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## Calculus 2

### Unit 5: Lesson 2

Second derivatives of parametric equations

# 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?

• 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.
• 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.
• 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.
• 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?
• Why do we write (d^2)y/dx^2 when algebraically It's (d^2)y/(dx)^2?
• 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
• 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.
• 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"?
• 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)
• 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 ?