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I get the first derivative correct, but what's wrong with my attempt to find the second derivative?
http://www.badongo.com/pic/421533
http://www.badongo.com/pic/421533
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a) the bottom line is wrong
should be [tex] \frac {cos(t)-1}{(1-cos(t))^2}[/tex]
and b) you need to further simplify to get the answer.
EDIT:
What is [tex]1-cos^2 (39)[/tex] what is [tex](1-cos(39))^2[/tex]
Thus.
[tex](1-cos(t))(1+cos(t)\not= 1-cos{^2}(t)[/tex]
OK, but this is the right answer, just that I have to simplify?a) the bottom line is wrong
should be [tex] \frac {cos(t)-1}{(1-cos(t))^2}[/tex]
Yep, expand the denominator, and then simplify from there. Your answer is correct apart from the obvious mistake, but it's often better to present the most simple form.OK, but this is the right answer, just that I have to simplify?
If you can get this I'd be surprisedThe answer is supposed to be -1/a(1-(cos(t))^2)
Whatever I do, I don't get an answer including a.
Just goes to show everyones human, even the editors and authors of text books.Schrodinger's Dog is correct. There appears to be a factor of y in the denominator of the middle term. This is a typo.
No, this is not correct. The red part above is dy'/dt. Plug this into the formula for the second derivative with respect to x.How about this one:
x=(cos t)^3
y=(sin t)^3
dy/dx = -tan t
dy'/dt = -1/(cos t)^2
Right?
I can simply derivate the dy/dx function even if it's a function of t and not x?
That's my fault I could not read his text book or the question properly, I think I need an eye test, apologies.Yes, sorry, I wasn't really looking at this properly the first time, and just glanced at Schrodinger's Dog's attempt and it looked right; however it is not. What he was doing was taking [tex]\frac{d^2y/dt^2}{d^2x/dt^2}[/tex], however this is of course not correct, since this does not equal d^{2}y/dx^{2}. You should follow your textbook. The correct way to approach the second derivative is to use the chain rule; letting y'=dy/dx: [tex]\frac{dy'}{dt}=\frac{dy'}{dx}\frac{dx}{dt}[/tex] which then gives the formula for the second derivative of y wrt x:[tex]\frac{d^2y}{dx^2}=\frac{dy'/dt}{dx/dt}[/tex].
Sorry for advising incorrectly!