If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Second derivatives (parametric functions)

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.G (LO)
,
CHA‑3.G.3 (EK)
Sal finds the second derivative of the function defined by the parametric equations x=3e²ᵗ and y=3³ᵗ-1.

Want to join the conversation?

  • hopper cool style avatar for user JPhilip
    Why can't you calculate x''(t) and y''(t) separately, and then divide y''(t)/x''(t) to find the answer? I tried doing it, and I got 3e^t/4. I set the two answers equal to each other and got 1 = 9. Why doesn't this method work? Thanks!
    (42 votes)
    Default Khan Academy avatar avatar for user
    • leaf orange style avatar for user John
      Here is an answer on stackexchange that is beautifully simple, it "just" uses the chain rule, and that is the insight I was missing.


      http://math.stackexchange.com/questions/49734/taking-the-second-derivative-of-a-parametric-curve


      I was getting stuck thinking of it as:
      "Second derivative of y with respect to t"
       dy2      d   {  dy }
      ---- = ---- { --- }
      dt2 dt { dt }


      But we're not doing that, we're looking for
      Second derivative of y with respect to x:
            dy2      d   {  dy }
      ---- = ---- { --- }
      dx2 dx { dx }

      I can't quite express it yet (I'm wrestling with the notation) but it the stackexchange link looks pretty solid so I wanted to share that now. I will update this if I figure it out way to write it (bothers me that I can't just bang something out to explain it :-) ).
      (24 votes)
  • old spice man green style avatar for user Refael Ben-Shlomo
    How are parametric functions different than vector valued functions?
    Why are the differentiation techniques different?
    Both have functions representing the x and y components of the curve.
    In the vector valued functions we just take the second derivative of each of the parts.
    But in the parametric function the technique is entirely different.
    (14 votes)
    Default Khan Academy avatar avatar for user
    • female robot grace style avatar for user tyersome
      The most significant difference between what we are doing here and the vector valued functions is that here we are interested in the derivatives with respect to x. In the preceding material, we were interested in the the derivatives with respect to t.
      (10 votes)
  • spunky sam green style avatar for user Not Friedrich Gauss
    is there a video where sal defines dy/dx = dy/dt/dx/dt ? I feel like the explanation of just replacing the y isnt enough, i want to know mathematically why that works.
    (5 votes)
    Default Khan Academy avatar avatar for user
  • female robot grace style avatar for user loumast17
    Any easy way to explain where the derivation formula for the parametric equations comes from? and the second derivative too.

    EDIT

    Nevermind, I got it. Just in case someone stumbles upon this and wants to know,

    We know f(x)=y, g(t)=x and h(t)=y, so this means f(g(t))=h(t). Taking the derivative of this uses the chain rule so f'(g(t))g'(t)=h'(t) and since g(t)=x f'(x)g'(t)=h'(t). We fant f'(x) or dy/dx and using algebra to move everything around gets us dy/dx=h'(t)/g'(t). Another way of writing this is d/dx(y)=(d/dt(y))/(d/dt(x)) which leads into taking the second derivative. Like it shows in the video, the first case is taking the derivative of y, so if we want to take the derivative of dy/dx, just replace all ys with dy/dx. And so on for further derivatives.
    (7 votes)
    Default Khan Academy avatar avatar for user
  • starky ultimate style avatar for user Jesse
    I'm having an incredibly hard time with the practice questions with this, specifically when you change the positive exponentiation into negative (or negative exponentiation into positive) when you divide a fraction by a fraction by multiplying one by its reciprocal, and when you don't. Is there a longer lesson about this, or a rule of thumb someone can point me to?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Michał  Krysiak
    Why do we write (d^2)y/dx^2 when algebraically It's (d^2)y/(dx)^2?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Roy Kimbrell
    I don't understand how the chain rule is applied to obtain the second derivative of a parametric equation system. The first derivative chain rule can be illustrated by

    dy/dt = dy/dx dx/dt

    since y = y(x) and x = x(t)

