Precalc question

Question

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

Answer's 67.572141357 minutes, but I don't know how to do it

Answer 1

Hey Flora,

Larry probably says the same thing in his video but my sound is not working so I couldnt check. Anyways, here is another explanation.

To start, let's create the sinusoidal function that represents the situation. To make our lives easier let's say that at time zero the cake is put in the oven. Then we have a cosine function with the following properties:

amplitude = (300 - 240) / 2 = 30

period = 20

Horizontal Shift = h (we will solve for this)

Vertical Offset = 240 + 30 = 270

This gives us the function:

T(t) = amplitude * cos(2πt/period + h) + Vertical Offset

= 30 cos(π t/10 + h) + 270

where T is the temperature at time t. We know that the cake is put in the oven at 270 degrees, and we said this is when t=0 so we can solve for h:

270 = 30 cos(h) + 270

0 = cos(h)

h = π/2

Since we want to shift our graph right, we must subtract this shift:

T(t) = 30 cos(π t/10 - π/2) + 270

Now the problem says the oven needs to be greater than or equal to 280° for 30 minutes. To find this we first need to figure out when it is 280°. Set the function equal to 280 and solve:

280 = 30 cos(π t/10 - π/2) + 270

1/3 = cos(π t/10 - π/2)

(arccos(1/3)+π/2)(10/π) = t

t = (10/π)arccos(1/3)+5

Since the function is periodic we know that t is really:

t = (10/π)arccos(1/3) + 5 + 20k

where k is any integer (non-decimal) value. However, this equation only gets us the time when it hits 280 degrees while the temperature is falling. To get the temperatures when it is rising, we need to find the symmetric point on the other side of the "hill" of the graph. Since we subtracted π/2 from the input after we multiplied it by the horizontal scaling factor, our actual shift is:

(π/2)(10/π) = 5

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

t₀ = (10/π)arccos(1/3) + 5 ≈ 8.9 seconds

This means that temperature is about 3.9 seconds after hitting 300 degrees, so it hits 280 degrees when the temperature is rising 3.9 seconds before t=5. This gives us a second equation for time:

t_r = (10/π)arccos(1/3) - 2.8 + 20k

t_{c =} (10/π)arccos(1/3) + 5 + 20k

So the oven first hits 280 degrees:

t_r = (10/π)arccos(1/3) - 2.8 ≈ 1 seconds

We know then there are 7.9 seconds of appropriate temperature before the oven cools too much. We now want to know how long it takes to warm up to 280 degrees again. This occurs when k=1 in t_r.

t_r = (10/π)arccos(1/3) - 2.8 + 20 ≈ 21 seconds

or 21 - 7.9 = 13.1 seconds after it cools off too much.

Now we have the pattern where there are 7.9 seconds of proper temperatures followed by 13.1 seconds of improper temperatures. So how many cycles of 7.9 seconds does it take to get to 30 seconds?

30 / 7.9 ≈ 3.8 cycles of 7.9 seconds

Ok, this has been a lot of work but now we get to put this all together. Here is our pattern:

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 seconds

To get the answer you were given you have to store more than 1 decimal place during your calculations (which I didn't do to help simplify the explanation). Hope this helps!

-Nathan

Answer 2