Tough Puzzle

[tedz0r]

here is a little problem solving quizz that u guys may find interesting...

A sine function is such that when u differentiate it, the resulting function has an amplitude twice the original function. If the original function passes through (PI/6,0) and the derivative passes through (7PI/6,0) find a possible sine function.

[zaza]

SHouldn't you be posting this in the "homework" section of the forums?

[tedz0r]

yea and on a different site aswell, this one's probably a little too advanced for u guys anyway

[krtxmrtz]

here is a little problem solving quizz that u guys may find interesting...

A sine function is such that when u differentiate it, the resulting function has an amplitude twice the original function. If the original function passes through (PI/6,0) and the derivative passes through (7PI/6,0) find a possible sine function.

If f(x) = A + Bsin(Cx + D) with A, B, C and D being constants can be admitted as a sine function, then it's elementary. Otherwise I see no solution... but it all depends on what you understand by a sine function.

[malik641]

I racked my brains out on this...I can't think of anything, but I have a question for you. Is this a Differential Equation problem? If so, I'm going to have to pull my notes from last semester.

[meboz]

i dont believe there is a solution to this problem

[meboz]

the problem reduces to a pair of simultaeneous equations as follows

0 = a*sin(pi/6) [from f(x)]
0 = 2a*cos(7*pi/6) [from f'(x)]

in that case, a is always 0, a very boring solution, correct me if im wrong please

[jemidiah]

Since it's been long enough that the homework should have been turned in by now...

Impossible.

If the derivative has an amplitude that is twice the original function's, then in the original function the sine function must have an argument who's derivative is 2. Since this is to be a regular sinusoidal function (you could conceivably be mean and make some very strange argument who's derivative is still 2), that means that the function must be like so: b(sin(cx+d)). c determines the period of the function, which is necessarily pi. Now, flash forward to the derivative.

It involves a cosine function who's period is also pi. The cosine returns back to it's position at y=1 ever pi that x changes, whereas the sine returns back to it's position at y=0 ever pi that x changes. Therefore, since the x-values in question are a multiple of pi away from each other, there is no way that cos(f(x)) and sin(f(x)) at those points could culminate in y-values that equal each other.

This is all negated if you are allowed to add a constant to the b(sin(cx+d)) equation, a+b(sin(cx+d)).

[krtxmrtz]

This is all negated if you are allowed to add a constant to the b(sin(cx+d)) equation, a+b(sin(cx+d)).

Right, that's what I said in my post above.

[jemidiah]

Yarp, 'twas.

I guess I just wanted to intuitively kill it, but it crept back up at the end because I was forgetting that constant, in which case it does become "elementary"