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I just don't seem to be able to understand how to get the second derivative of a parametric equation. Can someone walk me through it?

Mathematics
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Okay, my parametric equation is x=3+3t^2, y+3t^2+t^3. to find dy/dx you divide (dy/dt)/(dx/dt). Which I can do no problem.
For the first derivative I got for 1-((3t^2)/(4t)). Which the online homework problem says is correct.
To find the second derivative I have to find d/dt (dy/dx). I separated the two parts and got 0 for the one, of course, and then used the quotient rule on the -3t^2/4t part. Is this correct?

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Other answers:

a paramentric eqution is also a vector equation: r(t) = right?
Yes
x=3+3t^2 , y=3t^2+t^3 x'=6t , y=6t+3t r'(t) = < 6t , 6t+3t^2 > which is what youve got for the tangent
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:
1 Attachment
a tiff? really?
you got that in a jpeg?
Lol, here it is in jpg
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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
Sorry, I starting typing the first, which is pretty easy when I was actually going for a second problem. But this is the spot I always get confused. With that dy/dx I would lose the one and use quotient rule to solve for -3t^2/4t, wouldn't I? But when I do that it says I got the problem wrong
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

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