Water flows onto a flat surface at a rate of 5 \(cm^3/s\) forming a circular puddle 10 mm deep. How fast is the radius growing when the radius is: (a) 1 cm? (b) 10 cm? (c) 100 cm?
We know that the volume of a cylinder is \(V = \pi r^2h\). Using the product rule, \(\frac {d(uv)}{dt} = u\frac{dv}{dt} + v\frac{du}{dt}\), we can differentiate time (t) where \(u = r^2\) and \(v = h\).
\[ \frac{dV}{dt} = \pi (h \frac{d}{dt}r^2 + r^2\frac{d}{dt}h) = \pi (2rh\frac{dr}{dt} + r^2\frac{dh}{dt}) \]
We also know that \(\frac{dV}{dt} = 5\) \(cm^3/s\), h is constant at 10 mm deep and \(\frac{dh}{dt} = 0\) as there is not a change in height as time increases
This leaves us with \(\frac{dr}{dt} = \frac {5}{2 \pi r}\)
Lets create a function that can house this equation and solve fr our three scenarios.
speed_formula <- function(radius) {
return (5/(2 * pi *radius))
}speed_formula(1)## [1] 0.7957747speed_formula(10)## [1] 0.07957747speed_formula(100)## [1] 0.007957747