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wcaprar
 3 years ago
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?
wcaprar
 3 years ago
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?

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wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, 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.

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0For the first derivative I got for 1((3t^2)/(4t)). Which the online homework problem says is correct.

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0To 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?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1a paramentric eqution is also a vector equation: r(t) = <x(t),y(t)> right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1x=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

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0Lol, I mixed two problems in what I explained, but yes, go on

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1r''(t) is just the next derivative. r''(t) = <6, 6+6t> if im reading it right

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1http://tutorial.math.lamar.edu/Classes/CalcII/ParaTangent.aspx might help

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1bout 3/4 of the way down it goes into 2nd derivative

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0Right, but the question is asking for it in a certain form. Here is the question:

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1you got that in a jpeg?

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0I just sliced a portion of the window to get the picture

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1soo dy/dx = 1+ t/2 stick this in the derivative thing for: 1/2 right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dt}(\frac{dy}{dx})\to \frac{d}{dt}(1+t/2)\]

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, you're right. But what I have is a dy/dx = 13t^2/4t

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1or the right dy/dx .. lol

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry, 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
 3 years ago
Best ResponseYou've already chosen the best response.1reduce it by t and you got 3t/4 which is simpler to derive

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.1not reduce ..... simplify

wcaprar
 3 years ago
Best ResponseYou've already chosen the best response.0Oh my Gosh!! I can't believe I wasn't seeing that! Thank you!
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