Homework SourseFor the curve given by rt 13t3 i t2j 2tk find the unit ta
For the curve given by r(t) = 1/3t^3 i + t^2j + 2tk find the unit tangent vector, the unit normal vector, and the curvature Evaluate the limit or show that it does not exist lim_(x, y) rightarrow (0,0) x^2y/2y^2 - 3x^4
Solution
8 ) Given that
r(t) = (1/3)t3i + t2j + 2tk
a ) We know that
Unit tangent vector T(t) = r\'(t) /II r\'(t) II
r\'(t) = (1/3).3t2i + 2tj + 2k [ since d/dx(xn)=n.xn-1 ]
r\'(t) = t2i + 2tj + 2k
II r\'(t) II = [ ( t2)2 + (2t)2 + 22 ]1/2
= [ t4 + 4t2 + 4 ] 1/2
Hence,
Unit tangent vector T(t) = r\'(t) /II r\'(t) II
T(t) = [ t2i + 2tj + 2k ] / [ t4 + 4t2 + 4 ] 1/2
b ) We know that
Unit normal vector N(t) = T\'(t) /II T\'(t) II
T(t) = [ t2i + 2tj + 2k ] / [ t4 + 4t2 + 4 ] 1/2
Unit normal vector N(t) = { [ t2i + 2tj + 2k ] / [ t4 + 4t2 + 4 ] 1/2}\' / II [ t2i + 2tj + 2k ] / [ t4 + 4t2 + 4 ] 1/2 II
