Final answer:
To determine the point on the curve where the normal plane is parallel to the given plane, calculate the derivatives of the curve, find the tangent vector, and use cross product with the normal vector of the given plane to solve for t, then substitute t into the curve's equation.
Step-by-step explanation:
The question asks at what point on the curve defined by x = t³, y = 9t, z = t⁴ the normal plane is parallel to the plane 6x + 18y − 8z = 5. To solve this problem, we need to find the normal vector of the given plane, which is (6, 18, -8) and then find a point on the curve where the normal vector to the curve at that point is parallel to the normal vector of the given plane.
Firstly, we calculate the derivative of the position vector of the curve with respect to t to find a tangent vector: dx/dt = 3t², dy/dt = 9, dz/dt = 4t³. The tangent to the curve at a point is (3t², 9, 4t³), so the normal of the curve is perpendicular to this vector.
For two vectors to be parallel, their cross product must be zero. We can say that the cross product of tangent to the curve and the normal to the given plane must be zero. After solving for t, we find the appropriate value for t that satisfies this condition, and then plug it back into the curve equations to find the point (x, y, z) on the curve.