Turning points question

**JuriamboXD** · 08-04-12, 01:18 PM

Any assistance with this question would be greatly appreciated as I really need some help (i'm not particularly good at maths and my lecture notes don't cover this topic particularly well)

For the function x(t) = Ae^(-(alpha)t)sin((omega)t), where A, (alpha) and (omega) are constants, find an expression that is obeyed at the turning points of x(t). Are they periodic?

Thank you

**AlphaNumeric** · 08-04-12, 01:24 PM

The definition of a turning point is when the first derivative vanishes. So what's the derivative of the expression you just gave.

And to help you write everything here's the LaTeX code for the expression, $x(t) = Ae^ {-\alpha t}\sin(\omega t)$ . Quote this post and then copy and paste the various bits around to write out the expression you get for the derivative.

**JuriamboXD** · 08-04-12, 01:40 PM

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**JuriamboXD** · 08-04-12, 01:42 PM

Originally Posted by AlphaNumeric

The definition of a turning point is when the first derivative vanishes. So what's the derivative of the expression you just gave.

And to help you write everything here's the LaTeX code for the expression, $x(t) = Ae^ {-\alpha t}\sin(\omega t)$ . Quote this post and then copy and paste the various bits around to write out the expression you get for the derivative.

this is my first time using this forum, and thanks for such a quick response! I'm not entirely sure how to derive that expression, would I be using the chain rule or product rule or both?

**AlphaNumeric** · 08-04-12, 01:44 PM

The expression I gave was for the function you gave in your initial post. To compute the derivative of it, which is what you need to do to answer the question, will involve both the product (otherwise known as Leibnitz) rule and the chain rule (unless you know the derivative of $e^{-\alpha t}$ and $\sin \omega t$ off the top of your head).

**JuriamboXD** · 08-04-12, 01:57 PM

Originally Posted by AlphaNumeric

The expression I gave was for the function you gave in your initial post. To compute the derivative of it, which is what you need to do to answer the question, will involve both the product (otherwise known as Leibnitz) rule and the chain rule (unless you know the derivative of $e^{-\alpha t}$ and $\sin \omega t$ off the top of your head).

Is the derivative, Ae^(-(alpha)t)((omega)cos((omega)t) - (alpha)sin((omega)t))? (sorry I don't understand how to use LaTeX code yet) If that is right, then how do I find out if it is periodic or not?

**JuriamboXD** · 08-04-12, 02:14 PM

Turning points are where the derivative = 0 aren't they? But I don't understand how I use that in my answer/ what I do once I put that derivative (assuming it's right) equal to zero

**prometheus** · 08-04-12, 02:47 PM

The way I would handle this problem is to do the following:
1) Write your expression you need to find the turning points of. Call this $f(t)$
2) Compute the derivative of this expression, $\frac{df}{dt}$ .
3) Set this equal to zero, $\frac{df}{dt} = 0$
4) Solve $\frac{df}{dt} = 0$ for t. In general there will be more than one solution to this. If there are an infinite number of solutions the function may be periodic, but I would guess you need to show that the separation of the turning points is constant as well.
5) plug this value(s) of t back into f(t) to find the value of the expression at the turning point.
6) Job done.

**AlphaNumeric** · 08-04-12, 03:14 PM

Originally Posted by JuriamboXD

Is the derivative, Ae^(-(alpha)t)((omega)cos((omega)t) - (alpha)sin((omega)t))? (sorry I don't understand how to use LaTeX code yet) If that is right, then how do I find out if it is periodic or not?

The result is $Ae^{-\alpha t}(\omega \cos \omega t - \alpha \sin \omega t)$ , yes.

So this is x'(t), so if x'(t) = 0 then we have $Ae^{-\alpha t}(\omega \cos \omega t - \alpha \sin \omega t) = 0$ . So the question is whether the solutions to this are periodic. The value of x'(t) is not periodic because $e^{-\alpha t}$ decays, it isn't periodic. However, the values of t which satisfy x'(t) = 0 ARE periodic. This is because the $e^{-\alpha t}$ term doesn't affect where the roots are. The roots satisfy x'(t) = 0, so $Ae^{-\alpha t}(\omega \cos \omega t - \alpha \sin \omega t) = 0$ . We know A isn't zero so it can be dropped. We also know $e^{-\alpha t}$ isn't zero and it too is an overall factor so now we're down to $\omega \cos \omega t - \alpha \sin \omega t = 0$ . You can state that this is periodic immediately because you know $\cos \omega t$ and $\sin \omega t$ repeat every $\frac{2 \pi}{\omega}$ radians but you can rearrange the expression to get $\omega \cos \omega t = \alpha \sin \omega t$ so we get $\frac{\omega}{\alpha} =\tan \omega t$ . We know tan is periodic so the values which solve that are periodic, just add pi to any root to get another root. Job done

If you need anything I've said elaborated on just say which bit you don't get and I or someone else will explain it.