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6 marks) Find the length of the curve 7(t) =< 3.t3,t2 > for 0 < t < 2...

Question

6 marks) Find the length of the curve 7(t) =< 3.t3,t2 > for 0 < t < 2

6 marks) Find the length of the curve 7(t) =< 3.t3,t2 > for 0 < t < 2



Answers

Find the length of the curve.

$ r(t) = \langle 2t, t^2, \frac{1}{3} t^3 \rangle $ , $ 0 \le t \le 1 $

So if we want to find the arc length of this curd from negative 5 to 5, what we're going to need to do is apply the following integral. So we're going to integrate from so a to B oh, so B d x by d t squared plus dy by d t squared plus DZ by d t squared. And then we have the swear root of all of that And then we have d t on the outside. So if we go ahead and find these, then plug everything in from there, we should be able to find our arc length. So let's go ahead and do that. So, um, if we take the derivative and actually let me screw this down a little bit So this is supposed to be X o X is t. So if we take the derivative of this with respect to T uh, that will just give us one then this here is going to be wise. We have wise. It was just three co scientists. We have d y by D. T is equal to Well, the derivative of coastline is negative science, So this would be negative. Three scientific and then this is going to be easy. So Z is in 23 sign t and then DZ by D. T, while the derivative of sine is co science that this would be three co sign teeth and now we just need to come down here and plug these in O N r A. And B is supposed to be our interval start in our interval end. So is 95 B is five. So let's plug all that in now. So we have negative 5 to 5 of the square root. Also, uh, ex's derivative is one so once where it's just one plus the derivative of y squared, which would be nine signs square teeth, the derivative of Z squared is going to be nine co sign square tea and then we have d t. Now if we factor out a nine from this, we didn't know if sine squared plus Coastline square and we know from trigonometry or maybe even pre calculus, that sine squared. Plus, cosine squared is equal to one just pythagoras. So that means this is just going to be equal to nine. So it was going to be the integral from negative 5 to 5 Love the square root of one plus nine, which is 10 DT. And now to integrate this, it would just be this route 10 t evaluated from negative 5 to 5. So if we plug into five the Route 10 five, then minus minus five. All those counts, I only get 10. So that would be 10. Route 10. So the length of this curve, which would be 10 times route 10.

In this question we've given the equation for a curve represented by X equal to 6 70 Y equals six cost In an interval of tea from -5 x two. two Plus 5. Way too. In this range we have to find the length of the given curve. Mhm. So from the given any question you can write the X by DT is six because T And the Y by BT is -6. Sign T. Now if we want to find the length of the ark in the time in the interval of T the length of the ark that I am learning by S. Which will be equal to integration of The lower limit of indigenous from -5 way to to the higher limited is plus five way too. And then square out of B. S ability hold square is six costea Whole square plus the usability. Whole squares minus six scientist whole square DT. So this will give us s equals integration minus by by two two plus by way too. Six D. T. Further simplification of this will give us six integration t from minus by way too two plus my way to. So if you put the up unlimited. Lower limited at six times We'll get six times. Bye bye to minus minus by way too. So that will give us s equals 65. And this is the answer for this question

Okay, and this problem will be looking at how to find the arc length of a vector space are vector in space. Um So just recall that the formula to do that the way we do that is by finding the integral from our upper and lower bounds of the magnitude of the derivative um of the vector. Or another way to say that is the square root of the sum of the squares of the derivatives of the component of vector. So it sounds complicated, but once you see it in action, it's not too bad. So what we'll go ahead and do with this problem, This is the vector we want to find the arc length of between zero and 1. So we'll go ahead and start evaluating this. We know that this our plane From 0 to 1 is going to be the square root of the sum of the squares of the derivatives of each component. Okay, so our first component is one. There's nothing written there, so it's one derivative of one is zero and zero square to zero. Uh So we really have the zero. Yeah, here the derivative of t squared is two T. We're gonna square the derivative, we get to t squared. And the derivative the derivative of t cubed is three T squared. So we'll square that. And of course we're finding this integral with respect to T. Okay, so we'll go ahead and start simplifying and evaluating this integral. Yeah. Okay, so zero square 20 to T squared is for two squared, the square of three, T squared is nine to the fourth. Okay, so now we want to try to simplify this. This isn't exactly easy to integrate as it's written, but we can do some factoring and make it a little bit more bearable. I can factor out a T squared from both terms, Some left with four plus 90 squared. Um And the squirt of T squared is T. So you can just pull that guy out of my radical four plus nine two squared. Here we go. And why is this nice? Because now I can do some new substitution. Uh I know that that if I let you Equal four plus 90 square then Do you is going to be equal to 18 T. D. T. 19 squared is 18 T. 20 to 40 And so Do you over 18 is equal to T. T. T. In ways that mace. Because I have the T. D. T. In my integral right now I don't want to forget to substitute my bounds. Alright if I'm doing a new substitution I need to make sure I substitute those. Um So when I plug in zero for T. Here I get a lower bound of four plus zero. So lower bound of four. When I plug in one I get four plus nine. So an upper bound of 13. So now I'll complete my use substitution. I have this 1/18. I can just pull to the outside of my integral. My new bounds are now for and 13 I remember I replaced four plus 19 squared with you. And just to make it simpler I'm just going to write it as you know the one half instead of squared of you. And then T. D. T. Is now my do you It was d. over 18. But are you worried about the 18? Um so this should say 13. Sorry about that. All right. And so now I can go ahead and evaluate this integral. This is equal to 1:18. Okay you do the one half. I will add one to no power that will make it three house. And of course I'll divide by 3/2 which is the same as multiplying by 2/3. I'm evaluating this from four 2 13. Before I do that I'm just gonna simplify fractions 1/18 and two thirds. That's the same as one 27th. 1 27. You do the three house Evaluating that from 4 to 13. No I will plug in and they have 127th times 13 to the 3/2ves minus for 23 halves. I'm fine with that answer. As it is this is an acceptable answer. You could also 13-3 houses. It's so easy to evaluate but four to the three houses a little bit easier. So if you want to you could take it one step further. 423/2. of course that's the same as mm And there we have it in two forms and your answer. Mhm. And that is how you find the our claim

In this problem. First time writing our DST so our destinies can be written as to why. Pleasure to t g plus p school alky. Now I'm finding mode of R dash T, which is equal to under Rooty Square Plus two t Square plus t Square, Holy Square, which is equal to the square plus two now going forward and just calculating the value of L. So analytical to integration of 0 to 1 the squad plus two DT on integrating it. I get evaluation T Cube by three plus two t from 0 to 1 non simplifying it. Just putting the value I get the question edge one by three plus two which is equal to seven by three. So seven by three is our answer.


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