x = e^(2t), y=e^(3t)+8
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Hello, pourcarlos! Eliminate the parameter: . From [1]: . Substitute into [2]: . Therefore: .
oh great, thank you
Originally Posted by pourcarlos x = e^(2t), y=e^(3t)+8 An alternative to the perfectly correct solution provide by Soroban: and . Therefore .... How much of the curve is required will obviously depend on the values of t that are allowed. If then the curve starts at (1, 9) ....
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