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2016 Paper 1 Q4
D: 1516.0 B: 1502.9
differential equations differentiation curvature quotient rule circle chain rule geometric interpretation first order ODE

  1. Differentiate $\displaystyle \; \frac z {(1+z^2)^{\frac12}} \;$ with respect to \(z\).
  2. The {\em signed curvature} \(\kappa\) of the curve \(y=\f(x)\) is defined by \[ \kappa = \frac {\f''(x)}{\big({1+ (\f'(x))^2\big)^{\frac32}}} \,.\] Use this definition to determine all curves for which the signed curvature is a non-zero constant. For these curves, what is the geometrical significance of \(\kappa\)?

Solution:

  1. Let \(\displaystyle y = \frac z {(1+z^2)^{\frac12}}\) then \(\frac{d y}{d x} = \frac{(1+z^2)^{\frac12} - z^2(1+z^2)^{-\frac12}}{1+z^2} = \frac{(1+z^2)-z^2}{(1+z^2)^\frac32} = \frac{1}{(1+z^2)^\frac32}\)
  2. \(\kappa = \frac {f''(x)}{\big({1+ (f'(x))^2\big)^{\frac32}}}\) then \begin{align*} && \int \kappa \, dx &= \int \frac{f''(x)}{( 1 + (f'(x))^2)^{\frac32}} \, dx \\ && \kappa x &= \frac{f'(x)}{(1 + (f'(x))^2)^\frac12} + C \\ \Rightarrow && (\kappa x-C)^2 &= \frac{f'(x)^2}{1 + (f'(x))^2} \\ \Rightarrow && f'(x)^2((\kappa x - C)^2 - 1) &= -(\kappa x-C)^2 \\ \Rightarrow && f'(x) &= \frac{\kappa x - C}{\sqrt{1-(\kappa x - C)^2 }} \\ \Rightarrow && f(x) &= \frac{1}{\kappa} \sqrt{1 - (\kappa x - C)^2} \\ \Rightarrow && (\kappa y)^2 + (\kappa x - C)^2 &= 1 \\ \Rightarrow && y^2 + (x - C')^2 &= \frac{1}{\kappa^2} \end{align*} Therefore all the curves are circles and \(\kappa\) is the reciprocal of the radius.

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