- #1
kasse
- 384
- 1
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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Schrodinger's Dog said: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]
Schrodinger's Dog said:a) the bottom line is wrong
should be [tex] \frac {cos(t)-1}{(1-cos(t))^2}[/tex]
kasse said:OK, but this is the right answer, just that I have to simplify?
kasse said:The answer is supposed to be -1/a(1-(cos(t))^2)
Whatever I do, I don't get an answer including a.
cristo said:Schrodinger's Dog is correct. There appears to be a factor of y in the denominator of the middle term. This is a typo.
kasse said: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?
cristo said: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 d2y/dx2. 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!
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