roots

Caution: The equations have valid negative solutions, but there is an unremovable singularity in the equations if you try to move y2 or v2 through zero!

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Walter Roberson 2011-9-30

You cannot draw a graph between the bounds with the information given. The information given fixes exact values for y2 and V2. If you wish to draw a graph, you need at least one free variable, such as would be the case if you removed one of two equations and indicated which of the two variables' values that you wanted to range between.

But either way, if you were to fix v2 or y2, you would have a difficult time drawing a graph between the bounds: with the value of g that you have supplied, both v2 and y2 are necessarily complex. In order for them to avoid being complex, 2 + h - y1 must be less than or equal to 0: as h is 1/2 and y1 is 3/2, with the current values that is 2 + 1/2 - 3/2 which is 5/2 - 3/2 which is 1 which is unfortunately positive. If you were to reverse the values of h and y1, then 2 + 3/2 - 1/2 is even more positive, so that is not the solution either. For non-complex solutions you have the essential constraint that h <= y1 - 2 -- e.g., h <= - 1/2 (the negative of what you have now) would work.

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