Consider the vector function given below.
r(t) =
7t, 3 cos t, 3 sin t
(a) Find the unit tangent and unit normal vectors T(t) and N(t).
i found T(t) but I can't get N(t)
If you found T(t) = r'/|r'|, then you should be able to get
N(t) = T'/|T'|
Where do you get stuck?
To find the unit tangent vector T(t), you need to differentiate the vector function r(t) with respect to t:
r'(t) = (7, -3sin t, 3cos t)
Next, find the magnitude of the derivative vector:
|r'(t)| = √[(7)^2 + (-3sin t)^2 + (3cos t)^2]
= √[49 + 9sin^2 t + 9cos^2 t]
= √[58 + 9(sin^2 t + cos^2 t)]
= √[58 + 9]
= √67
Now, divide each component of the derivative vector by its magnitude to get the unit tangent vector T(t):
T(t) = (7/√67, (-3sin t)/√67, (3cos t)/√67)
To find the unit normal vector N(t), you need to differentiate the derivative vector r'(t) and then divide it by its magnitude:
r''(t) = (0, -3cos t, -3sin t)
|r''(t)| = √[(0)^2 + (-3cos t)^2 + (-3sin t)^2]
= √[0 + 9cos^2 t + 9sin^2 t]
= √[9(cos^2 t + sin^2 t)]
= √[9]
= 3
Finally, divide each component of the second derivative vector by its magnitude to get the unit normal vector N(t):
N(t) = (0/3, (-3cos t)/3, (-3sin t)/3)
= (0, -cos t, -sin t)