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a particle moving in the XY plane has velocity vector given by V of T is equal to all of this business and so using this notation it just means that the X component of velocity is as a function of time is 1 over T plus 7 and the Y component of velocity is a function of time is T to the fourth for time T greater than or equal to 0 at T equals 1 the particle is at the point 3 comma 4 so the first part is what is the magnitude of the displacement of the particle between time T equals 1 and T equals 3 and then we need to figure out its position I mean around to the nearest 10 so like always pause this video and I think you will have to use a calculator but pause this video and try to work through it on your own so we've done questions like this in one dimension but now we're doing it in two dimensions but the key is is to just break it up into the component dimensions so what we really want to do is let's find the displacement in the X direction so really just the change in X and then let's just find the displacement in the vertical direction or our change in Y and then we can use those essentially using the Pythagorean theorem to find the magnitude of the total displacement and also if we know the change in X and change in Y we just add the change in X to the 3 and we add the change in Y to the 4 to find the particles position at time T equals 3 so let's figure it out so change and X from T equals 1 to T equals 3 well that's just going to be the integral of the rate function in the X direction from time equal 1 to time equals 3 so in the X direction we have 1 over T plus 7 that's our X velocity as a function of time 1 over T plus 7 DT and what is this going to be equal to well you might want to do u substitution if you're unfamiliar but you might recognize that the derivative of T plus 7 is just 1 so you could think of this as 1 times 1 over T plus 7 and so we really can just take the antiderivative then with respect to T plus 7 so you get the natural log of the absolute value of T plus 7 and we are going to evaluate that at 3 and then subtract from that it evaluated at 1 so this is going to be a natural log of the absolute value of 10 which is this the natural log of 10 minus the natural log of the absolute value of 8 which is just the natural log of 8 which is equal to the natural log of 10 over 8 just using our logarithm properties which is equal to the natural log of 1.25 so I could get my calculator out in a second to calculate that actually let's just well I'll do that in a second and then let's figure out our change in Y our change in Y once again we're going to take the integral from 1 to 3 that's our change that's our time that the time over which we're thinking about the change and then what is the y component of our velocity was T to the fourth DT well this is going to be take the reverse power of a rule T to the fifth over 5 at 3 and 1 so this is 3 to the fifth over 5 is 243 over 5 minus 1 to the fifth over 5 minus 1/5 so this is equal to 242 over 5 which is what 48.4 48.4 now let me get my calculator out for this natural log of 1.25 1.25 natural log and we'll just round to two decimal places approximately 0.2 2 so this is approximately zero point two two so I figured out our change in X and our change in Y and actually just from that we can answer the second part of the question first what is the particles position at T equals 3 well it's going to be our position at T equals 1 where we to each of the components we add the respective change so we would add so this would be 3 plus our change in X from T equals 1 to T equals 3 and it would be 4 plus our change in Y so this is going to be equal to 3 plus our change in X well that's going to be approximately three point two two and four plus our change in Y that is what 52.4 this right over here is fifty two point four but we still have to answer the first question what is the magnitude of the disk well it's Pythagorean theorem I'll draw a very rough sketch of what's going on sometimes it's useful to visualize it so our initial position is at 3 comma 4 so 3 comma 4 so we're right over there and we figured out our change in X isn't much so our change in X is a positive 0 point 2 2 so our change in X we're barely moving in that direction and our change in Y is 48 point 4 so we have a dramatic change it really goes off the charts over here in that direction but if we wanted to add them together if we want to add those vectors together you could shift over your change in Y right over here and then find the hypotenuse the length of the hypotenuse would be the magnitude of the entire displacement and so let's do that so are the magnitude of the displacement it's going to be the square root of our change in x squared plus our change in Y squared once again this is just the Pythagorean theorem and what is this going to be I'll get my calculator out again for this okay that was our change in X let me square it and then plus we are going to have forty eight point four eight point four squared is equal to this right over here and then we take the square root of that so this is going to be see if we all right over there and there you go our total the magnitude of our total displacement is 48 if we round to the nearest tenth 48 point four so this is approximately 48 point four and we're done now one thing that you might be noting is it looks like our total displacement 48 point 4 is the same as our change in Y now the reason why it came out this way is because our change in Y was exactly 48 point 4 while the magnitude of our displacement was slightly more than 48 point 4 but when we round to the nearest tenth we got 248 point for the reason why they're so close is because our change in X was so samal we're talking about 0.22 is a change in X and our change in Y was so much that our hypotenuse was only slightly longer than our change in Y so that's why we got this result for this particular instance in general you're going to see the magnitude of the displacement is going to be larger than the magnitude of either X our change in X our change in Y alone