Answer:
-3x + 3z = 0
Explanation:
To find an equation of the plane that contains the curve with the vector equation R(t) = (t, t^3, t), we can use the fact that a plane can be defined by a point and a normal vector to the plane. We can choose any point on the curve as a point on the plane, say (0, 0, 0), and find a normal vector to the plane by taking the cross product of the tangent vectors to the curve at two different points.
To find the tangent vector to the curve at a point (t, t^3, t), we can take the derivative of the vector equation with respect to t:
R'(t) = (1, 3t^2, 1)
So, the tangent vector to the curve at (t, t^3, t) is (1, 3t^2, 1).
Now, we can find the normal vector to the plane by taking the cross product of the tangent vectors at two different points on the curve. Let's choose the points (0, 0, 0) and (1, 1, 1) on the curve:
Tangent vector at (0, 0, 0): R'(0) = (1, 0, 1)
Tangent vector at (1, 1, 1): R'(1) = (1, 3, 1)
The normal vector to the plane is the cross product of these two tangent vectors:
N = R'(0) x R'(1) = (-3, 0, 3)
Now, we can use the point-normal form of the equation of a plane to find the equation of the plane that contains the curve:
N · (r - P) = 0, where N is the normal vector to the plane, P is a point on the plane, and r is a point on the plane.
Substituting in the values we have, we get:
(-3, 0, 3) · (r - (0, 0, 0)) = 0
Simplifying this equation gives us:
-3x + 3z = 0
Therefore, the equation of the plane that contains the curve with the vector equation R(t) = (t, t^3, t) is -3x + 3z = 0.