26.6: Sample problems and solutions
-
- Last updated
- Save as PDF
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:
\[\begin{aligned} \frac{dN}{dt} &= b-2ct =0\\[4pt] \therefore t &= \frac{b}{2c}\end{aligned}\]
At that time, the number of customers will be:
\[\begin{aligned} N\left( t=\frac{b}{2c} \right) &=a+bt-ct^2\\[4pt] &=a+\frac{b^2}{2c} - \frac{b^2}{4c} = a+\frac{3b^2}{4c}\end{aligned}\]
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, \(x(t)\) :
\[\begin{aligned} x(t) &= \int v(t) dt = \int (at+bt^2) dt\\[4pt] &=\frac{1}{2}at^2 + \frac{1}{3}bt^3 + C\end{aligned}\]
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:
\[\begin{aligned} \Delta x &= x(t_1) - x(t_0)\\[4pt] &=\frac{1}{2}at_1^2 + \frac{1}{3}bt_1^3 + C - \frac{1}{2}at_0^2 + \frac{1}{3}bt_0^3 - C\\[4pt] &=\frac{1}{2}a(t_1^2-t_0^2) + \frac{1}{3}b(t_1^3-t_0^3)\end{aligned}\]