Is There a Simpler Way to Handle the Pi in Finding a Derivative?

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Discussion Overview

The discussion revolves around finding the derivative of the function y=10*sin(pi*(0.01x - 2.00t)). Participants explore different methods of handling the constant pi in the context of differentiation, specifically focusing on the application of the chain rule and the treatment of pi as a multiplier.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about whether it is correct to multiply pi into the parentheses when finding the derivative.
  • Another participant confirms that multiplying pi out is valid, referencing the distributive property.
  • A different participant suggests that treating pi as a multiplier is simpler and explains the application of the chain rule in this context.
  • Participants discuss the equivalence of different forms of the cosine function resulting from the derivative calculations.

Areas of Agreement / Disagreement

There is no clear consensus on the preferred method of handling pi, as participants present differing views on whether to multiply it out or treat it as a multiplier. The discussion remains unresolved regarding the simplicity of each approach.

Contextual Notes

Participants do not fully resolve the implications of their methods on the final derivative form, and there are varying interpretations of how to apply the chain rule with respect to pi.

paul11273
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I am having a conflict with two different ways of finding a derivative.
Here is the function:
y=10*sinpi(.01x-2.00t)

Yes, that pi is after sin, but not in the paranthesis. This is how the prof gave it to us. This may be my problem, how I am treating the pi. I figure it was factored out of the parenthesis. So, to find the partial derivative WRT t by hand I do this:

y=10*sin(.01pi*x - 2.00pi*t) I multiplied the pi into the ()
dy/dt = -2.00pi*10*cos(.01pi*x - 2.00pi*t) used the chain rule
dy/dt = -20pi*cos(.01pi*x - 2.00pi*t) final result

That is my result. I check this in Matlab by entering the following:

>> syms x t
>> diff(10*sin(pi*.01*x-pi*2*t),t)
ans =
-20*cos(1/100*pi*x-2*pi*t)*pi

So with that I am happy. Now the tricky question.
If I enter this same thing to my TI-89, I get:
-62.8319*cos(2pi*t - .031416*x)

Now...it just hit me that you can transpose the items inside the paranthesis of cosine, and it is the same result. Ok, duh. I don't want to delete everything I just typed. My next question...

Am I treating the pi correctly to begin with? Is it correct to multiply it into the () like that? If not, what should I do with it? Is there an easier way?
Thanks.
 
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You can certainly multiply it out - it's just the distributive property of numbers.
 
However, it is simpler just to treat [itex]\pi[/itex] as a multiplier:

The derivative of sin x is cos x so the derivative of [itex]sin(\pi x)[/itex] is [itex]cos(\pi x)[/itex] times the derivative of [itex]\pi x[/itex] (that's the "chain" rule) which is just [itex]\pi[/itex]: the derivative of [itex]sin(\pi x)[/itex] is [itex]\pi cos(\pi x)[/itex].

The partial derivative of [itex]sin(\pi(0.01x- 2.00t))[/itex], with respect to t, is just that times the derivative of 0.01x- 2.00t with respect to t: [itex]-2.00\pi cos(\pi (0.01x- 2.00t)[/itex].

Finally, the derivative of [itex]sin(\pi(0.01x- 2.00t))[/itex] is, of course, just 10 times that: [itex]-20.00\pi cos(\pi(0.01x-2.00t)[/itex].
 
Thanks

Thank you both for responding.
I see what Halls has done, and yes, I think that is simpler. It just didn't occur to me to put pi into the () one time, instead of multiplying to each term.
Thanks.
 

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