An engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min. How fast is the volume V of the cylinder changing when the diameter is 9…

Question

asked Jun 28, 2020 233k views

1 Answer

Ruben Sancho Ramos · Answer 1 · 2020-07-04T04:29:51+0000

Answer:

Explanation:

Given that an engine cylinder 13.7 cm deep is being bored such that the radius increases by 0.13 mm divided by min.

We convert everything in mm for uniformity

We have formula for volume as

$V=\pi r^2 h$

Given that
(dr)/(dt) =0.13

h is constant

$V= 3.14 (r^2) (13.7)\\(dV)/(dt) =6.28 (137) r dr/dt\\= 6.28(137)( (97))/(2) 0.13$

= 6954.67 mm/sec^3