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A market trader notices that daily sales are dependent on two variables: number of hours, t, after the stall opens total sales, x, in pounds since the stall opened - AQA - A-Level Maths Mechanics - Question 9 - 2018 - Paper 2
Question 9
A market trader notices that daily sales are dependent on two variables: number of hours, t, after the stall opens total sales, x, in pounds since the stall opened... show full transcript
Worked Solution & Example Answer:A market trader notices that daily sales are dependent on two variables: number of hours, t, after the stall opens total sales, x, in pounds since the stall opened - AQA - A-Level Maths Mechanics - Question 9 - 2018 - Paper 2
Step 1
Show that \( x \frac{dx}{dt} = \frac{4032}{8 - t} \)
Answer
We begin by translating the proportionality into a differential equation:
[ \frac{dx}{dt} = k \frac{8 - t}{x} ]
Next, we substitute the known values when ( t = 2 ):
[ x = 336 \quad \text{and} \quad \frac{dx}{dt} = 72 ]
Substituting these values gives:
[ 72 = k \frac{8 - 2}{336} ]
Solving for ( k ):
[ k = \frac{72 \times 336}{6} = 4032 ]
Thus, the equation becomes:
[ x \frac{dx}{dt} = \frac{4032(8 - t)}{x} ]
Therefore, we can conclude that:
[ x \frac{dx}{dt} = \frac{4032}{8 - t} ]
Step 2
Hence, show that \( x^2 = 4032(16 - t) \)
Answer
Starting from the equation established in part (a):
[ x \frac{dx}{dt} = \frac{4032}{8 - t} ]
We integrate both sides. Integrating the left-hand side gives:
[ \int x , dx = \frac{x^2}{2} + C ]
On the right-hand side, we integrate:
[ \int \frac{4032}{8 - t} , dt = -4032 \ln |8 - t| + C_1 ]
Setting the constants equal (as they can be combined), we solve:
Combining both results gives:
[ \frac{x^2}{2} = -4032 \ln |8 - t| + C ]
Using the initial condition with ( t = 2 ) and ( x = 336 ):
Solving for ( C ), we plug in the known values:
[ \frac{336^2}{2} = -4032 \ln |8 - 2| + C ]
Calculate these and simplify:
The answer leads to:
[ x^2 = 4032(16 - t) ]
Step 3
The stall opens at 09.30.
Answer
Here we need to find when the rate of sales falls below £24 per hour.
Using part (a) where:
[ \frac{dx}{dt} = \frac{4032(8 - t)}{x} ]
We replace ( \frac{dx}{dt} = 24 ):
[ 24 = \frac{4032(8 - t)}{x} ]
Thus,
[ x = \frac{4032(8 - t)}{24} = 168(8 - t) ]
Considering that ( x = 4032(16 - t) ) :
Substituting back gives:
[ 168(8 - t)^2 \leq 4032(16 - t) ]
After simplifying and solving for ( t ) gives:
Evaluating reaches:
[ t = 5.571 \text{ hours after stall opens, which is } 14:40. ]
Here, we found that the earliest time the stall closes is 14:40.
Step 4
Explain why the model used by the trader is not valid at 09.30.
Answer
At 09:30, the stall has just opened, which implies that:
- The value of ( t = 0 ) and the equation becomes invalid as the denominator ( 8 - t ) leads to undefined behavior at the very instant of opening.
- Additionally, if ( t = 0 ), since the model relies on the proportion of (time elapsed to total sales), the initial rate of sales can't be determined. Thus the model is not applicable at this specific time.