easier than working with square roots. So plug and teeth now is a good way to do this. But I did the doubt product the other way to show you that either ways equally valid. But since you only care about one point, why not just plug into you right now to make our lives easier? So v cross a will Just use our little terminate trick here at this specific point, we put our first vector in the second row and our second factor in the third row that will do our little determinant for the eye component. Delete their own. Cohen. We get negative five minus four minus J hat. Delete the road elite The column. We get negative one plus two and then for a K hat. We get deleting their own column two minus negative five. So two plus five or this man's We simplify this out. V cross, eh? Is the following negative five minus four is negative. Nine. Negative negative. One times negative. One plus two. It's just one and kay hat coefficient. Seven. So the magnet sued of a seven is just won over Groot 30 times The magnitude of the cross, eh? Or magnet suit. Avi Cross, eh is just the square root. Oh, 81 plus one plus 49 or square root of 1 31 So this gives us Hey, Savannah cools squared of 1 31 divided by screen of 30. Or just this whole thing under a square

Problem. 45 a one is equal toe I plus two J minus two two. Okay, eight and ginger off one is equal to three. Eight in general is able to be dot a over the north off V, which is equal to p notes of A which is equal to, uh, the in Jin Jin and the normal the main making off the engine ship vectors times the magnitude of a vector time. Suppose I That's data. So three is equal to one times one plus four plus four times courtside. It's theater, so co sign data is equal to one and scientific data should be quick to zero. So at the ain or is equal to the magnitude off be cross a over B, which is equal to be over the magnitude of me cross Kroc, which is equal to the magnitude off t gross A which is equal to the magnitude off team. The magnitude of a so I'd feta. So a norm is equal to one plus three plus at one time. Three times Scientist A, which is equal to the year. So a deeper to deal, so anything


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