Final answer:
To find the unit tangent and unit normal vectors, we first need to find the velocity and acceleration vectors.
Step-by-step explanation:
To find the unit tangent and unit normal vectors, we first need to find the velocity and acceleration vectors. The velocity vector is found by taking the derivative of the position function, which is r(t) = 2tî + 5cos(t)ĵ + 5sin(t)k. Taking the derivative, we get v(t) = 2î - 5sin(t)ĵ + 5cos(t)k. The acceleration vector is found by taking the derivative of the velocity vector, which is a(t) = 0î - 5cos(t)ĵ - 5sin(t)k
Next, we need to find the magnitude of the acceleration vector at a specific time. Substituting t = 2 into the acceleration function, we get a(2) = 0î - 5cos(2)ĵ - 5sin(2)k. The magnitude can be found using the formula |a(2)| = sqrt((0)^2 + (-5cos(2))^2 + (-5sin(2))^2) = sqrt(0 + 25cos^2(2) + 25sin^2(2)) = sqrt(25) = 5 m/s^2.
Finally, the unit tangent vector t(t) is found by dividing the velocity vector by its magnitude, t(t) = (v(t))/|v(t)|. The unit normal vector n(t) is found by dividing the acceleration vector by its magnitude, n(t) = (a(t))/|a(t)|. Hope this helps!