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a paramentric eqution is also a vector equation:
r(t) = right?

Yes

Lol, I mixed two problems in what I explained, but yes, go on

r''(t) is just the next derivative.
r''(t) = <6, 6+6t> if im reading it right

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

bout 3/4 of the way down it goes into 2nd derivative

Right, but the question is asking for it in a certain form. Here is the question:

a tiff? really?

you got that in a jpeg?

Lol, here it is in jpg

I just sliced a portion of the window to get the picture

\[\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]

soo
dy/dx = 1+ t/2 stick this in the derivative thing for: 1/2
right?

\[\frac{d}{dt}(\frac{dy}{dx})\to \frac{d}{dt}(1+t/2)\]

Yeah, you're right. But what I have is a dy/dx = 1-3t^2/4t

or the right dy/dx .. lol

reduce it by t and you got -3t/4 which is simpler to derive

not reduce ..... simplify

Oh my Gosh!! I can't believe I wasn't seeing that! Thank you!

youre welcome