anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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.
anonymous
  • anonymous
For the first derivative I got for 1-((3t^2)/(4t)). Which the online homework problem says is correct.
anonymous
  • anonymous
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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amistre64
  • amistre64
a paramentric eqution is also a vector equation: r(t) = right?
anonymous
  • anonymous
Yes
amistre64
  • amistre64
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
anonymous
  • anonymous
Lol, I mixed two problems in what I explained, but yes, go on
amistre64
  • amistre64
r''(t) is just the next derivative. r''(t) = <6, 6+6t> if im reading it right
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx might help
amistre64
  • amistre64
bout 3/4 of the way down it goes into 2nd derivative
anonymous
  • anonymous
Right, but the question is asking for it in a certain form. Here is the question:
1 Attachment
amistre64
  • amistre64
a tiff? really?
amistre64
  • amistre64
you got that in a jpeg?
anonymous
  • anonymous
Lol, here it is in jpg
1 Attachment
anonymous
  • anonymous
I just sliced a portion of the window to get the picture
amistre64
  • amistre64
\[\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]
amistre64
  • amistre64
soo dy/dx = 1+ t/2 stick this in the derivative thing for: 1/2 right?
amistre64
  • amistre64
\[\frac{d}{dt}(\frac{dy}{dx})\to \frac{d}{dt}(1+t/2)\]
anonymous
  • anonymous
Yeah, you're right. But what I have is a dy/dx = 1-3t^2/4t
amistre64
  • amistre64
or the right dy/dx .. lol
anonymous
  • anonymous
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
amistre64
  • amistre64
reduce it by t and you got -3t/4 which is simpler to derive
amistre64
  • amistre64
not reduce ..... simplify
anonymous
  • anonymous
Oh my Gosh!! I can't believe I wasn't seeing that! Thank you!
amistre64
  • amistre64
youre welcome

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