Question

Second Derivative for Parametric Equations

Answer 1

http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx

Answer 2

Can someone please explain the notation for the second derivative?

Answer 3

well you have
y dependent on t, and x dependent on t
so y(t) and x(t)
to take th derivative wrx we do
\[\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]
ok now we have a function of x, lets call it f'(x)
now if we want the derivative of it again we do the same thing(sort of like with Cartesian coordinates), except this time your y=f'(x)
\[\frac{d^2(f'(x))}{d^2x}=\large\frac{\frac{d\frac{dy}{dt}}{dt}}{\frac{dx}{dt}}\]

Answer 4

sorry that took so long I got lost in the latex

Answer 5

sorry that should day
\[\frac{d^2(f(t))}{dx}=\frac{\frac{df'(t)}{dt}}{\frac{dx}{dt}}=\frac{\frac{d\frac{dy}{dt}}{dt}}{\frac{dx}{dt}}\]

Answer 6

say*

Answer 7

I hope I didn't confuse you more.

Answer 8

\[\large \frac{d}{dx}y\qquad=\qquad \frac{d}{dt}y\cdot \frac{dt}{dx}\qquad=\qquad \frac{dy/dt}{dx/dt}\]

Narin do you understand where the first derivative is coming from?
The notation can be a little tricky.
It'll help you understand where that second derivative is coming from if you understand the first.

Answer 9

Now imagine a world without this notation....god bless Leibnitz

Answer 10

yeah i understood the first derivative and teh notation for it. i just couldn't figure out how paul'snotes created the formula for the second derivative

Answer 11

one sec while i gather what i'm confused about

Answer 12

zzr0ck, what does \[d^2 (f(t)) / dx \] mean? like i know that \[d(f(t)) / dx\] is just the derivative of f(t) (or y) with respect to t. but what does the second "d" in the numerator do/mean?

Answer 13

with respect to x*

Answer 14

can u help me zepdrix?

Answer 15

Mmm I dunno.
That notation is rather confusing, since both x and y are functions of t.

\[\large \frac{d}{dx}\left(\color{royalblue}{\frac{dy}{dx}}\right) \qquad=\qquad \frac{d}{dt}\left(\color{royalblue}{\frac{dy}{dx}}\right)\cdot \frac{dx}{dt} \qquad=\qquad \frac{\dfrac{d}{dt}\left(\color{royalblue}{\dfrac{dy/dt}{dx/dt}}\right)}{dx/dt}\]

So the square is telling us to differentiate twice.
But more specifically what it's doing is...

We find dy/dy as the ratio of parametric derivatives, taking the derivative of the top and bottom with respect to t.

Then we take the derivative of the top and bottom with respect to t again,
So we end up with the `2nd derivative of y` on top, and the `1st derivative of x on the bottom (with respect to t).

I just realized that I didn't use the square at all in my explanation.. hmm woops :\
Maybe I can give that another try.

Answer 16

`We find dy/dy` should be `We find dy/dx`*

Answer 17

yeah sorry I was trying to show that the original y was a function of t, to make the point...bad idea

Answer 18

The square tells us to differentiate twice.
In this first step, see how we lose one of the d/dx's and apply the derivative?
I know mixing Leibniz Notation with primes is kinda sloppy lol.
But I was just hoping this might help :3
\[\large \frac{d^2}{dx^2}y(t) \qquad=\qquad \frac{d}{dx}y'(t) \qquad=\qquad \frac{y''(t)}{dx/dt} \qquad=\qquad \frac{\left(\dfrac{d^2y(t)}{dt^2}\right)}{dx/dt}\]

Answer 19

Okay, I think I might understand things now:

http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent_files/eq0030MP.gif

While dy/dx states that y has been derived with respect to x, dy/dx is still a function of t. Thus when trying to figure out the derivative of dy/dx with respect to x, we should treat dy/dx in the same way we treated y(t) when we wanted to find y(t)'s derivative with respect to x earlier. That is, take the function that you want to derive with respect to x and derive it with respect to t. Then, take the differentiator (which was x when we did the first derivative and is also x in the second derivative) and derive that with respect to t also. Finally, take the ratio of the two. You guys probably didn't see this but if you did, am I right?

Answer 20

Yah that sounds right! :)

I probably shouldn't have skipped this step, it seems like it made the most sense to you,\[\large \frac{\dfrac{d}{dt}\left(\color{royalblue}{\dfrac{dy/dt}{dx/dt}}\right)}{dx/dt} \qquad=\qquad \frac{\dfrac{d}{dt}\left(\color{royalblue}{\dfrac{dy}{dx}}\right)}{dx/dt}\]