## Derivation of parametric function

I get the first derivative correct, but what's wrong with my attempt to find the second derivative?

http://www.badongo.com/pic/421533
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 a) the bottom line is wrong should be $$\frac {cos(t)-1}{(1-cos(t))^2}$$ and b) you need to further simplify to get the answer. EDIT: What is $$1-cos^2 (39)$$ what is $$(1-cos(39))^2$$ Thus. $$(1-cos(t))(1+cos(t))\not= 1-cos{^2}(t)$$
 Mentor You have a mistake in the denominator: $$(1-cost)^2 \not=1-cos^2t$$

## Derivation of parametric function

 Quote by Schrodinger's Dog a) the bottom line is wrong should be $$\frac {cos(t)-1}{(1-cos(t))^2}$$ and b) you need to further simplify to get the answer. EDIT: What is $$1-cos^2 (39)$$ what is $$(1-cos(39))^2$$ Thus. $$(1-cos(t))(1+cos(t)\not= 1-cos{^2}(t)$$

The answer is supposed to be -1/a(1-(cos(t))^2)
Whatever I do, I don't get an answer including a.

 Quote by Schrodinger's Dog a) the bottom line is wrong should be $$\frac {cos(t)-1}{(1-cos(t))^2}$$
OK, but this is the right answer, just that I have to simplify?

 Quote by kasse OK, but this is the right answer, just that I have to simplify?
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.

 Quote by kasse The answer is supposed to be -1/a(1-(cos(t))^2) Whatever I do, I don't get an answer including a.
If you can get this I'd be surprised

Your first answer is better than this.

$$\frac{cos(t)-1}{(1-cos(t))^2}$$ correct.

the simplified answer is:-

$$\frac {1}{(cos(t)-1)}$$

Can you work out how you would get this simplification from expanding the denominator in your first answer? -1 as the numerator, is correct but it can be made even simpler.
 Well, my book says: second derivative = ((cos t)(1-cos t) - (sin t)(sin t)) / ((1-cos t)^2*a(1-cos t)) = -1/(a(1-cos t)^2) Heres the page http://www.badongo.com/pic/422090 (example 6) Mystic...
 OK I'll check my result with someone else, but I'm pretty sure on this one. a is a constant and since 3 sin(29)/3 cos(29) is the same as saying sin/cos I don't see why it belongs in the second order derivative. Try integrating the result of your first answer? $$\frac {cos(t)-1}{(1-cos(t))^2}$$ What do you end up with? I checked it on a maths web site and:- $$\frac{d}{dx} \frac {sin (t)}{1-cos(t)} = \frac {1}{(cos(t)-1)}dx$$
 Mentor Schrodinger's Dog is correct. There appears to be a factor of y in the denominator of the middle term. This is a typo.

 Quote by cristo Schrodinger's Dog is correct. There appears to be a factor of y in the denominator of the middle term. This is a typo.
Just goes to show everyones human, even the editors and authors of text books.

I also asked someone else where I work, since I work in a med physics dept, there's usally someone around who still remembers doing calculus without breaking into a cold sweat and they said I was right.
 Cool, I beat the textbook!
 Well kind of, but you made a typo of your own, you just needed to simplify after expanding the denominator (1-cos(t))^2 which is (1-cos(t))(1+cos(t)), of course you can check this by multiplying. $$\frac {cos(t)-1}{(1-cos(t))^2} => \frac {cos(t)-1}{(1-cos(t))(1+cos(t))} => \frac {-1}{1-cos(t)} => \frac {1}{cos(t)-1}$$
 How about this one: x=(cos t)^3 y=(sin t)^3 dy/dx = -tan t (dy/dx)' = -1/(cos t)^2 Right? I can simply derivate the dy/dx function even if it's a function of t and not x?
 I think the book is right after all. Formula nr. (5) is correct, and that's why you also have to multiply with a(1-cos t) in the denominator, since dx/dt = a(1-cos t)
 Mentor 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 $$\frac{d^2y/dt^2}{d^2x/dt^2}$$, 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: $$\frac{dy'}{dt}=\frac{dy'}{dx}\frac{dx}{dt}$$ which then gives the formula for the second derivative of y wrt x:$$\frac{d^2y}{dx^2}=\frac{dy'/dt}{dx/dt}$$. Sorry for advising incorrectly!

Mentor
 Quote by kasse 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?
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.

 Quote by cristo 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 $$\frac{d^2y/dt^2}{d^2x/dt^2}$$, 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: $$\frac{dy'}{dt}=\frac{dy'}{dx}\frac{dx}{dt}$$ which then gives the formula for the second derivative of y wrt x:$$\frac{d^2y}{dx^2}=\frac{dy'/dt}{dx/dt}$$. Sorry for advising incorrectly!
That's my fault I could not read his text book or the question properly, I think I need an eye test, apologies.

also I simply typed the equation in the first part in and got the first and second derivative without a. Which is my answer, but not it seems what the question is asking.

In other words since the first derivative simplified is

$$\frac {a(sin-(t))}{a(1-cos(t))} = \frac{sin-(t)}{1-cos(t)}$$

I simply did a derivation of the second part instead of the first.

In other words my answer is the answer you get if you ignore the constant, which of course is not what the question was asking but then I misread it completely anyway. Like I say eye test time

in real life if I had to derive this particular one.

$$\frac{d^2y}{dx^2} \frac{a(cos(x))}{a(sin(x))}$$ as it's the same as saying $$\frac{a(2)}{a(4)} = \frac{2}{4}$$ if you see what I mean.

I'd ignore the constant a. But this aint real life and that's not what the question wants you to do. Oops apologies.
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