AP®︎/College Calculus BC
- 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)
The position of a particle moving in the xy-plane is given by the position vector (-3t³+4t²,t³+2). Sal analyzes it to find the acceleration vector of the particle at time t=3.
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- Why can we get the velocity vector by taking the derivative of each of the components in the position vector? What if the position vector is (t, t+2), then if we take the derivative of both t and t+2, we will get velocity vector (1, 1).
But it doesn't seem to be right, because we know the derivative of y=t+2 is 1 for all x values, we can write it as y=1 (x∈R), is a horizontal line rather than a single point we just calculated. What went wrong?(4 votes)
- Remember that when you calculate the first derivative, you are not calculating the rate of change of y with respect to x (dy/dx) but rather the rate of change of the position vector's x and y coordinates with respect to time. Then you calculate the second derivative to find out how that rate of change is changing over time. the actual path of motion assymptotically approaches the line y=-3x, but the particle is moving ever faster and faster along its trajectory with an increasing acceleration.(3 votes)
- Umm, question, if i will be using the general formula(e.g. V=diplacement/t, a=change in V/t) in position vectors, will the answers be the same or not, my bet is that it won't, but i just need some confirmation(2 votes)
- Very late answer but just in case someone else was wondering the same thing, no. Those two formulae assume a critical detail: velocity and acceleration are constant. However, here, both velocity and acceleration aren't constants, as they are functions of time.(2 votes)
- How is the derivative in this form useful? Wouldn't it be better to find the second derivative with respect to x instead of respect to t? And if it is better, how would we do that? If it is not, how do you use the vector form of the second derivative to analyze the function?(1 vote)
- I believe that we should be taking the second derivative with respect to t as its x-coordinate here is a function of t. Here, 'x' represents the amount moved along the x-axis as a function of t, while the y-coordinate 'y' represents the amount moved along the y-axis as a function of t. The second derivative here tells us that the acceleration is in the upper left direction of the Cartesian plane at t=3 (as it has a negative x coordinate and a positive y coordinate, and leftward is taken as negative according to convention) with a magnitude of 46 units/s^2 in the negative x-direction and 18 units/s^2 in the positive y direction. The function essentially just tells us about the motion of a particle in two dimensions with respect to time, which is why we take the second derivative with respect to time.
I hope this answers the question you were asking, and that it's still useful, if at all. I am answering this based off what I understood from the video, so if anyone finds any discrepancies, feel free to correct me.(3 votes)
- Are the velocity and acceleration vectors in standard position, i.e. are their tails on the origin?(2 votes)
- Sal mentionnedit in an earlier video how these derivative vector functions aren't with their tails on the origin but rather on the point which is correspondent to them on their normal graph. So what I'm trying to say is that if a function r(t) gives the position in the (x,y) plane as time progresses, then the derivative vectors will have their tales on the (x,y) position at which it was at that specif time. I hope this helps though anyone feel free to phrase this concept better.(1 vote)
- What would dy/dx and d^2y/dx^2 of r(x,y) actually represent in this example? And the same with v(x,y) and a(x,y) ?
I can't get my head around the "meaning" of dy/dx with vector valued fonctions / parametric functions.(1 vote)
- A vector valued function gives a curve in the Cartesian plane, just like a 'standard' function. So it will represent the slope of the curve, like we're used to with derivatives.(3 votes)
- At the beginning I tried to solve it by myself, and acceleration described by vector didn't sound logical to me, so I rewrote velocity function as a square root of sum of both directions squared (based on Pythagorean theorem) to get velocity as a scalar. Then I calculated rest using scalar–based functions and result seems quite ok.
Is that acceptable approach?(1 vote)
- I'm not too sure what you mean, as I have not yet learned much physics. However, I do know that acceleration and velocity are both vectors because they contain information about their respective directions. Velocity as a scalar is simply speed, which does not show direction. Calculating with scalars should have given you a scalar result, which is incorrect (I believe), as we are trying to find the acceleration vector.
Hope that I helped, and correct me if I'm wrong.(1 vote)
- [Voiceover] A particle moves in the xy plane so that at any time t is greater than or equal to zero its position vector is, and they give us the x component and the y component of our position vectors, and they're both functions of time. What is the particle's acceleration vector at time t equals three? Alright, so our position, let's denote that it's a vector valued function, it's gonna be a function of time. It is a vector. And they already told us that the x component of our position is negative three t to the third power plus four t-squared, and the y component is t to the third power plus two. And so you give me any time greater than or equal to zero, I put it in here and I can give you the corresponding x and y components. And this is one form of notation for a vector, another way of writing this, you might be familiar with engineering notation. It might be written like, or sometimes people write this as unit vector notation. Negative three to the third plus four t-squared times the unit vector in the horizontal direction, plus t to the third plus two, times the unit vector in the vertical direction. This is just denoting the same thing. This is the x component, this is the y component. This is the component in the horizontal direction, this is the component in the vertical direction, or the y component. Now the key realization is if you have the position vector, well the velocity vector's just going to be the derivative of that. So V of t is just going to be equal to r prime of t, which is going to be equal to, well you just have to take the corresponding derivatives of each of the components. So let's do that. So if we wanna take the derivative of the x component here with respect to time, we're just gonna use the power rule a bunch. So it's three times negative three, so it's negative nine t-squared and then plus two times four is eight, so plus eight t to the first. So plus eight t. And then over here for the y component. So the derivative of t to the third with respect to t is three t-squared. Three t-squared, and then the derivative of two is just a zero, so I actually have space to write that three t-squared even bigger. Three t-squared, alright. And if we wanna find the acceleration function, or the vector valued function that gives us acceleration as a function of time, well that's just going to be the derivative of the velocity function with respect to time. So this is going to be equal to, this is going to be equal to, let me give myself some space. And so the x component, well I just take the derivative of the x component again, and let me find a color I haven't used yet, I'll use this green, so let's see. Two times negative nine, negative 18 times t to the first power plus eight, derivative of eight t is just eight if we're taking the derivative with respect to t. And then here in the orange, derivative of three t-squared, so it's two power rule here, over and over again. Two times three is six t to the first power, just six t. So this is, we've just been able to by taking the derivative of this position vector valued function twice, I'm able to find the acceleration function. And now I just have to evaluate it at t equals three. So our acceleration at t is equal to three is equal to, so in green it's going to be negative eighteen times three plus eight, comma, and then we're gonna have six times three, six times three. And so what does this simplify to? Well this is going to be equal to, let's see, negative 18 times three is negative 54, negative 54 plus eight is negative 46. Negative 46, and then six times three is eighteen. Did I do that arithmetic right? So this is negative 54 plus eight. So negative 54 plus four would be negative 50, plus another four would be negative 46. Yep, there you have it. Negative 46, comma eighteen. That is its acceleration vector at t is equal to three.