A researcher models the radial gradient of a potential field V V\,V using the equation dVdr=1r2r2−16\displaystyle \frac{dV}{dr} = \frac{1}{r^2 \sqrt{r^2 - 16}}drdV=r2r2−161 for r>4r > 4r>4. To solve for VVV, the researcher considers the transformation y=secϕy = \sec \phiy=secϕ.
(i) Express y y\,y in terms of cosϕ\cos \phicosϕ.
(ii) Hence, show that dydϕ=secϕtanϕ\displaystyle \frac{dy}{d\phi} = \sec \phi \tan \phidϕdy=secϕtanϕ.
(iii) Show that for 0<ϕ<π2\displaystyle 0 < \phi < \frac{\pi}{2}0<ϕ<2π, y2−1y=sinϕ\displaystyle \frac{\sqrt{y^2-1}}{y} = \sin \phiyy2−1=sinϕ.
(i) Use the substitution r=4secϕr = 4 \sec \phir=4secϕ to show that for r>4r > 4r>4, the integral ∫1r2r2−16 dr\displaystyle \int \frac{1}{r^2 \sqrt{r^2 - 16}} \, dr∫r2r2−161dr can be written as k∫cosϕ dϕ k \int \cos \phi \, d\phi\,k∫cosϕdϕ where k k\,k is a constant to be determined.
(ii) Hence, show that ∫1r2r2−16 dr=r2−1616r+C\displaystyle \int \frac{1}{r^2 \sqrt{r^2 - 16}} \, dr = \frac{\sqrt{r^2 - 16}}{16r} + C∫r2r2−161dr=16rr2−16+C.
