Consider the following vector function. r(t) = 3sqrt(2) t , e^(3t),e^(-3t)
(a) Find the unit tangent and unit normal vectors T(t) and N(t).
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)
can you simplify the unit tangent vector?
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)>
To find the unit tangent vector T(t) and unit normal vector N(t), we need to follow the steps given below:
Step 1: Calculate the derivative of the vector function r(t) to find the velocity vector v(t).
Step 2: Calculate the magnitude of the velocity vector v(t).
Step 3: Divide the velocity vector v(t) by its magnitude to obtain the unit tangent vector T(t).
Step 4: Take the derivative of the unit tangent vector T(t) to get the acceleration vector a(t).
Step 5: Divide the acceleration vector a(t) by its magnitude to get the unit normal vector N(t).
Let's calculate each step for the given vector function.
Step 1: Calculate the derivative of r(t):
r'(t) = (3sqrt(2), 3e^(3t), -3e^(-3t))
Step 2: Calculate the magnitude of the velocity vector v(t):
|v(t)| = √[(3sqrt(2))^2 + (3e^(3t))^2 + (-3e^(-3t))^2]
= √[18 + 9e^(6t) + 9e^(-6t)]
Step 3: Divide the velocity vector v(t) by its magnitude to get the unit tangent vector T(t):
T(t) = v(t) / |v(t)|
= (3sqrt(2) / √[18 + 9e^(6t) + 9e^(-6t)], 3e^(3t) / √[18 + 9e^(6t) + 9e^(-6t)], -3e^(-3t) / √[18 + 9e^(6t) + 9e^(-6t)])
Step 4: Calculate the derivative of the unit tangent vector T(t):
T'(t) = [d/dt (3sqrt(2) / √[18 + 9e^(6t) + 9e^(-6t)]), d/dt (3e^(3t)/ √[18 + 9e^(6t) + 9e^(-6t)]), d/dt (-3e^(-3t)/ √[18 + 9e^(6t) + 9e^(-6t)])]
Step 5: Divide the acceleration vector a(t) by its magnitude to get the unit normal vector N(t):
N(t) = T'(t) / |T'(t)|
By following these steps with the given vector function, you should be able to obtain the unit tangent vector T(t) and unit normal vector N(t) for any value of t.