Answer:
Step-by-step 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
[tex]V=\pi r^2 h[/tex]
Given that [tex]\frac{dr}{dt} =0.13[/tex]
h is constant
[tex]V= 3.14 (r^2) (13.7)\\\frac{dV}{dt} =6.28 (137) r dr/dt\\= 6.28(137)\frac{ (97)}{2} 0.13[/tex]
= 6954.67 mm/sec^3
