Consider the following vector function. r(t) = 3sqrt(2) t , e^(3t),e^(-3t)

Question

(a) Find the unit tangent and unit normal vectors T(t) and N(t).

Answer 1

To find the unit tangent vector, we first find the derivative of r(t):

r'(t) = <3sqrt(2), 3e^(3t), -3e^(-3t)>

The magnitude of r'(t) is:

|r'(t)| = sqrt((3sqrt(2))^2 + (3e^(3t))^2 + (-3e^(-3t))^2)
= sqrt(18 + 9e^(6t) + 9e^(-6t))

So the unit tangent vector is:

T(t) = r'(t) / |r'(t)|
= <3sqrt(2), 3e^(3t), -3e^(-3t)> / sqrt(18 + 9e^(6t) + 9e^(-6t))

To find the unit normal vector, we first find the derivative of T(t):

T'(t) = <0, 9e^(3t), 3e^(-3t)> / sqrt(18 + 9e^(6t) + 9e^(-6t)) -
<3sqrt(2), 3e^(3t), -3e^(-3t)> (1/2)(18 + 9e^(6t) + 9e^(-6t))^(-1/2)(18e^(6t) - 18e^(-6t))

Simplifying, we get:

T'(t) = <3sqrt(2)/sqrt(18 + 9e^(6t) + 9e^(-6t)), 3e^(3t)/sqrt(18 + 9e^(6t) + 9e^(-6t)),
-3e^(-3t)/sqrt(18 + 9e^(6t) + 9e^(-6t))>

So the unit normal vector is:

N(t) = T'(t) / |T'(t)|
= <3sqrt(2)/sqrt(18 + 9e^(6t) + 9e^(-6t)), 3e^(3t)/sqrt(18 + 9e^(6t) + 9e^(-6t)),
-3e^(-3t)/sqrt(18 + 9e^(6t) + 9e^(-6t))>
/ sqrt((3sqrt(2)/sqrt(18 + 9e^(6t) + 9e^(-6t)))^2 + (3e^(3t)/sqrt(18 + 9e^(6t) + 9e^(-6t)))^2
+ (-3e^(-3t)/sqrt(18 + 9e^(6t) + 9e^(-6t)))^2)

Answer 2

can you simplify the unit tangent vector?

Answer 3

To simplify the unit tangent vector T(t), we can factor out 3sqrt(2) from the numerator, and then factor out 3 from the denominator, as follows:

T(t) = <3sqrt(2), 3e^(3t), -3e^(-3t)> / sqrt(18 + 9e^(6t) + 9e^(-6t))
= 3sqrt(2) / 3 * <1, e^(3t), -e^(-3t)> / sqrt(2 + e^(6t) + e^(-6t))
= <1/sqrt(2), e^(3t)/sqrt(2), -e^(-3t)/sqrt(2)>

So the simplified unit tangent vector is T(t) = <1/sqrt(2), e^(3t)/sqrt(2), -e^(-3t)/sqrt(2)>