To find the equations of the normal plane and osculating plane of the curve defined by the parametric equations x = 5t, y = t^2, z = t^3 at the point (5, 1, 1), we first need to determine the parameter value t that corresponds to this point.
From the equations:
- 5t = 5 → t = 1
- y = t^2 → 1 = 1^2
- z = t^3 → 1 = 1^3
So, the point corresponds to t = 1.
Next, we find the tangent vector at this point by calculating the derivative of the position vector:
r(t) = (5t, t^2, t^3)
Taking the derivative gives us:
r'(t) = (5, 2t, 3t^2)
At t = 1, the tangent vector is:
r'(1) = (5, 2, 3)
Next, we need to find the normal vector, which is given by the cross product of the tangent vector and the second derivative of the curve:
First, compute the second derivative:
r”(t) = (0, 2, 6t)
Thus, at t = 1, we have:
r”(1) = (0, 2, 6)
The normal vector N is:
N = r'(1) × r”(1)
Calculating this cross product, we find:
N = (5, 2, 3) × (0, 2, 6) = (12, -30, 10)
Now we can write the equation of the normal plane. The general equation for the plane is given by:
N (x – x_0) + N (y – y_0) + N (z – z_0) = 0
Substituting in our normal vector and point:
12(x – 5) – 30(y – 1) + 10(z – 1) = 0
This simplifies to:
12x – 30y + 10z – 66 = 0
Now, for the osculating plane, the normal vector is given by r'(1), r”(1). The equation for the osculating plane is given by:
r'(1) (x – 5) + r”(1) (y – 1) + r”(1) (z – 1) = 0
Using our previously calculated vectors, the equation becomes:
5(x – 5) + 2(y – 1) + 6(z – 1) = 0
This simplifies to:
5x + 2y + 6z – 36 = 0
In conclusion, we have the equations:
- Normal Plane: 12x – 30y + 10z – 66 = 0
- Osculating Plane: 5x + 2y + 6z – 36 = 0