    Then

    dy/dx = dy/dt // dx/dt

    But I don't see how a "second application" of the chain rule results in the expression

    d²y/dx² = d/dt[dy/dx] // dx/dt
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leafers ultimate style avatar for user Gavin Yu
    In the next video (second derivatives of vector valued functions), we calculated x''(t) and y''(t), and combined, we get a vector that can be understood as acceleration.
    Is there also a practical use or good analogy that explains what d^2y/dx^2 "means"?
    Thanks in advance!
    (2 votes)
    Default Khan Academy avatar avatar for user
    • stelly green style avatar for user The #1 Pokemon Proponent
      Think of( d²y)/(dx²) as d/dx [ dy/dx ]. What we are doing here is: taking the derivative of the derivative of y with respect to x, which is why it is called the second derivative of y with respect to x. For example, let's say we wanted to find the second derivative of y(x) = x² - 4x + 4. We'd first take the derivative of y(x) which is 2x - 4 and then take the derivative of this to get the second derivative, 2.
      (1 vote)
  • blobby green style avatar for user fatimadorgham142
    At , I couldn't understand why should we differentiate dy/dx and then divide it by dx/dt, couldn't we just differentiate the whole dy/dx again ?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Walt Williams
    Has there been a mistake here? When performing the second derivative Sal properly placed the dy/dx from previous process above the dx/dt derivative, BUT then simplified the upper with the lower term without taking the second derivative. The correct second derivative should be 1/24*e^t

    I watched this video to refresh my memory on how to setup the second parametric derivative.
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

- [Voiceover] So here we have a set of parametric equations where x and y are both defined in terms of t. So if you input all the possible t's that you can into these functions and then plot the corresponding x and y's for each t, this will plot a curve in the x-y plain. What I wanna do in this video is figure out the first derivative of y with respect to x and the second derivative of y with respect to x. And in both cases it's going to be in terms of t. So let's get to it. So first lets find the first derivative of y with respect to x. First derivative of y with respect to x. And we've seen this before in other videos, where this is going to be the derivative of y with respect to t over the derivative of x with respect to t. And so this is going to be equal to, well, what is the derivative of y with respect to t? Dy, dt is equal to. Let's see, the derivative of either the 3t with respect to 3t is just e to the 3t. And then the derivative of 3t with respect to t is going to be three, so I can say times three like that, or I can put that three out front. And then the derivative of negative one, well, a constant doesn't change no matter what you do to your t, so that's just going to be zero. So that's dy, dt. So it's going to be equal to 3e to the 3t, all of that over, well, what's the derivative of x with respect to t? Derivative of x with respect to t is equal to, well, we're gonna have the three out front, and so the derivative of e to the 2t with respect to 2t is going to be e to the 2t and then we're going to take the derivative of 2t with respect to t, which is just two, so this is gonna be 6e to the 2t. 6e to the 2t. And lets see, we could simplify this a little bit. I'll now go to a neutral color. This is equal to, so this is gonna be one half, that's three over six, e to the 3t minus 2t. 3t minus 2t. And I'm just using exponent properties right over here. But if I have three t's and I take away two of those t's, I'm just gonna have a t. So this is just going to simplify to a t right over here. So now that we've now figured out the first derivative of y with respect to x in terms of t, now how do we find the second derivative? How do we find the second derivative of y with respect to x now? And I'll give you a hint! We're going to use this same idea. If you wanna find the rate of change of something with respect to x, you find the rate of change of that something with respect to t and divide it by the rate of change of x with respect to t. So what this is going to be, we wanna find the derivative of the first derivative with respect to t, so let me write this down. We wanna take the derivative, with respect to T in the numerator, of the first derivative, which I will put in blue now, of dy, dx, all of that over dx, dt. If it doesn't jump out at you why this is the same thing that we did before, I encourage you to pause the video and think about it. Think about what we did over here the first time. When we wanted to find the derivative of y with respect to x, we found the derivative of y with respect to t and then divided that by the derivative of x with respect to t. Here, we wanna find the second derivative of y with respect to x. Actually, let me just write it down out here a little bit clearer. What we really wanna do is we wanna find the derivative with respect, let me write it this way: When we wanted to find the derivative with respect to x of y, that was equal to derivative of y with respect to t over the derivative of x with respect to t. Now we wanna find the derivative with respect to x of the first derivative with respect to x. And so everywhere we saw a y here, replace that with the first derivative. So this is going to be equal to, in the numerator the derivative with respect to t of dy, dx. Notice this was derivative with respect to t of y. In fact, let me write it that way just so you can see it. So if I clear this out, if I clear that out we're gonna get, this is the derivative with respect to t of y. So hopefully you see, before we had a y there, now we have a dy, dx. Dx, dt. Now this might seem really daunting and complicated except for the fact that these are actually fairly straight things to evaluate. Taking the derivative with respect to t of the first derivative, well, that's just taking the derivative with respect to t of this, and this is pretty easy. This is the derivative, it's just gonna be one half. And the derivative with respect to t of e to the t is just e to the t. And so that's going to be over the derivative of x with respect to t, which we already saw is 6e to the 2t. 6e to the 2t. We can write this as, one half divided by six is one over 12, and then e to the t minus 2t, which is equal to, we could write this as 1/12th e to the negative t, or we could write this as one over 12 e to the t. And we're done.