Final answer:
The query involves finding a point on a curve where the tangent is parallel to a plane. We determine the velocity vector for the curve and set its dot product with the plane's normal vector to zero to find the specific parameter value. Substitute this value back into the curve to get the required point.
Step-by-step explanation:
The student's question involves finding a point on a parametric curve in three-dimensional space, where the tangent line to the curve at that point is parallel to a given plane. The curve is represented by the vector function r(t) = <4 cos(t), 4 sin(t), e^t> and the plane is described by the equation 3x + y = 1. To solve this, we first need to find the velocity vector r'(t) which is the derivative of r(t) with respect to t. This gives us the direction of the tangent at any point on the curve.
Once the velocity vector r'(t) is obtained, the condition for parallelism to the plane needs to be established. Specifically, we need to ensure that the directional vector of the tangent line (velocity vector) does not have a component in the direction normal to the plane.
The normal vector to the given plane is <3, 1, 0>. We can then use the dot product between the velocity vector and this normal vector and set it equal to zero which will provide us with the specific t value at which the tangent line is parallel. With this t value, we can substitute back into the original curve equation r(t) to find the desired point.