00:01
So the question asked to compute the following derivatives, and we're given d over dt within parentheses of t -cubbed i minus t2t, j minus 2k times t -i minus t -square -j minus t -qubukk.
00:21
And when we're doing a multiplication sign, we're basically multiplying t -terms from i, j, and k.
00:26
So essentially, this is the same thing as doing the derivative, so d of d -t, of this whole thing within.
00:36
We're going to have t to the fourth because t cubed plus t goes t to the fourth or t to the fourth i which is a vector then we do minus two t times minus t squared which can be plus two t cubed so plus two t cubed j and then we're doing minus two k times minus two t cubed k which is going to be plus two t cubed okay so now we're going to do derivative of each individual term and turn this into a vector so doing the derivative of our first term in which we're given t to the fourth because this is going to relate to a vector in which i is the x coordinate.
01:28
So this is the x coordinate.
01:30
This is the y coordinate is the z coordinate.
01:32
T to the fourth, same thing.
01:34
We'll subtract one and multiply by our power and then subtract one.
01:38
We get 4t cubed.
01:41
J...