00:01
In this problem, we have to use our knowledge of differentiation to optimize a specific function we're given.
00:09
And the function that we're given deals with various things like distance and speed.
00:17
So we're given this function we'll call e of v equals a, v cubed, times l over v minus u.
00:25
Well, we can simplify that a little bit to say e of v equals a, v cubed, times l, l, all over v minus u.
00:33
What we want is we want the derivative equal to zero.
00:37
So we'll take the derivative of this function and we'll set it equal to zero.
00:42
We'll have 2 a v cubed times l minus 3a u lv squared equal to 0.
00:50
We can simplify a little bit and cancel out some terms.
00:54
We'll have 2v cubed minus 3 u v squared equal to 0 and then we can factor out a common fact essentially.
01:03
We have v squared and we'll multiply that by the quantity to v minus 3u equal to 0...