# 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.

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