00:01
In this problem, we have volume as a function of r given as k times r power 4, r0 minus r with r lying between r0 by 2 and r0.
00:21
Now, we want to find the interval in which v is increasing or decreasing.
00:26
So for that, we will find first the critical points.
00:40
So to find the critical point, we have v dash of r must be 0.
00:48
V dash of r is nothing but k into r power 4 into minus 1 plus k into r0 minus r into 4r cube.
01:02
This must be 0.
01:03
Now, taking k times r cube common, we have minus r plus 4 times r0 minus r.
01:17
This must be 0.
01:20
R cannot be 0 because it is lying between r0 by 2 and r0.
01:24
So we have simply minus r plus 4r0 minus 4r must be 0 or 4r0 must be equals to 5r or r is equals to 4r0 by 5.
01:42
So we have on the number line, we have this as r0 by 2, this as r0.
01:52
We have 4r0 by 5 somewhere here.
01:57
And let us check the signs of v dash of r.
02:02
So let us take point here.
02:07
Let us take, say, 3r0 by 5, which will be somewhere here.
02:24
So at 3r0 by 5, we substitute v dash at 3r0 by 5.
02:35
This one from here, it will be simply kr cube.
02:41
This is minus r.
02:43
And this becomes minus again, 4 times...