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### Course: AP®︎/College Calculus BC>Unit 9

Lesson 2: Second derivatives of parametric equations

# Second derivatives (parametric functions)

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!
• 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 :-) ).
• 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.
• 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.
• 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.
• 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.
• 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.
• 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?
• To help me understand how the second derivative method is derived, I took the reciprocals and looked at it as "the derivative of y with respect to x, but since both are in terms of t, we chain y to x, through t." So the second derivative looks the way it does because it's the first derivative of y with respect to x, differentiated -- necessarily in terms of t -- then just chained back to x again. Is that a proper way to look at it?
• Why do we write (d^2)y/dx^2 when algebraically It's (d^2)y/(dx)^2?
• 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
• 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?
• 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.
• 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 ?
• 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.
• 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