00:01
For this problem, we are asked in part a to use a differential to approximate the change in the volume of a right circular cylinder if the height changes from 2 to 2 .1 centimeters and the radius changes from 0 .5 to 0 .49 centimeters.
00:18
So to begin, we have that the volume of our right circular cylinder v equals pi times r squared times h.
00:27
We have that r -not the initial radius is two we have that delta r equals 0 .1 we have that h -not equals 0 .5 and delta h equals negative 0 .01 putting this together then we can approximate the change in the volume using our differential so dv would be equal to first partial derivative with respect to r, so it would be 2r -py, or let me change how i write this here, it would be 2 -py -r -h -d -r.
01:13
Then we'd have plus partial derivative with respect to h, which would be pi -r -squared, d -h.
01:22
Now plugging in our r -not, our h -0, and our deltas, we'd have that this is 2 times pi times 2 times 0, or one half times dr, so times 0 .1.
01:37
And then we'd have plus pi times r not squared, so pi times 2 squared, times dh, which is negative 0 .01.
01:46
So we'll find then dv will be equal to.
01:49
So the first term, we can write as 0 .2 pi.
01:54
And the second would be negative 0 .04 pi...