00:01
The sides of a cube increase so that its volume increases at a rate t cubic inches per minute.
00:17
Now, the volume of a cube with side length s is v is equal to s cube.
00:29
Let the side length of the cube be s of t at time t.
00:35
As the volume increases at a rate of p cubic inches per minute, so it can be expressed as dv by dt is equal to p, where dv by dt is the rate change of the volume with respect to time.
00:57
Next, express the surface area of the cube as a function of side length s, which can be expressed as a is equal to 6s square.
01:15
Now since it has been given that q is related to side length q is equal to 6s square, that implies s square is equal to q by 6.
01:38
Now differentiate both sides with respect to time t.
01:43
So, we have 2s into ds by dt is equal to 1 by 6 into dq by dt.
01:59
Now find ds by dt which is the rate at which the side length of the cube is changing with respect to time t.
02:12
To do that, differentiate the volume formula with respect to t...