Asked by Anonymous

Can somebody explain how the derivative of (cost-sint)i and (cost+sint)j equal -2sinti and 2costj. Should it be (-sint-cost)i and(-sint-ccost)j?

Answers

Answered by oobleck
Close. It should be (-sint-cost)i and(-sint+cost)j
No idea how they came up with their answers.

I will note that cost-sint = √2 sin(π/4 - t) but that may be of no help.
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