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Help me with the algebra in this http://i.imgur.com/avF62lx.png I set the derivatives equal to each other, but how do I get t out of two separate trig functions?
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set both the velocities equal to each other and solve \(t\) : \[x'(t) = y'(t)\]
\[-6\pi \sin(\pi t) = 2\pi \cos(\pi t)\] solve \(t\)
\[-3\sin(\pi t)=\cos(\pi t)\] I can take the arc cos of both sides but i'm still left with a variable inside of sin
calculator exercise i guess
\[-3\tan(\pi t)=1\\ \tan(\pi t)=-\frac{1}{3}\]\[\pi t=\arctan(-\frac{1}{3})\]
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