Answer to Question #117461 in Calculus for Olivia

Question #117461
Consider the parametric curve described by r(t) =〈2 sin(3t), √4t, 2cos(3t)〉. Find the unit tangent vector of r(t).
1
Expert's answer
2020-05-25T20:48:43-0400

Let k\vec k be the tangent vector and e\vec e the unit tangent vector, then


k=drdt\vec k=\frac {d\vec r}{dt}

e=kk\vec e=\frac {\vec k}{|k|}

Find the tangent vector

k=drdt=d(2sin(3t))dt,d4tdt,d(2cos(3t)dt=6cos3t,2,6sin3t\vec k=\frac {d\vec r}{dt}=〈\frac{d(2 sin(3t))}{dt}, \frac{d\sqrt 4t}{dt}, \frac{d(2cos(3t)}{dt}〉=〈6cos3t, 2, -6sin3t〉

Now we find the module of the k\vec k

k=(6cos3t)2+22+(6sin3t)2=40=210|\vec k|=\sqrt{(6cos3t)^2+2^2+(-6sin3t)^2}=\sqrt{40}=2\sqrt{10}e=6cos3t,2,6sin3t210=31010cos3t,1010,31010sin3t\vec e=\frac{〈6cos3t, 2, -6sin3t〉}{2\sqrt{10}}=〈\frac{3\sqrt{10}}{10} cos3t, \frac{\sqrt{10}}{10}, -\frac{3\sqrt{10}}{10}sin3t〉

Answer: 31010cos3t,1010,31010sin3t〈\frac{3\sqrt{10}}{10} cos3t, \frac{\sqrt{10}}{10}, -\frac{3\sqrt{10}}{10}sin3t〉







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