Question
A cell phone starts charging at 07:00. While being charged, the percentage of power, \( P \), in the phone is modelled by the function:
\[ P = 100 – 60a^{-t} \]
where \( t \) is the number of hours after 07:00.
(a) Find the percentage of power in the phone at 07:00.
The percentage of power in the phone reaches 75% at 08:00.
(b) Find the value of \( a \).
(c) Draw the graph of \( P = 100 – 60a^{-t} \) on the following set of axes.
(d) State a mathematical reason why the model predicts the percentage of power in the phone will never reach 100%.
▶️Answer/Explanation
Detailed solution
(a) Finding the Percentage of Power at 07:00
Substituting \( t = 0 \) into the given equation:
\[ P = 100 – 60a^{-0} \]
Since \( a^0 = 1 \):
\[ P = 100 – 60(1) \]
\[ P = 40 \]
So, the percentage of power in the phone at 07:00 is 40%.
(b) Finding the Value of \( a \)
Given that \( P = 75 \) when \( t = 1 \), we substitute these values into the equation:
\[ 75 = 100 – 60a^{-1} \]
Rearranging for \( a \):
\[ 60a^{-1} = 100 – 75 \]
\[ 60a^{-1} = 25 \]
\[ a^{-1} = \frac{25}{60} \]
\[ a = \frac{60}{25} = 2.4 \]
So, the value of \( a \) is 2.4.
(c) Graph of \( P = 100 – 60(2.4)^{-t} \)
The function is an exponential growth model approaching 100%. The graph should start at \( P = 40 \) when \( t = 0 \) and gradually increase towards 100% asymptotically.
(d) Mathematical Reason Why \( P \) Never Reaches 100%
- The function has a horizontal asymptote at \( P = 100 \), meaning the value of \( P \) gets closer and closer to 100 but never actually reaches it.
- Mathematically, the equation is written as \( 100 – 60a^{-t} \), where \( 60a^{-t} \) is always a positive number.
- Since \( 60a^{-t} \) never becomes 0, \( P \) can never equal exactly 100.
……………………………Markscheme……………………………….
(a)
$P = 100 – 60a^{-0}$
40
(b)
$75 = 100 – 60a^{-1}$
$a = 2.4$
(c)
Correctly drawn exponential curve approaching $P = 100$
(d)
Asymptote at $P = 100$
$P$ never reaches 100