Question

An airplane moves in a trajectory given by r(t)=4ti+tj+t 2

k. Given this trajectory, the airplane will intersect the surface z−x−y 2twice. (a) Find the time, t, when the airplane intersects the surface the 2nd time. (b) Find the two unit tangent vectors to the airplane's trajectory at the time in (a). (c) Determine the line tangent to the airplane's trajector at the 2nd time it hits the surface.

Answer 1

To find the time when the airplane intersects the surface for the second time, set z(t) equal to -x(t) - y(t)^2 and solve for t. The airplane intersects the surface for the second time at t = 0. The unit tangent vectors at t = 0 are V(t) = 4i + j + 2k and V(t) = 4i + j - 2k. To determine the line tangent to the trajectory at the second time it hits the surface, use the derivative of the trajectory equation at t = 0.

Step-by-step explanation:

(a) To find the time when the airplane intersects the surface for the second time, we need to equate the z-coordinate of the trajectory to -x-y^2 and solve for t.

Setting z(t) = -x(t) - y(t)^2, we get t^2 = -3t, which gives us t = 0 or t = -3.

Since time cannot be negative, the airplane intersects the surface for the second time at t = 0.

(b) To find the unit tangent vectors at t = 0, we differentiate the trajectory equation with respect to t twice to get a(t). Plugging in t = 0, we find that the unit tangent vectors are V(t) = 4i + j + 2k and V(t) = 4i + j - 2k.

(c) To determine the line tangent to the trajectory at the second time it hits the surface, we use the derivative of the trajectory equation at t = 0. The line tangent to the trajectory is given by r(t) = r(0) + V(t)(t - 0), where r(0) is the position vector at t = 0 and V(t) is the unit tangent vector at t = 0.