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### Course: AP®︎/College Calculus BC > Unit 9

Lesson 5: Solving motion problems using parametric and vector-valued functions- Planar motion example: acceleration vector
- Planar motion (differential calc)
- Motion along a curve: finding rate of change
- Motion along a curve: finding velocity magnitude
- Motion along a curve (differential calc)
- Planar motion (with integrals)
- Planar motion (with integrals)

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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.

## Want to join the conversation?

- 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?(8 votes)
- Instead of adding delta x, delta y to 3, 4, could you find the actual position function by adding +Cx and +Cy to the corresponding antiderivatives? Then solve for the constants by substituting in (1,3)? Guessing this would be a more general solution to what we do here(4 votes)
- Yep! That works too. And yeah, that would be the more general case, which should be used if the problem asked you for a position vector without giving any initial conditions.

However, once you find the position vectors, you'll need to find the position of the particle at t = 3, then subtract from its position at t = 1 (For both x and y), and then find the magnitude. It's a lengthier process, but will still give you the answer.(4 votes)

- 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.(5 votes) - 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?(2 votes)- x does not equal ln|t+7|. x equals ln|t+7|+C , sal didnt add the C when finding the definite integral (the
*change*in x) since the C would be canceled anyway

we can find C by plugging 3 for x and 1 for t:

3 = ln|1+7| + C

C = 3 - ln|8|

thus, x = ln|t+7| + 3 - ln(8)

if you plug 1 you do get 3

same goes for y

does this help?(1 vote)

- 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?(2 votes)
- 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)

- The question should be rewritten to the magnitude of the total displacement, or the average displacement, I think. Am I wrong?(1 vote)
- You could rephrase it as the magnitude of total displacement between t = 1 and t = 3. That works too! In the question, it is implied that we need the total displacement anyway.

Average displacement is a weird concept here. Remember what average velocity meant? it was the total displacement over total time. See that to calculate average velocity (f'(x)), you needed total displacement (f(x)). So, to calculate average displacement, you'll need the total of a quantity whose derivative is displacement. That quantity is called absement. So, to calculate average displacement, we'd need total absement, and we clearly don't have that. Hence, average displacement isn't what we're after here(2 votes)

- 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)

## Video transcript

- [Instructor] 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 as a function of the time is one over T plus seven, and the Y component of velocity as a function of time is T to the fourth for time T greater than or equal to zero. At T equals one, the particle is at the point three comma four. So the first part is what is the magnitude of the displacement of the particle between time T equals
one and T equals three? And then we need to
figure out its position, and we need to round to the nearest 10th. So like always, pause this video, and I think you'll 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 that 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 three, and we add the change in Y to the four to find the particle's position at time T equals three. So let's figure it out. So change in X from T equals
one to T equals three, well, that's just gonna be the integral of the rate function in the X direction from time equal one to time equals three, so in the X direction, we
have one over T plus seven. That's our X velocity
as a function of time, one over T plus seven DT, and what is this going to be equal to? Well, you do might want
to do u-substitution if you're unfamiliar,
but you might recognize that the derivative of T
plus seven is just one, so you could think of this as one times one over T plus seven, and so we really can just
take the antiderivative then with respect to T plus seven, so you get the natural
log of the absolute value of T plus seven, and we are going to evaluate that at three and then subtract from that. It evaluated at one, so this is going to be the natural log of the absolute value of 10, which is just the natural log of 10 minus the natural log of
the absolute value of eight, which is just the natural log of eight, which is equal to the
natural log of 10 over eight, 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 gonna take the
integral from one to three. That's the time over which
we're thinking about the change, and then what is the Y
component of our velocity? Well, it's T to the fourth DT. Well, this is going to be... Take the reverse power rule
T to the fifth over five at three and one, so this is three to the fifth
over five is 243 over five minus one to the fifth
over five minus 1/5, so this is equal to 242 over five, 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 0.22, so this is approximately 0.22. 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 particle's
position at T equals three? Well, it's going to be our
position at T equals one, where to each of the components we add the respective change, so we would add... So this would be three plus our change in X from T equals one to T equals three, and it would be four plus our change in Y. So this is going to be equal
to three plus our change in X. Well, that's going to
be approximately 3.22, and four plus our change
in Y, that is, what, 52.4. This right over here is 52.4, but we still have to
answer the first question. What is the magnitude of the displacement? Well, it's the 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 three comma four, so three comma four, so we're right over there, and we figured out our
change in X isn't much. So our change in X is a positives 0.22, so our change in X, we're
barely moving in that direction, and our change in Y is 48.4, 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 the magnitude of the displacement is
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 48.4, 8.4, 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... Right over there, and there you go. The magnitude of our
total displacement is 48, if we round to the nearest 10th, 48.4. So this is approximately 48.4, and we're done. Now, one thing that you might be noting is it looks like our total
displacement, 48.4, 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 48.4 while the magnitude of our displacement was
slightly more than 48.4, but when we round to the
nearest 10th, we got to 48.4. The reason why they are so
close is because our change in X was so small. We're talking about
0.22 as the 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 our change in X or our change in Y alone.