Consider two functions, \(f(x)\) and \(g(x)\text{.}\) We can find the total area between these two curves over an interval \(a \le x \le b\) by integrating the absolute value of the difference between these two functions. In other words,
\begin{equation*}
\text{ Total Area } = \displaystyle\int_a^b \abs{f(x)-g(x)}\,dx\text{.}
\end{equation*}
If \(f(x) \ge g(x)\) over that interval, then the absolute value can be dropped. However, if \(g(x) \ge f(x)\) over that interval, then the absolute value can be dropped and the order of subtraction reversed.
In other circumstances, we may be interested in the net area between \(f(x)\) and \(g(x)\text{.}\) For example, if \(f(x)\) and \(g(x)\) both represent rates of change of some quantity, then the net area
\begin{equation*}
\text{ Net Area } = \displaystyle\int_a^b \left(f(x)-g(x)\right)\,dx
\end{equation*}
will give us the net difference between the two quantities over an interval. The net area may be negative, while the total area will always be positive.
To begin, you should view the region between the two curves and determine how the intersection points depend on the value of \(c\text{.}\) Create a plot of \(y = x^2 - c^2\) and \(y=c^2 - x^2\) using your choice of nonzero \(c\) and adjust the view as necessary to the \(x\)-coordinates of the two points of intersection.
Maple will not be able to plot these two curves on the \(xy\)-plane without substituting in some value of \(c\text{,}\) since \(c\) would be viewed as an unexpected third variable in the equation of each curve.
Use the solve() (or fsolve()) command to find the \(x\)-coordinates of the two intersection points of \(y = x^2 - c^2\) and \(y = c^2 - x^2\text{.}\) These \(x\)-coordinates should be dependent on \(c\text{.}\)
Set up an equation where the integral giving the area between \(y = x^2 - c^2\) and \(y = c^2 - x^2\) is equal to 576 and solve the equation for \(c\) using the solve() (or fsolve()) command.
Plot both velocity functions on the same axes for a duration of \(2\) minutes (use \(0 \leq t \leq 120\text{,}\) since \(t\) is in seconds). Using your graph, try to guess which runner makes it the farthest distance in \(2\) minutes.
Make sure that you defined the exponential function properly in the two velocity functions. You can use the exp() function for \(e^x\) or use the palettes toolbar. You cannot use the letter βeβ on the keyboard for the exponential function.
Find the net area between \(v_A(t)\) and \(v_B(t)\) on the interval \([0,120]\text{.}\) This corresponds to the net difference in displacement during the race. Which runner made it the farthest distance in \(2\) minutes? By how many metres?