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## Differential Calculus

### Course: Differential Calculus>Unit 6

Lesson 4: Planar motion

# Planar motion example: acceleration vector

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.

## Want to join the conversation?

• 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?
• 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.
• 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
• 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.
• 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.
• Are the velocity and acceleration vectors in standard position, i.e. are their tails on the origin?
• 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.