Given that u(x) is always positive and u'(x) < 0, I need to find values of x so that f(x) and g(x) are increasing. f(x) = [u(x)]^2 and g(x) = u(u(x)).
for f(x) is increasing when f'(x) > 0. so f'(x) = 2u(x) => 2u(x) > 0. would f(x) always be increasing since 2u(x) will always be increasing (...
well then, thanks again quasar. so was it irrelevant to ask to find the rate of change when the radius was 2, meaning its a constant rate of change when the volume and surface area decrease proportionally?
thanks quasar, with that information ill try the problem.
\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}
Since \frac{dV}{dt}=-3A
Then -3(4{\pi}r^2)=\frac{dV}{dr}\frac{dr}{dt}
But now it seems if i differentiate the volume with respect to the radius ill get the equation for surface...
This is the problem as it appears in the text. "As a spherical raindrop evaporates, its volume changes at a rate proportional to its surface area A. If the constant of proportionality is 3, find the rate of change of the radius r when r=2." My first question is does the constant of...