Let k be the tangent vector and e the unit tangent vector, then
k=dtdr
e=∣k∣k Find the tangent vector
k=dtdr=〈dtd(2sin(3t)),dtd4t,dtd(2cos(3t)〉=〈6cos3t,2,−6sin3t〉
Now we find the module of the k
∣k∣=(6cos3t)2+22+(−6sin3t)2=40=210e=210〈6cos3t,2,−6sin3t〉=〈10310cos3t,1010,−10310sin3t〉 Answer: 〈10310cos3t,1010,−10310sin3t〉
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