Find the normal vectors to r(t) = <$at, cos t> at t = (π / 4) and t = (3π / 4).
Recall the definition of unit tangent vector and normal vector.
The unit tangent vector is T(t) = r'(t) / ||r'(t)||.
The unit normal vector to r(t) is
1rpHlqKjYRtdcq8jmbRj55+GDssUsrQSMqV/HtFJ6TB7UAF5HgEP5R9FQWnUzhmcrG+DpITM8nWqw/dPqFLMxzen17zNnbQGbqAE/uTAMCnYNkDBEyg6pdJdyiMLg8fUewdOWvUL+yFOCmTeSUoae/0mcZKTMfwAHwsTlO05VBlhEDYY0DdFOEiBPuo=Find r'(t).
r'(t) = <nc1ItEz0kR4=, -sin t>
r'(t) =
Find the magnitude ||r'(t)||.
||r'(t)|| = √(iSba6t70dtA= + sin2t).
Thus we have the unit tangent vector for r(t).
T(t) = r'(t) / ||r'(t)||
= (1/√(iSba6t70dtA= + sin2t))<nc1ItEz0kR4=, -sin t>
In order to find the normal vector, we need to compute T'(t).
Find T'(t) using the product rule.
T'(t) = (d / dt)((1/√(iSba6t70dtA= + sin2t))<nc1ItEz0kR4=, -sin t>)
= (1/√(iSba6t70dtA= + sin2t))(d / dt)<nc1ItEz0kR4=, -sin t> + <nc1ItEz0kR4=, -sin t>(d / dt)(1/√(iSba6t70dtA= + sin2t))
(d / dt)<$a, -sin t> = <1Wh3cvJ2xF4=, -cos t>
(d / dt)(1/√($b + sin2t)) = FmAhxBkQqN0=sin t cos t /(2(iSba6t70dtA= + sin2t)3/2)
Thus we have the normal vector for r(t).
T'(t) = (1/√($b + sin2t))(d / dt)<$a, -sin t> + (d / dt)(1/√($b + sin2t))<$a, -sin t>
= (1/√($b + sin2t))<0, -cos t> - 2sin t cos t /(2($b + sin2t)3/2)<$a, -sin t>
Find the normal vectors to r(t) at t = (π / 4) and t = (3π / 4).
T'(π / 4) = (1/√($b + (1/√2)2))<0, -√2/2> - 2(1/√2)2 /(2($b + (1/√2)2)3/2)<$a, -√2/2>
= (√2/√$c)<0, -√2/2> - (√2/($c√$c))<$a, -√2/2>
= <IHLqJALES1Y=√2 / (SFgqQUkJGdg=√$c), iXKaZpK/a+Q= / (SFgqQUkJGdg=√$c)>