91Ó°ÊÓ

In our exercise, the river's velocity is expressed as a fourth-degree polynomial rather than a simpler quadratic one. This polynomial is described by the equation: \[ v_{R} = v_{0}\left(1 - \frac{x^4}{a^4}\right), \] where:

• \(v_{R}\): velocity of the river at point \(x\),
• \(v_{0} = 9 \, \text{mi/h}\): maximum velocity of the river,
• \(a = 0.5 \, \text{mi}\): reference distance along the riverbank.

This equation explains that the velocity diminishes as the swimmer moves from the center of the river (where \(x = 0\)) to the edge (where \(x = a\)). This non-linear equation describes how the force of the river's flow varies across its width, which is more complex than a straightforward linear or quadratic model, capturing the subtleties of natural water currents.

To find how far downstream a swimmer drifts, we need to compute a definite integral. This involves an integral from 0 to \(a\) of the river's velocity function: \[ D = \int_0^a v_0 \left(1 - \frac{x^4}{a^4}\right) \, dx. \] In this exercise, the definite integral represents the total effect of the river's velocity on the swimmer over the span \(x\) from 0 to \(a\). The definite integral essentially adds up all of the tiny pieces of area under the curve described by our velocity function, which, in this case, gives us the total drift distance downstream. Substituting the given values:

• \(v_0 = 9\),
• \(a = 0.5\),

our integral becomes: \[ D = \int_0^{0.5} 9 \left(1 - 16x^4\right) \, dx. \] Solving this allows us to calculate the drift caused by the gentle advance of the current as it nudges the swimmer downstream.

Once we have set up our definite integral, it's time to compute it to find how far the swimmer is carried downstream. We've simplified the integral expression as much as possible and reached the stage where we need to execute the integration: \[ D = 9 \int_0^{0.5} (1 - 16x^4) \, dx. \] After integrating, we substitute the limits 0 and 0.5 into our integrated equation: \[ D = 9 \left[ x - \frac{16}{5}x^5 \right]_0^{0.5}. \] Evaluating this gives us: \[ D = 9 \left(0.5 - 0.00625\right) = 9 \times 0.49375. \] Doing the arithmetic, the drift comes out to be approximately 4.44 miles. This result illustrates the entire downstream drift the swimmer experiences due to the nuanced river velocity—showing how an integral can translate a velocity function across a distance into a meaningful real-world measurement.