Answer:
Explanation:
To find the normal plane and osculating plane, we first need to find the required derivatives.
x = 5t, y = t^2, z = t^3
dx/dt = 5, dy/dt = 2t, dz/dt = 3t^2
So, the velocity vector v and acceleration vector a are:
v = <5, 2t, 3t^2>
a = <0, 2, 6t>
Now, let's evaluate them at t = 1 since the point (5, 1, 1) is given.
v(1) = <5, 2, 3>
a(1) = <0, 2, 6>
The normal vector N is the unit vector in the direction of a:
N = a/|a| = <0, 1/√10, 3/√10>
Using the point-normal form of the equation for a plane:
normal plane equation = 0(x-5) + 1/√10(y-1) + 3/√10(z-1) = 0
Simplifying this equation we get:
5x + 2y + 3z = 30
The osculating plane can be found using the formula:
osculating plane equation = r(t) · [(r(t) x r''(t))] = 0
where r(t) is the position vector, and x is the cross product.
At t = 1, the position vector r(1) is <5, 1, 1>, v(1) is <5, 2, 3>, and a(1) is <0, 2, 6>.
r(1) x v(1) = <-1, 22, -5>
r(1) x a(1) = <12, -6, -10>
v(1) x a(1) = <-12, 0, 10>
Substituting these values into the formula, we get:
osculating plane equation = (x-5, y-1, z-1) · <12, -6, -10> = 0
Simplifying this equation we get:
3x - 15y + 5z = 5
Therefore, the equations for the normal plane and osculating plane at (5, 1, 1) are:
(a) 5x + 2y + 3z = 30
(b) 3x - 15y + 5z = 5