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# 26.6: Sample problems and solutions

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}

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