PROFESSOR: An object moves in a path defined by the vector value function r of t equal to 4ti plus t squared j where t measures time in seconds. Part A, find the tangential component of a celebration and the normal component of a celebration as functions of t. And part B, find the tangential component of celebration and the normal component of acceleration at time t equals negative 3. So if we've determined the functions for part A, we can evaluate that for part B.

Now, the first thing we need to do is identify what the velocity and acceleration functions are because the tangential component of acceleration is equal to-- there are many formulas, but the one I'm going to use is that the tangential component of acceleration is equal v.a divided by the magnitude of v. So v of t is the derivative of r of t. That would be 4i plus 2tj, which would make the acceleration the derivative of the velocity 2j.

All right. Now to go with that, we need the velocity or the magnitude-- the magnitude of the velocity. So the magnitude of v of t is going to be equal to the square root of 4 squared plus 2t quantity squared. It's actually our speed, right? That would be equal to 2 square root of 4 plus t squared in that square root. Now fitting this into our equation, v dotted with a is going to be equal to 4t-- so that's our numerator-- divided by 2 square root plus t squared in that radical. And that would simplify so that are tangential component is 2t over the square root of 4 plus t squared in that radical.

Now we also need the normal component of acceleration. So the normal component of acceleration is defined as the square root of a large quantity the magnitude of acceleration squared minus the tangential component of acceleration squared. Now our acceleration-- the magnitude of acceleration is equal to 2. So going back to our function of 2j-- a of t equals 2j.

The magnitude of that is just 2. So this is equal to the square root of 2 squared 4 minus 2t over the square root of 4 plus t squared in that square root. And then square that quantity, and then in the large radical we have.

Now squaring the tangential component and combining these two with a common denominator, we end up with the square root of a fraction 16 over 4 plus t squared, so that the normal component of acceleration is equal to 4 over the square root of 4 plus t squared in that radical in the denominator. Now, if we evaluate for part B, the tangential component at a value of negative 3, the tangential component at negative 3 is going to be equal to negative 6 over the square root of 13, and then rationalizing that is equal to negative 6 square root of 13 all over positive 13. All right. And then evaluating the normal component at t equals negative 3. That would be equal to 4 over the square root of 13, which is equal to 4 square root of 13 all over 13.