Question

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 9.7 cm question mark Use 3.14 for pi. Round to the nearest hundredth.

Answer 1

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

$V=\pi r^2 h$

Given that $\frac{dr}{dt} =0.13$

h is constant

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

= 6954.67 mm/sec^3