point) The pressure P (in pounds per square foot), in a pipe varies over time. Four times an hour, the pressure oscillates from a low of 40 to a high of 300 and then back to a low of 40. The pressure at time t = 0 is 40. Let the function P = f(t) denote the pressure in pipe at time t minutes.
Find a possible formula for the function P = f(t) described above.
mean = 340/2 = 170
amplitude = 300-170 = 130
min at t = 0 so
y = 170 - 130 cos(w t) which is 40 at t = 0
what is w?
when t = 15 min , w t = 2 pi
so
w 15 = 2 pi
w = (2/15) pi
so
y = 170 - 130 cos(2 pi t/15)
To find a possible formula for the function P = f(t) described above, let's analyze the given information.
We know that the pressure oscillates four times in an hour, which means the entire oscillation cycle takes 60 / 4 = 15 minutes.
We also know that the pressure oscillates from a low of 40 to a high of 300 and then back to a low of 40.
Based on this information, we can define the function f(t) as follows:
1. For the first 15 minutes, the pressure increases from a low of 40 to a high of 300 linearly. This can be represented by the equation:
P = mt + b
where m is the slope, b is the y-intercept, and t is the time in minutes.
Since the pressure starts at 40 and reaches 300 in 15 minutes, we can plug these values into the equation:
40 = m * 0 + b (at t = 0, P = 40)
300 = m * 15 + b (at t = 15, P = 300)
Solving these two equations simultaneously, we find:
b = 40
m = (300 - 40) / 15 = 20
Therefore, for the first 15 minutes, the equation becomes:
P = 20t + 40
2. For the next 15 minutes, the pressure decreases from a high of 300 back to a low of 40 linearly. This can be represented by the equation:
P = -mt + b
where -m is the slope (negative because pressure decreases), b is still the y-intercept, and t is the time in minutes.
Similar to the previous step, we can plug the values into the equation:
300 = -20 * 15 + b (at t = 30, P = 300)
40 = -20 * 0 + b (at t = 30, P = 40)
Solving these two equations simultaneously, we find:
b = 300 + (20 * 15) = 600
m = (40 - 600) / 15 = -40
Therefore, for the next 15 minutes, the equation becomes:
P = -40t + 600
3. We repeat steps 1 and 2 for the entire hour, resulting in the following formula:
P = 20t + 40 (0 ≤ t < 15)
P = -40t + 600 (15 ≤ t < 30)
P = 20t - 600 (30 ≤ t < 45)
P = -40t + 1200 (45 ≤ t < 60)
Notice that we alternate between positive and negative slopes to represent the oscillation between low and high pressures.
Therefore, a possible formula for the function P = f(t) is:
P = 20t + 40 for 0 ≤ t < 15
P = -40t + 600 for 15 ≤ t < 30
P = 20t - 600 for 30 ≤ t < 45
P = -40t + 1200 for 45 ≤ t < 60
To find a possible formula for the function P = f(t) described above, we can break it down into two parts: the oscillating part and the constant part.
1. Oscillating part:
The pressure oscillates between a low of 40 and a high of 300, and it goes through this cycle four times per hour. So, each cycle takes 60/4 = 15 minutes.
During each cycle, the pressure starts at the low of 40, increases to the high of 300, and then returns to the low of 40.
To represent this oscillating behavior, we can use the sine function, which oscillates between -1 and 1. By scaling and shifting the sine function, we can achieve the desired range (from 40 to 300) and time period (15 minutes).
Let's first scale the sine function to the desired range:
p(t) = A*sin(t)
The amplitude A represents half the range, so we need 300 - 40 = 260 / 2 = 130. Therefore, the formula now becomes:
p(t) = 130*sin(t)
Now, we need to shift the sine function to start at the low of 40 instead of 0. We can accomplish this by subtracting 40 from the formula:
p(t) = 130*sin(t) - 40
2. Constant part:
The pressure at time t = 0 is given as 40, which represents the constant component of the function. So, we simply add this constant component to the oscillating part:
P = p(t) + 40
Combining everything together, a possible formula for the function P = f(t) is:
P = 130*sin(t) - 40 + 40
Simplifying it further:
P = 130*sin(t)