Main content
AP®︎/College Calculus BC
Course: AP®︎/College Calculus BC > Unit 9
Lesson 2: Second derivatives of parametric equationsSecond 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?
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!(41 votes)
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 :-) ).(26 votes)
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)
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)
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)
Here's a quick, concise explanation of why it works (it's a little less hand wavy): http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx
Another way you can think of it is that you're multiplying by dt/dt, which cancels out to be one (this is because differentials never equal 0, though they approach it, meaning the operation is defined)
On a side note, it's great that you want to see where the formula is coming from instead of just accepting that it works.(7 votes)
- 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) 
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)- Have you looked at the material on exponents?
I think the following section will be helpful:
https://www.khanacademy.org/math/pre-algebra/pre-algebra-exponents-radicals/pre-algebra-negative-exponents/v/negative-exponents(2 votes)
- Why do we write (d^2)y/dx^2 when algebraically It's (d^2)y/(dx)^2?(2 votes)
It's called Leibniz notation, and the simplest answer I can give you is that we just do it because it is the convention. The wikipedia page has a nice proof on it if you're still interested:
https://en.wikipedia.org/wiki/Leibniz%27s_notation(4 votes)
what's with the second derivative being written that way? d2y/dx2
is that just convention? what does it mean?
why not dy2/dx2 or something?(2 votes)
I went through some answers on StackExchange and here's one which explains the notation in a decently intuitive manner.
So, the second derivative is just (d/dx)(dy/dx). Now, dy/dx is just "change of y with respect to x". But, d/dx of that would imply "change of dy/dx with respect to x". So, we're calculating the change of something with respect to x which itself is changing with respect to x. So, that's where x^2 comes from. However, we are only considering one change overall (the other one) so the d doesn't become d^2.
To be honest, it's a pretty weird notation which we've come to accept. The explanation I provided is more to convince yourself that its true than it actually being true.(4 votes)
- 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)- Think of it like this:
d/dt[dy/dx] // dx/dt
= d²y/dx * (1/dt // dx/dt)
= d²y/dx * 1/dx (The dts cancel)
= d²y/dx²
Note: dxs, dys, and dts don't actually work like fractions, but it can be helpful to think of them as such.(2 votes)
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)
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)
- At3:30, 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)
Notice that 𝑑𝑦∕𝑑𝑥 is written in terms of 𝑡.
In order to differentiate 𝑑𝑦∕𝑑𝑥 directly we need to write 𝑡 in terms of 𝑥.(1 vote)
3:30Video 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.