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# Planar motion (with integrals)

To analyze planar motion where the rate vector is given, we need to find the displacement in each direction separately. Then we will either use that to find the new position, or to find the magnitude of the displacement using the Pythagorean theorem.

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• I seem to recall being able to do this in Calc III 15 years ago by taking the integral of y*dx. The explanation was basically that the vast majority of integrals in single variable calculus are set up that way.

In the case of this problem, y would be t^4, and x would be 1/(t+7), so that dx/dt would be ln|t+7| and dx = ln|t+7|dt.
It follows that ydx would be (t^4)ln{t+7|dt.

For this problem, of course, that seems like a difficult integral to find. I was going to try to confirm it on my own without asking, but I ran into trouble here.

Can anyone confirm for me that, in theory, at least, what I am describing is a valid method for solving a problem like this one? I suppose it might give you the magnitude of the displacement of the particle, but not specifically how much movement occurs in the x and y directions.

Thanks for any insight that anyone is able to provide. • If we wanted to find the distance that the particle travelled, we could integrate using the arc length formula, as the length of the curve IS the distance travelled. This seems a little bit different from the first question, since the magnitude of the displacement isn't necessarily equal to the distance travelled, but the magnitude of the start point and end point, regardless of distance travelled.. correct? • t=1 x=3 y=4
x=ln(t+7) y=t^5/5
but if you plug in 1 you dont get those cordinates? why?
those coordinates are given and position vector should give then in t=1, what am i missing? • I have a question related to this topic: why is it that when I take the definite integral from a to b of [dx/dt] dt and [dy/dt] dt, and get ∆x and ∆y respectively and then plug them into the following square root((∆x)^2 +(∆y)^2), I don't get the exact same answer in my calculator as when I enter in the definite integral from a to b of [square root((dx/dt)^2 +(dy/dt)^2)] dt, but I still get very similar answers. For example if a is 3, b is 8, and dx/dt is 3x^3 while dy/dt is 12x^2, from the former method I'll get 3582.07015 which rounds to 3582.1 an answer khan accepts for such a problem, but from the latter method I'll get 3600.628589 which rounds to 3600.6 which Khan marks as incorrect. Which method is providing a more accurate answer and which method is providing the answer that the college board will accept? Will college board accept both answers and methods?
(1 vote) • Hey Daniel,
I think the second method is totally wrong, you see when dealing with integrals you should be careful with the order of operations, Taking an integral of a function from a to b then squaring the result is totally different from squaring the function and then taking the integral they both give different answers unless it is an exceptional case of course !
//maybe try to find an exception if you are up to the challenge!
You can try it with simple functions like f(x) = x from 0 to 10
Taking the integral and the squaring yields 2500
but squaring the function and then integrating yields 333.3

In fact I think the only operation where the order doesn't matter is when adding integrated functions and when multiplying by constants !

In the question sal is solving we take the integral from a to b to figure out Δx and Δy and then add their squares and then take a square root to figure out the magnitude and the order does matter here you have to figure out the integral first which gives you a scalar quantity.
It is surprising that the answers you got were similar !

I hope this was helpful to you, if you didn't understand anything you can reply :)
(1 vote)
• I have question about the second part, shouldn't the position be (3+change in x between 0 to 3, 4+change in y between 0 to 3), but in the video, Sal added the change in position between 1 to 3.
(1 vote) 