Question

Find the 2nd derivative. Parametric curves.

Answer 1

Find.\[\frac{ d^2y }{ dx^2 }\]
\[x=e^t, y=te^{-1}\]

Answer 2

So, far I have: \[x' = e^t, y' = e^{-t}(1-t), \frac{ dy }{ dx }=(1-t)e^{-2t}\]

Answer 3

\[\frac{ d^2y }{ dx^2 }=\frac{ \frac{ d }{ dt }*(y') }{ x' }\]

Answer 4

what is d/dt?

Answer 5

try writing dy/dx in terms of x and y

Answer 6

Did you solve for t in one of the two equation and plugged it to the other equation?

Answer 7

Like what you used to do in systems of equations.

Answer 8

You want to get rid of the t-variable here.

Answer 9

for d/dt?

Answer 10

You want to get your equation to look like y=f(x). To do that you need to combine both equations x=x(t) and y=y(t)

Answer 11

I have no clue where the d/dt came from

Answer 12

Any ideas @wio ?

Answer 13

I think that \[ \Large
\frac{d^2y}{dx^2} = \frac{d\left(\frac{dy}{dx}\right)}{dx} = \frac{\frac{d\left(\frac{dy}{dx}\right)}{dt}}{\frac{dx}{dt}}
\]

Answer 14

We already know \[\Large
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]

Answer 15

So, I think we just need to replace \(y\) with \(\frac{dy}{dx}\).

Answer 16

The source of the d/dt equation

Answer 17

Well, that is what I was explaining.

Answer 18

yea but what do we plug in for d/dt?

Answer 19

\[
\frac{d(\ldots)}{dt}
\]Is the derivative of \(\ldots\) with respect to \(t\).
Just like \(\frac{dy}{dt}\).

Answer 20

It's not a value, it's an operation you perform.

Answer 21

\[\frac{dy}{dx} = z = (1-t)e^{-2t}\]

then
\[\frac{dz}{dx}= \frac{dz}{dt} \times \frac{dt}{dx}\]

Answer 22

Can you differentiate \(dy/dx\) with respect to \(t\)?
That is what \(\frac{d(dy/dx)}{dt}\) means.

Answer 23

Would I need to differentiate \[(1-t)e^{-2t}\] ?

Answer 24

Yes

Answer 25

\[\frac{d^2 y}{dx^2} = \frac{d}{dt}((1-t)e^{-2t}) \times (x')^{-1}\]

Answer 26

How would I simplify \[(-1)e^{-2t}+(1-t)(-2e^{-2t})\]?

Answer 27

re-write it again in terms of x and y

\[x=e^{-t} => x^2=e^{-2t}\\x-y=(1-t)e^{-t}\]

Answer 28

\[x(x-y)=(1-t)e^{-2t}\]

Answer 29

use those

Answer 30

Did you divide by \(x'\) yet?

Answer 31

not yet, i am attempting to simplify as previously posted first

Answer 32

Factor out the \(e^{-2t}\)

Answer 33

Okay, got it, the 2nd derivative is \[(2t-3)e^{-3t}\]

Answer 34

Thanks all :)