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wcaprar
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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?
 2 years ago
 2 years ago
wcaprar Group Title
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?
 2 years ago
 2 years ago

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wcaprar Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

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

wcaprar Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
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
 2 years ago

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

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

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
a tiff? really?
 2 years ago

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

wcaprar Group TitleBest ResponseYou've already chosen the best response.0
Lol, here it is in jpg
 2 years ago

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

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

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

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

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
or the right dy/dx .. lol
 2 years ago

wcaprar Group TitleBest ResponseYou've already chosen the best response.0
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
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
reduce it by t and you got 3t/4 which is simpler to derive
 2 years ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
not reduce ..... simplify
 2 years ago

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

amistre64 Group TitleBest ResponseYou've already chosen the best response.1
youre welcome
 2 years ago
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