A set of plastic spheres are to be made with a diameter of 1cm. If the manufacturing process is accurate to 1mm, what is the propagated error in the volume of the spheres?
The formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \] where \(r\) is the radius of the sphere. The relationship between the radius and diameter \(d\) of the sphere is \(r = \frac{d}{2}\). So the volume in terms of diameter is: \[ V = \frac{4}{3} \pi \left(\frac{d}{2}\right)^3 = \frac{4}{3} \pi \frac{d^3}{8} = \frac{\pi d^3}{6} \]
The propagated error in the volume \(V\) when the diameter \(d\) has an error \(\Delta d\) can be found using the derivative of \(V\) with respect to \(d\), multiplied by \(\Delta d\): \[ \frac{dV}{dd} = \frac{\pi}{6} \cdot 3d^2 = \frac{\pi d^2}{2} \] \[ \Delta V = \frac{dV}{dd} \cdot \Delta d = \frac{\pi d^2}{2} \cdot \Delta d \]
Given:
The propagated error in the volume \(\Delta V\) is: \[ \Delta V = \frac{\pi (0.01)^2}{2} \cdot 0.001 \]
d <- 0.01 # diameter
delta_d <- 0.001 # error in diameter
# propagated error in volume
delta_V <- (pi * d^2 / 2) * delta_d
delta_V## [1] 1.570796e-07