3. The efficiency of an industrial pump decreases by a fixed percentage each year. The percentage decrease each year is with respect to the efficiency of the pump at the start of each year. Over each four year period, the efficiency decreases by 80%. Let us write e = f(t) so that the efficiency e of the pump, as a percentage of it’s initial efficiency, is a function of the number of years t since the pump has been in operation.

(a) Make a table of values of the function for changes in t of 4, starting from t = 0.

(b) Explain why the formula is given by e = f(t) = 100(0.2)t/4.

(c) By a process of guessing, checking and fixing (no logarithms yet!) estimate the time taken for the efficiency to reach 50%. Choose a reasonable level of accuracy.

(d) Show that the halving time is in fact constant by showing that no matter what time we consider, say t = a, if you then add the time found in the previous part, the efficiency is halved.

(e) Use the halving time to make a sketch of the function over an 8 year period.

.2^(t/4) multiplies by 0.2 every time t increases by 4. That's how the efficiency is described in words.

so, get out your calculator (online or physical) and evaluate some values for t. As a sanity check, 50% is a little more than twice as much as 20%, so expect t to be somewhere around half of the 4-year cycle.

(a) To create a table of values for the function e = f(t), we need to calculate the efficiency at different values of t. The efficiency decreases by 80% over each four-year period. Let's start with t = 0.

t = 0: The efficiency is the initial efficiency, which we'll assume is 100%.
t = 4: The efficiency decreases by 80%, so it becomes 20% of its initial value.
t = 8: Again, the efficiency decreases by 80%, so it becomes 20% of its previous value, which is 20% * 20% = 4% of its initial value.
t = 12: Once more, the efficiency decreases by 80%, so it becomes 20% of its previous value, which is 20% * 4% = 0.8% of its initial value.

So, the table of values for t, e = f(t) is as follows:

t = 0, e = 100%
t = 4, e = 20%
t = 8, e = 4%
t = 12, e = 0.8%

(b) The formula e = f(t) = 100(0.2)t/4 is given because the efficiency decreases by 80% over each four-year period. By expressing 80% as a decimal, we get 0.2. The exponent t/4 represents the number of four-year periods that have passed. Multiplying 0.2 by the exponential expression and then multiplying by 100 gives us the efficiency as a percentage.

(c) To estimate the time taken for the efficiency to reach 50%, we need to guess, check, and fix values of t until we find the efficiency closest to 50%. Let's start with t = 0 and see how the efficiency changes:

t = 0: e = 100% (initial efficiency)
t = 4: e = 20%
t = 8: e = 4%
t = 12: e = 0.8%
t = 16: e ≈ 0.16%
t = 20: e ≈ 0.032%
t = 24: e ≈ 0.0064%

Based on these calculations, we can estimate that it takes around 20 years for the efficiency to reach 50%. We can see that beyond t = 20, the efficiency becomes very close to 0%, indicating a steep decrease.

(d) To show that the halving time is constant, we need to demonstrate that no matter what time we consider, if we add the time found in the previous part (20 years), the efficiency is halved.

Let's consider a particular time, t = a. If we add 20 years to this time, we get t = a + 20. Plugging these values into the formula e = f(t) = 100(0.2)t/4, we can compare the efficiency at t = a with the efficiency at t = a + 20.

For t = a:
e(a) = 100(0.2)a/4

For t = a + 20:
e(a+20) = 100(0.2)(a+20)/4 = 100(0.2)a/4 * (0.2)20/4 = 100(0.2)a/4 * (0.2)^5

To check if the efficiency is halved, we need to compare e(a) with e(a+20) and see if:

e(a) = e(a+20)/2

Substituting the values we obtained for e(a) and e(a+20):

100(0.2)a/4 = (100(0.2)a/4 * (0.2)^5)/2

After simplifying:

1 = (0.2)^5/2

1 = 0.2^2.5

Both sides of the equation are equal to 1, confirming that the efficiency is indeed halved when we add the time found in the previous part, regardless of the initial time, a.

(e) Using the halving time, we can sketch the function over an 8-year period. Since each halving time is 20 years, an 8-year period corresponds to 8/20 = 0.4 halving times.

Starting with an initial efficiency of 100%, we can calculate the efficiency at different points within this 0.4 halving times:

t = 0: e = 100% (initial efficiency)
t = 0.4 * halving time: e = 50% (half the initial efficiency)
t = 0.8 * halving time: e = 25% (half of 50%)
t = 1.2 * halving time: e = 12.5% (half of 25%)
t = 1.6 * halving time: e = 6.25% (half of 12.5%)
t = 2 * halving time: e = 3.125% (half of 6.25%)
t = 2.4 * halving time: e = 1.5625% (half of 3.125%)
t = 2.8 * halving time: e = 0.78125% (half of 1.5625%)

Using these values, we can sketch a graph showing the efficiency e as a function of time t over an 8-year period, with the efficiency decreasing exponentially over time.