3 - t 3 + (-1 + 2 3 a ) te-2 t 3 (2) y ( t ) contains two...

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From part ( a ) and problem 18, we get y ( t ) = ae - 2 t 3 + ( - 1 + 2 3 a ) te - 2 t 3 (1) Part ( b ) is kind of trick. Here we don’t need to take the derivative. When do you need to take the derivative? Usually, when we want to find the minimum or maximum value of y ( t ), we need to take the derivative and set to derivative equal to zero, try to solve for t , then plug into y ( t ) and try to get the minimum or maximum value of y ( t ). But for this problem, it asks us to find the critical value of a that separates solutions that become negative from those that are always positive, so we look at y ( t ) y ( t ) = ae - 2
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Unformatted text preview: t 3 + (-1 + 2 3 a ) te-2 t 3 (2) y ( t ) contains two terms and the rst term is ae-2 t 3 and the second term is (-1 + 2 3 a ) te-2 t 3 . And we also know that the rst term is positive as long as a > 0 and second term is positive as long as-1 + 2 3 a > a > 3 2 . Hence we know that y ( t ) will be positive as long as a > 0 and a > 3 2 So the critical value is a = 0 or a = 3 2 . But for this problem when t is really big, the second term will dominates the solution. So the critical value is a = 3 2 ....
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