# So Lost ? Derivative of Exponential and Trig Functions.

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• May 5th 2013, 02:50 PM
elittlewood
So Lost ? Derivative of Exponential and Trig Functions.
If y=Acos(kt)+Bsin(kt) , where A, B and k are constants, show that d^2y/dx^2+K^2y=0

I just have no idea where to go with the question.

Help would really be appreciated :)
• May 5th 2013, 02:59 PM
ManosG
Re: So Lost ? Derivative of Exponential and Trig Functions.
You must take the second derivative of y. Can you do that or you want help?
• May 5th 2013, 02:59 PM
emccormick
Re: So Lost ? Derivative of Exponential and Trig Functions.
Your sine and cosine are cyclical and $\frac{dy}{dx}[g(x)+f(x)]=\frac{dy}{dx}g(x)+\frac{dy}{dx}f(x)$, so you can break $\frac{d^2y}{dx^2}+K^2y=0$ up, do the first and second derivatives, and see where it goes.

And by break up, I mean the y=Acos(kt)+Bsin(kt) part...
• May 5th 2013, 03:28 PM
elittlewood
Re: So Lost ? Derivative of Exponential and Trig Functions.
Okay ... Im going to take the derivatives and see what i get ...

i ended up with... y''=-AK2cos(Kt)-BK2sin(Kt)

Is this right?
• May 5th 2013, 03:53 PM
elittlewood
Re: So Lost ? Derivative of Exponential and Trig Functions.
I don't see how your supposed to deal with the other part. ...
• May 5th 2013, 04:10 PM
Plato
Re: So Lost ? Derivative of Exponential and Trig Functions.
Quote:

Originally Posted by elittlewood
If y=Acos(kt)+Bsin(kt) , where A, B and k are constants, show that d^2y/dx^2+K^2y=0

I just have no idea where to go with the question.

You wrote $y=A\cos(k{\color{red}t})+B\sin(k{\color{red}t})$

If that is correct then $\frac{dy}{dx}=0$ so $\frac{d^2y}{dx^2}=0$.

Did you post it correctly?
• May 5th 2013, 04:45 PM
elittlewood
Re: So Lost ? Derivative of Exponential and Trig Functions.
Yes i did post it correctly.... i don't know what you are supposed to do to prove that d^2y/Dx^2 + k^2y = 0 ... ?
• May 5th 2013, 04:56 PM
Plato
Re: So Lost ? Derivative of Exponential and Trig Functions.
Quote:

Originally Posted by elittlewood
Yes i did post it correctly.... i don't know what you are supposed to do to prove that d^2y/Dx^2 + k^2y = 0 ... ?

Well my goodness, if there is no $x$ in the definition then the derivative $\frac{d}{dx}$ is the change in $x$ and that is zero.
This is a trick question..
• May 5th 2013, 05:00 PM
ManosG
Re: So Lost ? Derivative of Exponential and Trig Functions.
you take the second derivative of y with respect to x or t (it depends on what is your variable) $\frac{d^2 y}{d x^2} + k^2 y = -Ak^2 cos(kx) - Bk^2 sin(kx) + Ak^2 sin(kx) + Bk^2 cos(kx) = 0$
• May 5th 2013, 05:05 PM
Plato
Re: So Lost ? Derivative of Exponential and Trig Functions.
Quote:

Originally Posted by ManosG
you take the second derivative of y with respect to x or t (it depends on what is your variable) $\frac{d^2 y}{d x^2} + k^2 y = -Ak^2 cos(kx) - Bk^2 sin(kx) + Ak^2 sin(kx) + Bk^2 cos(kx) = 0$

• May 5th 2013, 05:38 PM
elittlewood
Re: So Lost ? Derivative of Exponential and Trig Functions.
I determined the second derivative... but again ... i don't know what to do with the x as brought up by Plato .
• May 5th 2013, 05:43 PM
emccormick
Re: So Lost ? Derivative of Exponential and Trig Functions.
Here is the thing that I still see here:

If $y=A\cos(kt)+B\sin(kt)$ , where A, B and k are constants, show that $\frac{d^2y}{dx^2}+K^2y=0$
OK: $\frac{d^2}{dx^2}y+K^2y=0$ Thanks to y not being a function of x, the derivative zeros that. So then you still have:

$K^2\left[A\cos(kt)+B\sin(kt)\right]=0$ to deal with. It really does not make sense as a question unless there is some part missing that makes the solution something along the lines of what ManosG posted. Without a derivative to cancel out that part, it is simply not true that it = 0.
• May 5th 2013, 05:48 PM
Plato
o
Quote:

Originally Posted by elittlewood
I determined the second derivative... but again ... i don't know what to do with the x as brought up by Plato .

If the original posting is correct, then there is no x in the definition.
Thus, the derivative with respect to x is zero.
As I said before, it is a trick question.

What don't you get about that?
• May 5th 2013, 06:06 PM
emccormick
Re: So Lost ? Derivative of Exponential and Trig Functions.
What is the source of the question?

I ask because if something can be either true or false it generally starts with "If possible...." This one just says "show that" which usually means it is true and you are doing a proof. However, the way it has been stated seems to not allow for it to be true, so not a proof. Therefore, there must be a problem somewhere.

The only two places I see a possible problem are either in the source or how it has been copied. If it is the source, not much can be done other than to complain or assume they meant to write it differently. If it is how it is copied, all it would take would be something like a notation for the set of problems this is on, a missing f(t) type statement, or just those words, "If possible..."
• May 5th 2013, 06:10 PM
ManosG
Re: So Lost ? Derivative of Exponential and Trig Functions.
Yes Plato, you may have a point, but the exercise does not say "find the t for which d^2y/dx^2+K^2y=0 ", it says "show that d^2y/dx^2+K^2y=0 ", so it makes no sense the "K^2y=0". I believe, he has written false the equation and the right answer is to take the second derivative (Happy)(Happy)
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last