# 26.6: Sample problems and solutions

- Page ID
- 19576

Exercise \(\PageIndex{1}\)

You find that the number of customers in your store as a function of time is given by: \[\begin{aligned} N(t) = a+bt-ct^2\end{aligned}\] where \(a\), \(b\) and \(c\) are constants. At what time does your store have the most customers, and what will the number of customers be? (Give the answer in terms of \(a\), \(b\) and \(c\)).

**Answer**-
We need to find the value of \(t\) for which the function \(N(t)\) is maximal. This will occur when its derivative with respect to \(t\) is zero: \[\begin{aligned} \frac{dN}{dt} &= b-2ct =0\\ \therefore t &= \frac{b}{2c}\end{aligned}\] At that time, the number of customers will be: \[\begin{aligned} N\left( t=\frac{b}{2c} \right) &=a+bt-ct^2\\ &=a+\frac{b^2}{2c} - \frac{b^2}{4c} = a+\frac{3b^2}{4c}\end{aligned}\]

Exercise \(\PageIndex{2}\)

You measure the speed, \(v(t)\), of an accelerating train as function of time, \(t\), to be given by: \[\begin{aligned} v(t)=at+bt^2\end{aligned}\] where \(a\) and \(b\) are constants. How far does the train move between \(t=t_0\) and \(t=t_1\)?

**Answer**-
We are given the speed of the train as a function of time, which is the rate of change of its position: \[\begin{aligned} v(t)=\frac{dx}{dt}\end{aligned}\] We need to find how its position, \(x(t)\), changes with time, given the speed. In other words, we need to find the anti-derivative of \(v(t)\) to get the function for the position as a function of time, \(x(t)\): \[\begin{aligned} x(t) &= \int v(t) dt = \int (at+bt^2) dt\\ &=\frac{1}{2}at^2 + \frac{1}{3}bt^3 + C\end{aligned}\] where \(C\) is an arbitrary constant. The distance covered, \(\Delta x\), between time \(t_0\) and time \(t_1\) is simply the difference in position at those two times: \[\begin{aligned} \Delta x &= x(t_1) - x(t_0)\\ &=\frac{1}{2}at_1^2 + \frac{1}{3}bt_1^3 + C - \frac{1}{2}at_0^2 + \frac{1}{3}bt_0^3 - C\\ &=\frac{1}{2}a(t_1^2-t_0^2) + \frac{1}{3}b(t_1^3-t_0^3)\end{aligned}\]