[ Follow Ups ] [ Post Followup ] [ Main Message Board ] [ FAQ ]

Posted by Brad Paul on October 19, 2002 at 09:30:02:

In Reply to: explanation please posted by Barry on October 18, 2002 at 22:22:53:

: [t (radians) such that 0 < t < (1/2)(pi) ]

: T(t) = (5 / x)[t + 2(cos t)]
: dT/dt = (5 / x)[1 – 2(sin t)]
: (5 / x)[1 – 2(sin t)] = 0
: 1 – 2(sin t) = 0
: 1 = 2(sin t)
: sin t = 1/2
: t = arcsin (1/2)
: t = (1/6)(pi) radians or 30 degrees
: Hence, (1/6)(pi) radians or 30 degrees.

: Could someone explain in detail how the "derivative" is arrived at. Thank you.
: I know the 1st derivative of (example) 3X^5 = 15x^4.
You have a function of t. In this function you have an x. How does x
change when you change t? Well it doesn't, x is just a symbol that
represents some number you don't know. And this number doesn't depend
on t. Now when you ask the question: How does T(t) change with respect
to t? You know that you can treat x just like a constant. This is why
what you have done is correct.

If your question is why is (d/dt)(cos(t))=-sin(t). Then you can use
everyones favored derivative equation:

Dx xn=n xn-1

to show it. How you wonder? Use a series representation of sin(t) and
cos(t). I will not do it here because it is easy to do but hard to
format. Just write out the first few terms of the series for cos(t)
then use everyones favored derivative equation and you will start to
see that you are generating the terms for the -sin(t) series.

Name:
E-Mail:

Subject: