# The mechanics at lincoln automotive are reboring a 6-in deepcylinder to fit a new piston. The machine they are usingincreases the cylinder's radius one-thousandth of an inch every 3min. How rapidly is the cylinder volume increasing when thebore(diameter) is 3.800 in

The mechanics at lincoln automotive are reboring a 6-in deepcylinder to fit a new piston. The machine they are usingincreases the cylinder's radius one-thousandth of an inch every 3min. How rapidly is the cylinder volume increasing when thebore(diameter) is 3.800 in
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Elberte

We know that the volume of a cylinder is given by V=pi*r^2*h, butfor this question, the height is fixed at 6 in. Our volumefunction becomes $V=6\cdot \pi \cdot {r}^{2}$, where both V and r are functions oftime. The "borer" is increasing the radius at the rate of0.001 in/3 min; that is the value of dr/dt.
$V=6\pi {r}^{2}\to \frac{dV}{dt}=12\pi r\frac{dr}{dt}$
Now, substitute   $\frac{dV}{dt}=12\pi \cdot 3.800\cdot \frac{0.001}{3}=0.0477522\dots \frac{{\in }_{3}}{min}$

###### Not exactly what you’re looking for?
So what Elberte was on the right track, but it is still wrong. Understand everything Elberte presented, but notice that at the end when plugging in r, instead of dividing the diameter by 2, he just plugged the entire diameter into the equation as if it were the radius. So look at the question and understand that the diameter is said to be 3.8, not the radius. Divide that by two when plugging into the radius.