Answer and Step-by-step explanation:
a) To find the curvature of the space curve r(t), we need to find the magnitude of the curvature vector. The curvature vector can be calculated using the formula:
Curvature = |(dT/dt)| / ||r'(t)||
where T is the unit tangent vector and r'(t) is the derivative of r(t).
First, we need to find the derivative of r(t):
r'(t) = (3cos t)i + 4j - (3sin t)k
Next, we find the magnitude of the derivative:
||r'(t)|| = √[(3cos t)² + 4² + (-3sin t)²]
= √[9cos² t + 16 + 9sin² t]
= √(25)
= 5
Now, we need to find the unit tangent vector T:
T = r'(t) / ||r'(t)||
= [(3cos t)i + 4j - (3sin t)k] / 5
Finally, we find the magnitude of the derivative of the unit tangent vector:
|(dT/dt)| = |[-3sin t)i + 0j - (3cos t)k] / 5|
= √[(-3sin t)² + 0 + (3cos t)²] / 5
= √(9sin² t + 9cos² t) / 5
= √(9) / 5
= 3 / 5
Now, we can calculate the curvature:
Curvature = |(dT/dt)| / ||r'(t)||
= (3 / 5) / 5
= 3 / 25
Therefore, the curvature of the space curve r(t) is 3/25.
b) To find the equation of the normal plane α to the curve at point P(0,0,3), we need to find the normal vector to the plane. The normal vector is given by the derivative of the unit tangent vector dT/dt evaluated at the point P.
First, we find the derivative of T:
(dT/dt) = [-3sin t)i + 0j - (3cos t)k]
Next, we substitute t = 0 into the derivative:
(dT/dt) = [-3sin 0)i + 0j - (3cos 0)k]
= [0i + 0j - (-3)k]
= [0i + 0j + 3k]
= 3k
Since we are evaluating at point P(0,0,3), the normal vector is parallel to the z-axis.
Therefore, the equation of the normal plane α to the curve at point P(0,0,3) is of the form x = 0, y = 0, and z is free.