Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 9 / b / iv

Retain the question's common crossing condition $y^{\prime }(0)=1$, and replace the slope-ratio condition of part (iii) by the specified initial slope. Then part (i) givesso $\alpha =7$. Substitution in part (ii) gives the rayIt starts at $(-7,0)$ with slope $1/2$, passes through $(0,9/2)$ with slope one, and reaches the receiver with slope $1/\sqrt{8}$. The final height isThe left-hand curvature is $y^{\prime \prime }=\frac{1}{3}(1-x{)}^{-4/3}>0$; the right-hand curvature is $y^{\prime \prime }=-\frac{1}{2}(1+x{)}^{-3/2}<0$. Thus the graph is increasing throughout, with a change from upward to downward curvature at the interface.

If the word “instead” were also taken to remove the common crossing condition, the initial slope alone would leave $\alpha$ undetermined: it would give $p=\mu (-\alpha )/\sqrt{5}$ for any selected source position. The numerical answer above uses the crossing condition stated before all four subparts.