Question

The temperature in Gavin’s oven is a sinusoidal function of time. Gavin sets his oven so that it has a maximum tem- perature of 300◦F and a minimum tempera- ture of 240◦ . Once the temperature hits 300◦ , it takes 20 minutes before it is 300◦ again. Gavin’s cake needs to be in the oven for 30 minutes at temperatures at or above 280 ◦. He puts the cake into the oven when it is at 270 and rising. How long will Gavin need to leave the cake in the oven?

Answer 1

It equals so divide 45 in 377

Answer 2

Let's begin by creating a sinusoidal function to reflect the circumstance. Let's pretend the cake is put in the oven at time zero to make our lives easy. Then there's the cosine function, which has the following characteristics:

The computation:

$\text{Amplitude} = \frac{(300 - 240) }{2} = 30\text{period = 20}\text{Horizontal Shift = h (we will solve for this)}\text{Vertical Offset} = 240 + 30 = 270$

As a result, we have the following function:

$T(t) = \text{amplitude} \text{ x } cos(\frac{2\pi t}{period} + h) + \text{Vertical Offset}$

$= 30 \text{ cos}(\frac{\pi t}{10} + h) + 270$

T denotes the temperature at t. We know the cake is baked at 270°F and that this is when t=0, therefore we can solve for h:

$270 = 30 \text{ cos(h)} + 270\\0 = \text{cos(h)}\\h = \frac{\pi }{2}$

We must subtract this shift from our graph in order to shift it right:

$T(t) = 30 \text{ cos}(\frac{\pi t}{10} - \frac{\pi }{2} ) + 270$

The oven temperature must now be greater than or equal to 280° for 30 minutes, according to the problem. To do so, we must first determine when the temperature is 280°. Set the value of the function to 280 and solve:

$280 = 30 \text{ cos}(\frac{\pi t}{10} - \frac{\pi }{2} ) + 270$

$\frac{1}{3} = cos(\frac{\pi t}{10} - \frac{\pi }{2} )$

$(\text{arccos}(\frac{1}{3} )+\frac{\pi }{2} )(\frac{10\pi }{\pi } ) = t$

$t = (\frac{10}{\pi } ) \text{ arccos}(\frac{1}{3} )+5$

We know that t is really: since the function is periodic.

$t = \frac{10}{\pi } \text{ arccos}(\frac{1}{3} ) + 5 + 20k$

where k is any non-decimal integer value. However, this equation only gives us the moment when the temperature reaches 280 degrees while it is falling. To obtain the temperatures as they arise, we must locate the symmetric point on the opposite side of the graph's "hill." Our real shift is: because we subtracted /2 from the input after multiplying it by the horizontal scaling factor.

$(\frac{\pi }{2} )(\frac{10}{\pi } ) = 5$

We then need the point where the falling temperature reaches 280 degrees (when k=0):

$t0 = (\frac{10}{\pi } )arccos(\frac{1}{3} ) + 5$ ≈ $8.9 seconds$

 

This suggests that it takes around 3.9 seconds for the temperature to reach 280 degrees after reaching 300 degrees, and it reaches 280 degrees 3.9 seconds before t=5. This yields a second-time equation:

tr = (10/π)arccos(1/3) - 2.8 + 20k

tc = (10/π)arccos(1/3) + 5 + 20k

As a result, the oven is preheated to 280 degrees:

$tr = (\frac{10}{\pi } )arccos(\frac{1}{3} ) - 2.8$ ≈$1 second$

Before the oven cools too much, we know there are 7.9 seconds of optimum temperature. How long does it take for the temperature to return to 280 degrees? When k=1 in tr, this happens.

tr = (10/π)arccos(1/3) - 2.8 + 20 ≈ 21 seconds or 21 - 7.9 = 13.1 seconds after it cools off too much.

We now have a pattern of 7.9 seconds of correct temperatures followed by 13.1 seconds of incorrect values. So, how many 7.9-second cycles does it take to reach 30 seconds?

$\frac{30}{7.9}$ ≈ $\text{3.8 cycles of 7.9 seconds}$

Okay, we've put in a lot of effort, but now it's time to put it all together. Here's how we did it:

1 sec | 7.9 sec | 13.1 sec | 7.9 sec | 13.1 sec | 7.9 sec | 13.1 sec | (7.9)(.8)

We can add these up to get our answer:

$1 + 7.9 + 13.1 + 7.9 + 13.1 + 7.9 + 13.1 + 6.3 = 70.32 \text{ seconds}$

Learn more about sinusoidal function here,

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