Word problem involving sinusoidal model

1. Jul 14, 2014

maxpan

1. The problem statement, all variables and given/known data

2. Relevant equations

$$y=30\sin(\dfrac{2\pi}{20}t)+270$$

General principal solutions:

$$t=\left(\dfrac{\arcsin(\dfrac{1}{3}) 20}{2\pi}\right)+20k, k\in \mathbb{Z}$$

$$t=1.08173+20k$$

General symmetry solutions:

$$t=\left(-\dfrac{(\arcsin(\dfrac{1}{3})-\pi)20}{2\pi}\right)+20k, k\in \mathbb{Z}$$

$$t=8.91827+20k$$

3. The attempt at a solution

What I did was I divived 30 by the difference between these two values of time to find how many such segments I need to accumulate 30 min of 280 degrees:

$$\dfrac{30}{8.91827-1.08173}=3.828$$

And then I can find the answer with a bunch of manual work, involving referring to the graph, but I just feel like this is a very ineffective approach.

Last edited: Jul 14, 2014
2. Jul 14, 2014

Simon Bridge

The problem statement did not come through.

3. Jul 14, 2014

maxpan

Thanks. Fixed it.

4. Jul 14, 2014

Simon Bridge

... this is the equation for the temperature of the oven right?
Did you try sketching this and drawing a horizontal line for the desired temperature?

So your strategy is to find out how much time T1 in one cycle y>280F, divide the cooking time by T1 to find out how many cycles you need, then multiply that by the period.

I think the main thing you have to watch for is the last fraction of a period... that what you wanted to refer to the graph for?

Anyway - that's the approach all right.

5. Jul 14, 2014

maxpan

Yes, well, I find where the 4th segment ends and then subtract (4-3.828) times segment from it.

$$t=8.91827+20*3$$ gives me coordinate where 4th segment ends and then I do the following:
$$t=(8.91827+20*3)-((4-3.828)*8.91827)$$

So, yes, I just wanted to know if there is a better way to solve it or maybe it is supposed to be such a graph-involving problem. Well, maybe with calculus though, but I'm not yet at that level.

6. Jul 14, 2014

maxpan

Well, actually not that much of "manual" work, just a less algebraic approach.

7. Jul 14, 2014

Simon Bridge

Well you could do it algebraically - but the graph approach is easier to think about.
AFAIK there is no fancy magic-wand approach - pretty much all alternatives involve headaches.