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).
717 exam-style questions on Edexcel A Level Maths Differentiation, covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change. Each one has a worked solution and a mark scheme showing where the marks go.