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Example: |
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| Person 1 is standing at some point. | ||
| Person 2 is standing 20 miles to the east of Person 1 ... | ||
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| At the same moment, Person 1 starts walking south at 4 miles per hour, | ||
| and Person 2 starts walking west at 2 miles per hour ... | ||
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| How long after they start walking, will the distance between them be a minimum? | ||
| OK, the first thing to do is to figure out the equation | ||
| for the distance between them. | ||
| We drop a coordinate axis on them (I hope it didn't hurt much). | ||
| It will make things easier if we put the origin at the place person 1 started ... | ||
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| So by arranging the axis this way, each person walks along one of the axis. | ||
| That is, for person 1 position, only the Y coordinate changes | ||
| for person 2, only the X coordinate changes. | ||
| You will see in a second that that makes the math easier. | ||
| Person 1 is moving south along the Y axis. | ||
| When person 1 starts walking, they are at (0,0). | ||
| At any time (T) after that, they are at: | ||
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(0, (0-4T)) |
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| Person 2 starts out at a point 20 miles east of person 1. That is, at (20,0) | ||
| Person 2 walks west at 2 miles per hour. | ||
| That means at any point in time they are at: | ||
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((20 - 2T), 0) |
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| If you look at the diagram above, | ||
| you can see that the distance between them is the hypotenuse of a triangle. | ||
| That means we can use the old Pythagorean distance formula! | ||
| So at any time T, the distance between them is: | ||
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| Multiplying this out ... | ||
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| Combining terms ... | ||
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| We want to find where the distance, D, is a minimum. | ||
| That means we need to check the point where the slope is zero and the endpoints. | ||
| One endpoint is where time = 0, that is, the starting point. | ||
| There really isn't another end point. Time can run forever here. | ||
| That means we need to calculate a limit as time approaches infinity. | ||
| To get the slope equals zero point, | ||
| we need the first derivative of the distance equation. | ||
| I find it easier to change roots to fractional exponents when doing this. | ||
| A square root is the same as raising the expression to the one half power. | ||
| So ... | ||
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| Don't forget that this is a power chain rule problem, | ||
| so we have to deal with an "inside part." | ||
| We need to find the point where D'=0, | ||
| so it would help to write this as a real fraction... | ||
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| When the numerator equals zero, the whole expression will equal zero. | ||
| As long as the denominator does not equal zero. | ||
| So ... | ||
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0 = 40T - 80 |
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T = 2 hours |
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| Check to make sure the denominator is not equal to zero at T = 2. | ||
| We only need to find the value under the square root radical. | ||
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20T2 - 80T + 400 |
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20(2)2 - 80(2) + 400 |
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80 - 160 + 400 |
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320 |
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| OK, the denominator is not equal to zero. | ||
| Now lets check the second derivative to make sure that this is a minimum... | ||
| Use the quotient rule, and hold on ... | ||
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| Do some simplifying ... | ||
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| Solve this mess for when T = 2 ... | ||
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| The second derivative is positive. | ||
| That means the T=2 hours point is a minimum. | ||
| So what is the distance between the two people at that point? | ||
| The distance formula was ... | ||
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| Plug in T = 2 and chug away. | ||
| You should get ... | ||
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D = 17.89 miles |
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| The endpoints are where T = 0 and where T approaches infinity. | ||
| At T = 0, the first two terms under the radical equal zero | ||
| so the distance is the square root of 400, so | ||
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D = 20 |
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| Where T approaches infinity, | ||
| the value will be controlled by the highest exponent term under the radical. | ||
| It takes a bit of fancy math to prove this one, | ||
| but it is pretty obvious that the distance approaches infinity. | ||
| So the minimum distance of 17.89 miles happens when T = 2. | ||
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copyright 2005 Bruce Kirkpatrick |
