00:01
In this problem we are asked to find a point p that lies on the curve r of t which equals to t squared times i plus 2t times j plus 2 t times k and that is closest to the point 4 .25.
00:25
Any point that lies on this curve is of the form t squared 2 t 2 t and now since we require this point to be closest to the given point let us find out the distance between both of these so we have d to be equal to square root of t squared minus 4 the whole squared plus 2 t minus 4 the whole squared plus 2 t minus 25 the whole squared plus 2 t minus 2 the whole squared.
01:07
Squaring on both sides we get t squared equals to expanding each of these we have t raised to the power 4 minus 8 t squared plus 16 expanding the second one we have 4 t squared minus 100 times t plus 625 plus 4 t squared.
01:30
Minus 8 times t plus 4.
01:34
So here, negative 8 t squared, positive 4t squared and positive 4 t squared get cancelled.
01:43
Combining the like terms we have d squared to be equal to t raised to the power 4 minus 1 .8 t plus 645.
01:53
So now let us find the first derivative of this function.
02:04
And let us denote this by capital d.
02:06
So we have d prime to be equal to 4 times t cubed minus 1 .8.
02:13
Equating this first derivative to 0 and solving for t, we obtain the critical points.
02:19
So we have 4 t cubed equals to 108, which implies that t cubed equals to 108 equals to 108 divided by 4, which is 27, which further implies that t cubed, equals to positive 3.
02:38
So now let us find the second derivative.
02:46
So we have d double prime to be equal to 4 times 3 t squared and this equals to 12 t squared which is clearly greater than 0 when t equals to 3...