Question 194837
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Let <b><i>r</i></b> represent the rate of the current.  Then <b>3 - <i>r</i></b> is the rate of the boat going upstream and <b>3 + <i>r</i></b> is the rate of the boat going downstream.  Then let <b><i>t</i></b> be the time taken to go upstream which means that the time to go downstream must be <b>8 - <i>t</i></b>

Using *[tex \LARGE d = rt], the upstream trip is then described by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6 = (3 - r)t]

And the downstream trip is described by:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6 = (3 + r)(8 - t)]

Solve each of these equations for <b><i>t</i></b>.

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{6}{3 - r}]

And

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8 - t = \frac{6}{3 + r}]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = 8 - \frac{6}{3 + r}]

Now we have two expressions equal to <b><i>t</i></b>, so set them equal to each other:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8 - \frac{6}{3 + r} = \frac{6}{3 - r}]

Put everything on the left:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8 - \frac{6}{3 + r} - \frac{6}{3 - r} = 0]

LCD is *[tex \LARGE (3 + r)(3 - r) = 9 - r^2], so:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{8(9-r^2)-6(3-r)-6(3+r)}{9-r^2} = 0]

Multiply both sides by *[tex \LARGE 9 - r^2] to get rid of the denominator:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8(9-r^2)-6(3-r)-6(3+r) = 0]

Remove parentheses and collect like terms:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 72 - 8r^2 - 18 + 6r - 18 - 6r = 0]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -8r^2 + 36 = 0]

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2 = \frac{36}{8}]

Only consider the positive root because we are reasonably certain that the current never went backwards during this trip, so:

*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r = \sqrt{\frac{36}{8}} = \frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}]

Checking the answer is left as an exercise for the student.

John
*[tex \LARGE e^{i\pi} + 1 = 0]
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