calc_tutorial_13_4_39

Question 1

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

A.

κ(t)

B.

T'(t)

C.

r'(t)

D.

r''(t)

Find r'(t).

r'(t) = <, -sin t>

r'(t) =

Find the magnitude ||r'(t)||.

||r'(t)|| = √( + sin²t).

Question 2

Thus we have the unit tangent vector for r(t).

T(t) = r'(t) / ||r'(t)||

= (1/√( + sin²t))<, -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/√( + sin²t))<, -sin t>)

= (1/√( + sin²t))(d / dt)<, -sin t> + <, -sin t>(d / dt)(1/√( + sin²t))

(d / dt)<3, -sin t> = <, -cos t>

(d / dt)(1/√(9 + sin²t)) = sin t cos t /(2( + sin²t)^3/2)

Question 3

Thus we have the normal vector for r(t).

T'(t) = (1/√(9 + sin²t))(d / dt)<3, -sin t> + (d / dt)(1/√(9 + sin²t))<3, -sin t>

= (1/√(9 + sin²t))<0, -cos t> - 2sin t cos t /(2(9 + sin²t)^3/2)<3, -sin t>

Find the normal vectors to r(t) at t = (π / 4) and t = (3π / 4).

T'(π / 4) = (1/√(9 + (1/√2)²))<0, -√2/2> - 2(1/√2)² /(2(9 + (1/√2)²)^3/2)<3, -√2/2>

= (√2/√19)<0, -√2/2> - (√2/(19√19))<3, -√2/2>

= <√2 / (√19), / (√19)>