8. After you eat something containing sugar, the pH or acid level in your mouth changes. This can be modelled by the function below, where L is the pH level and n is the number of minutes that have elapsed since eating.
L(n)= 6- 20.4n/n^2 +36
a) What is the initial pH level before you start eating?
b) Determine the average rate of change of the pH level from 2 minutes to 3.5 minutes.
c1) The average rate of change of the pH level from 3.5 to 5 minutes is approximately -0.1282/minute. Compare this to the average rate of change you calculated in.
c2). What does this mean with respect to the given situation?
d) Estimate the instantaneous rate of change at 3 minutes. Explain what the value represents in this situation.
I assume you meant L(n)= 6 - 20.4n/(n^2 +36) If not, then redo the calculations below
(a) L(0) = 6
(b) (L(3.5)-L(2))/(3.5-2) =
(c) look at the graph. L is decreasing, but more slowly
(d) either use a small interval such as [3,3.1] or just take the derivative to get the exact value.
Hi can you explain to me how to do d and explain how you got b
To solve these questions using the given function, let's go through each question step by step.
a) What is the initial pH level before you start eating?
The initial pH level refers to the pH level before any time has elapsed since eating. In this case, we can find the initial pH level by substituting n = 0 into the function:
L(0) = 6 - (20.4 * 0) / (0^2 + 36)
L(0) = 6 - 0 / (0 + 36)
L(0) = 6 - 0 / 36
L(0) = 6
Therefore, the initial pH level before you start eating is 6.
b) Determine the average rate of change of the pH level from 2 minutes to 3.5 minutes.
To find the average rate of change, we need to calculate the difference in the pH levels at these two times and divide it by the difference in time:
Average rate of change = (Change in pH level) / (Change in time)
Let's calculate it using the given function:
L(2) = 6 - (20.4 * 2) / (2^2 + 36)
L(2) = 6 - 40.8 / 40
L(2) = 6 - 1.02 = 4.98
L(3.5) = 6 - (20.4 * 3.5) / (3.5^2 + 36)
L(3.5) = 6 - 71.4 / 50.75
L(3.5) = 6 - 1.41 = 4.59
Change in pH level = 4.59 - 4.98 = -0.39
Change in time = 3.5 - 2 = 1.5
Average rate of change = (-0.39) / (1.5) = -0.26
Therefore, the average rate of change of the pH level from 2 minutes to 3.5 minutes is approximately -0.26/minute.
c1) The average rate of change of the pH level from 3.5 to 5 minutes is approximately -0.1282/minute. Compare this to the average rate of change you calculated in b).
The average rate of change from 3.5 to 5 minutes, -0.1282/minute, is different from the average rate of change calculated in question b), which was -0.26/minute.
c2) What does this mean with respect to the given situation?
The different rates of change indicate that the pH level in the mouth is changing at a slower rate from 3.5 to 5 minutes compared to the rate of change from 2 to 3.5 minutes. This suggests that the pH level is stabilizing or becoming less acidic as time progresses.
d) Estimate the instantaneous rate of change at 3 minutes. Explain what the value represents in this situation.
To estimate the instantaneous rate of change at 3 minutes, we need to find the derivative of the function L(n) with respect to n and substitute n = 3 into the derivative equation.
L(n) = 6 - 20.4n / (n^2 + 36)
To find the derivative, we apply the power rule and quotient rule:
dL/dn = (20.4*(n^2 + 36) - (6 - 20.4n)*2n) / (n^2 + 36)^2
Now, substitute n = 3 into the derivative equation:
dL/dn | (n=3) = (20.4*(3^2 + 36) - (6 - 20.4*3)*2*3) / (3^2 + 36)^2
dL/dn | (n=3) = (20.4*(9 + 36) - (6 - 61.2)*6) / (9 + 36)^2
Simplifying the equation:
dL/dn | (n=3) = (20.4*45 - (-55.2)*6) / 45^2
dL/dn | (n=3) = (918 - (-331.2)) / 2025
dL/dn | (n=3) = 1249.2 / 2025
The estimated instantaneous rate of change at 3 minutes is approximately 0.617/minute.
In this situation, the positive value of the instantaneous rate of change indicates that the pH level is increasing at a faster rate at 3 minutes. This means that the mouth is becoming more acidic.