Find the components of velocity and acceleration in the direction of a given unit vector at t=1.

Hey Alaks, good to see you again. To find the components of velocity and acceleration in the given direction, we first need to find the velocity and acceleration vectors, evaluate them at t=1, and then calculate their scalar projections onto the unit vector in the specified direction. The position of the particle is given by: x(t) = 2t + 3 y(t) = t^2 + 3t z(t) = t^3 + 2t^2 Step 1: Find the velocity vector v(t). The velocity vector is the first derivative of the position vector with respect to time. v(t) = (dx)/(dt) i + (dy)/(dt) j + (dz)/(dt) k (dx)/(dt) = (d)/(dt)(2t + 3) = 2 (dy)/(dt) = (d)/(dt)(t^2 + 3t) = 2t + 3 (dz)/(dt) = (d)/(dt)(t^3 + 2t^2) = 3t^2 + 4t So, the velocity vector is: v(t) = 2i + (2t + 3)j + (3t^2 + 4t)k Step 2: Evaluate the velocity vector at t=1. Substitute t=1 into the velocity vector: v(1) = 2i + (2(1) + 3)j + (3(1)^2 + 4(1))k v(1) = 2i + (2 + 3)j + (3 + 4)k v(1) = 2i + 5j + 7k Step 3: Find the acceleration vector a(t). The acceleration vector is the first derivative of the velocity vector (or the second derivative of the position vector) with respect to time. a(t) = (d^2x)/(dt^2) i + (d^2y)/(dt^2) j + (d^2z)/(dt^2) k (d^2x)/(dt^2) = (d)/(dt)(2) = 0 (d^2y)/(dt^2) = (d)/(dt)(2t + 3) = 2 (d^2z)/(dt^2) = (d)/(dt)(3t^2 + 4t) = 6t + 4 So, the acceleration vector is: a(t) = 0i + 2j + (6t + 4)k Step 4: Evaluate the acceleration vector at t=1. Substitute t=1 into the acceleration vector: a(1) = 0i + 2j + (6(1) + 4)k a(1) = 0i + 2j + (6 + 4)k a(1) = 0i + 2j + 10k Step 5: Find the unit vector in the given direction. The given direction vector is d = 2i + 3j + 4k. The magnitude of d is: |d| = sqrt(2^2 + 3^2 + 4^2) = sqrt(4 + 9 + 16) = sqrt(29) The unit vector u in this direction is: u = d|d| = (1)/(sqrt(29))(2i + 3j + 4k) Step 6: Calculate the component of velocity in the given direction. The component of velocity v(1) in the direction of u is given by the dot product v(1) · u: Component of velocity = (2i + 5j + 7k) · (1)/(sqrt(29))(2i + 3j + 4k) = (1)/(sqrt(29))((2)(2) + (5)(3) + (7)(4)) = (1)/(sqrt(29))(4 + 15 + 28) = (47)/(sqrt(29)) Step 7: Calculate the component of acceleration in the given direction. The component of acceleration a(1) in the direction of u is given by the dot product a(1) · u: Component of acceleration = (0i + 2j + 10k) · (1)/(sqrt(29))(2i + 3j + 4k) = (1)/(sqrt(29))((0)(2) + (2)(3) + (10)(4)) = (1)/(sqrt(29))(0 + 6 + 40) = (46)/(sqrt(29)) The component of velocity in the direction of 2i + 3j + 4k at t=1 is (47)/(sqrt(29)). The component of acceleration in the direction of 2i + 3j + 4k at t=1 is (46)/(sqrt(29)). Send me the next one 📸