The radius of a 10 inch right circular cylinder is measured to be 4 inches, but with a possible error of ±0.1 inch. What is the resulting possible error in the volume of the cylinder? Include units in your answer.

Question

anonymous · Answer

@freckles @Luigi0210 @Nnesha @Compassionate can u plz help? :)

anonymous · Answer

@iambatman

freckles · Answer

$$\Delta V \approx \Delta x \cdot \frac{dV}{dx}|_{x=a} $$
So we are given $$\Delta x =\pm 0.1 \ V=\pi r^2 h $$
$$\text{ We want to find } \Delta V $$
Hmmm I wonder if both r and h are changing...

freckles · Answer

It sounds like the radius is the one with the possible error right?

freckles · Answer

So we will assume h is constant and take derivative of V with respect to r 
And the equation I have above we will replace x with r instead

freckles · Answer

$$V=\pi h r^2 $$
So I assume you can find the d (r^2)/dr

freckles · Answer

$$\frac{dV}{dr}=\frac{d( \pi h r^2)}{dr}= \pi h \frac{d(r^2)}{dr} \ \frac{dV}{dr}=\pi h \frac{d (r^2)}{dr}$$