The number of people in a community library's reading room, \(P\), during the first \(t\) hours after opening is modelled by \(P = t^3 - 12t^2 + 36t\), for ...
Question 1 Report
The number of people in a community library's reading room, \(P\), during the first \(t\) hours after opening is modelled by \(P = t^3 - 12t^2 + 36t\), for \(0 \leq t \leq 8\).
- Find \(\frac{dP}{dt}\). (1)
- Evaluate this derivative when \(t = 1\). (1)
- Hence work out the equation of the tangent line where \(t = 1\). (2)
- Solve \(\frac{dP}{dt} = 0\) to find the stationary values of \(t\). (2)
- Use the second derivative to determine the nature of each stationary point. (2)
- State when the library should schedule extra staff, with a reason. (1)
Answer Details
Each part uses the derivative of the cubic footfall model. Evaluating the derivative at a point gives an instantaneous rate (or a tangent gradient there); setting the derivative to zero finds the times when footfall is momentarily unchanging; and the second derivative distinguishes a peak in footfall from a dip.
- \[ \frac{dP}{dt} = 3t^2-24t+36 \] [1 mark]
- At \(t=1\): \[ 3(1)^2-24(1)+36 = 15 \] [1 mark]
- The point on the curve at \(t=1\) is \(P(1)=1-12+36=25\). Using this point and gradient \(15\): \[ P-25=15(t-1) \] so \[ P=15t+10 \] [2 marks]
- \[ 3t^2-24t+36=0 \] so \[ 3(t-2)(t-6)=0 \] giving \[ t=2 \text{ and } t=6 \] [2 marks]
- \[ \frac{d^2P}{dt^2} = 6t-24 \]; at \(t=2\): \(-12\lt0\), a maximum; at \(t=6\): \(12\gt0\), a minimum. [2 marks]
- Staff should be scheduled at \(t=2\) hours after opening, since this is when footfall reaches its maximum of \(P(2)=8-48+72=32\) people, confirmed by the second derivative test in part (e). [1 mark]
The model has a maximum at \(t=2\) and a minimum at \(t=6\); scheduling extra staff makes sense for the peak, not the trough, so identifying which stationary point is which is exactly the purpose of the second derivative test.
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