# Thread:

1. ## Need Help with some Trig for a model (graph)

I attached a crude drawing of the function I'm trying to graph. I'm doing a problem with the model $\frac{dT}{dt}=k(T-T_m)$, where $T_m$ is a function shown in the graph, but I need help with the trigonometry here. Note that the graph I drew is sloppy, but you should be able to get the idea. It's just a transformed sine function.

$T_m(t)=asin(bt+c)+d$

The amplitude is a=30, and it's been moved up 80 units, so d=80. Since I know that b divides the normal period $2\pi$ so that $\frac{2\pi}{b}=period$. The period appears to be 24, so I think the $b=\frac{\pi}{12}$. I'm kind of confused about what the phase shift should be though.
Attached Thumbnails

2. The phase shift is where the function starts. As in, with sin(x), sin(0)=0; it `starts' halfway along an up-stroke.

Basically (assuming you haven't touched the period), if your function starts half-way along an up-stroke, your phase shift will be zero. If its half-way along a down stroke, it will be $\pi$. If it is at the top of an up-stroke, it will be $\pi/2$, which is essentially cos(x), and if it is at the bottom of a down-stroke it will be $3\pi/2$.

Basically, draw a sine curve, and take a ruler. Put the ruler on the vertical (y-)axis and move it right until what is to the right of your ruler looks like the curve you want. the x-value at this point is what you want.

This will be scaled when you add the period; if the phase-shift (worked out as-above) is c, and the period is d, then the equation you want is $sin(d(t+c)) = sin(dt+dc) = sin(dt+c^{\prime})$.

3. Good work so far! Now the sine function you've got there would normally start out going up from its midpoint, which in your case is 80. However, your graph has it starting at its low point (I'm assuming). So, the entire graph has been shifted to the right for 1/4 of a period. What does that mean for the phase shift?

[EDIT]: Didn't see Swlabr's post.
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