The potential energy VVV, in Joules, of a particle in a localized field is modelled by the function V(r)=(2r−5)4e−2rV(r) = (2r - 5)^4 e^{-2r}V(r)=(2r−5)4e−2r, where r r\,r is the distance from a fixed origin in centimeters.
Show that the rate of change of potential energy with respect to distance is given by
dVdr=K(2r−5)3(9−2r)e−2r \frac{dV}{dr} = K(2r - 5)^3(9 - 2r)e^{-2r} drdV=K(2r−5)3(9−2r)e−2rwhere K K\,K is a constant to be determined.
Hence find the exact coordinates of the two stationary points of the function V(r)V(r)V(r).
A second particle's potential energy is modelled by the function W(r)W(r)W(r), where
W(r)=3V(r−2) W(r) = 3V(r - 2) W(r)=3V(r−2)Determine the coordinates of the maximum stationary point for the function W(r)W(r)W(r).
133 exam-style questions on WJEC A Level Maths 3.6.5 Differentiation (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.