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Point on a Curve |
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| Jan20-09, 10:27 PM | #1 |
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Point on a Curve
1. The problem statement, all variables and given/known data
A particle moves along a path described by y = 4 - x^2. At what point along the curve are x and y changing at the same rate 2. Relevant equations Simple equations regarding derivatives. 3. The attempt at a solution It's been a while before I've done any related rates problems, could someone please let me know if this is correct: Since, x and y must be changing at the same rate (presumably with respect to time) x' = y' and y' = -2xx'. Therefore, -2x = 1 and x = -1/2. Placing my x value into the original equation yields 15/4. Hence, the point is (-1/2, 15/4). Thanks. |
| Jan20-09, 10:37 PM | #2 |
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SEEMS correct...
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| Jan20-09, 11:08 PM | #3 |
Recognitions:
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Of course, it's right. What could go wrong?
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| Jan20-09, 11:13 PM | #4 |
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Point on a Curve
Plenty, I could have made an incorrect assumption ultimately leading to false conclusions.
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| Jan20-09, 11:15 PM | #5 |
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Good answer but you do not need to assume that x and y are varying wrt an external parameter. The derivative y'(x) = dy/dx of y wrt x expresses the instantaneous rate of change of y wrt a change in x.
The points where y and x are changing at the same rate are those where y'(x)=1. |
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