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Second derivatives (vector-valued functions)

Sal finds the first and the second derivatives of the vector-valued function h(t)=(-t⁵-6,4t⁴+2t+1).

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  • leaf blue style avatar for user S
    Why would we want to find the second derivative of a vector-valued function?
    (9 votes)
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    • leafers seedling style avatar for user Tommi Rossi
      well, as sal pointed out, higher order derivatives give different things, an example being, in physics, derivatives of position with respect to time.

      p(t) = position, p'(t) = velocity, p''(t) = acceleration, p'''(t) = jolt or jerk, p''''(t) = jounce or snap etc.

      while jolt, snap and pop or whatever the higher ones are might seem kind of abstract, you can probably see the use in finding the acceleration of something.
      (15 votes)
  • mr pink green style avatar for user Jayson Matchado
    "H prime prime of t" is that really how it is called?
    (5 votes)
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  • starky ultimate style avatar for user nfleming65
    So it pretty much seems like vector valued functions are the same exact thing as parametric curves (I don't care about the "well, technically..." explanation, thank you).
    So is the derivative of a parametric curve WITH RESPECT TO T, found in the same manner as this? Because in the section on parametric curves, we only went over the derivatives of y with respect to x.
    Also, conceptually, what is the connection, if any, between the derivatives of the curve with respect to t vs. the derivatives of the y-component with respect to the x-component?
    (2 votes)
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    • leaf green style avatar for user kubleeka
      Yes, the derivative of the parametric curve with respect to the parameter is found in the same manner.

      If you have a vector-valued function r(t)=<x(t), y(t)> the graph of this curve will be some curve in the plane (y will not necessarily be a function of x, i.e. it may not pass the vertical line test.) Each point on the curve will have some value of t associated with it as well. You can think of r as the path traced out by a particle as it moves over time.

      In this case, dy/dx is a scalar function of t, and it gives the slope of the tangent line to r at the point associated with t.

      dr/dt is another vector-valued function, which we can interpret as the velocity of the particle tracing r.
      (6 votes)
  • blobby green style avatar for user Javier
    I thought i understand it but i can't. What is the difference between the second derivative of a vector ( acceleration w.r.t position) and the second derivative of a paremtric ecuation.
    As far as i've investigated they are both the same, since any vector can be expressed as an x coordinate and y coordinate with respect to another parameter.
    Therefore why for the parametric ecuations we have d(dy/dx)/dx while in the vectors we have (d^2y/dt^2)/(d^2x/dt^2)).
    If the parametric ecuation gives the slope of the tangent line to the point at t and the vector gives the same but they are calculated differently what is the difference between them?
    (3 votes)
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    • eggleston orange style avatar for user francisco.scorsato
      The difference is that when taking derivatives (or second derivatives) of parametric equations, you are actually taking the derivative of a parametric equation with respect to the other. (you do dy/dx, not dy/dt or dx/dt).

      However, when taking derivatives of vector-valued functions, you are actually seeing how the x and y components of said functions vary with respect to a third variable (generally, time t). And that is the difference: You are, in essence, taking dy/dt or dx/dt, in contrast to dy/dx.
      (3 votes)
  • leafers ultimate style avatar for user Gavin Yu
    In the previous video (Second derivatives of parametric functions) we found (d^2y)/(dx^2).
    In this video, we found (d^2x)/(dt^2) and (d^2y)/(dt^2). If we divide the latter by the former, we get (d^2y)/(d^2x).
    What is the difference between that and (d^2y)/(dx^2)?
    (2 votes)
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  • blobby green style avatar for user earl kraft
    how does one distinguish between a vector scenario where they should be viewed like position vectors, where the tail of the vector is at the origin vs. the vectors which can be translated?
    (2 votes)
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Video transcript

- [Voiceover] So I have a vector valued function h here, and when I say vector value, it means you give me a t, it's a function of t, and so you give me a t, I'm not just going to give you a number, I'm going to give you a vector. And as we'll see, you're going to get a two dimensional vector. You could view this as the x component of the vector and the y component of the vector. And you are probably familiar by now, that there is multiple notations for even a two dimensional vector. For example, you could use what's often viewed as engineering notation here, where the x component is being multiplied by the horizontal unit vector. So you might see something like that, where that's the unit vector, plus the y component. Four t to the fourth, plus two t, plus one, is multiplied by the vertical unit vector. So these are both representing the same thing, it just has a different notation. And sometimes you'll see vector valued functions with an arrow on top to make it explicit that this is a vector valued function. Sometimes you'll just hear people say, well, let h be a vector valued function, and they might not write that arrow on top. So now that we have that out of the way, what we're interested in is, well let's find the first and second derivatives of h with respect to t. So, let's first take, let's take the first derivative, h prime of t, was as you'll see, it's actually quite straight forward. You're just going to take the respective components, with respect, take the derivative of the respective components with respect to t. So the x component, with respect to t, if you were to take the derivative, with respect to t, what are you going to get? Well we're going to use the power rule right over here, five times the negative one or times the negative. You're going to get negative five times t to the five, minus one power, so t to the fourth power. The derivative with respect to t of negative six, well that's just zero, so that's the rate of change of the x component, with respect to t. And now we go to the y component. So we're going to do the same thing, the derivative with respect to t, is going to be, and once again, we just use the power rule, four times four is 16t to the third power. Derivative of two t is just two, and then derivative of a constant, well that's zero, we've already seen that. So there you have it. So this is the rate of change of the x component with respect to t, this is the rate of change of the y component, with respect to t. And one way to do it, vectors can represent many, many, many different things, but the type of, a two dimensional vector like this, you can imagine this being, h of t being a position vector in two dimensions. And then if you're looking at the rate of change of position with respect to time, well then this would be the velocity vector. And then if we were to take the derivative of this with respect to time, well we're going to get the acceleration vector. So if we say h prime, prime of t, what is that going to be equal to, h prime, prime of t? Well, we just apply the power rule again. So four times negative five is equal to negative 20t to the four minus one, so t to third power. And then we have three times 16 is 48t squared, and then the derivative of two is just zero. And so there you have it, for any, if you view t as time, for any time, if you view this one as position, this one as velocity and this is acceleration, you could, this would now give you the position, velocity and acceleration. It's important to realize that these vectors could represent anything of a two dimensional nature.