Mathematical Immaturity

14.15 Exercises

• If two plane curves with Cartesian equations $y = f(x)$ and $y = g(x)$ have the same tangent at a point $(a, b)$ and the same curvature at that point, prove that $|f''(a)| = |g''(a)|.$
• Use the result of Exercise 6 (b)
• Let the two plane curves be described by the positional vectors $\mathbf{q}(x)$ and $\mathbf{r}(x)$ where: $$ \begin{align*} \\ \mathbf{q}(x) &= x\,\mathbf{i} + f(x)\,\mathbf{j} \\ \\ \mathbf{r}(x) &= x\,\mathbf{i} + g(x)\,\mathbf{j} \end{align*} $$ Their respective velocity and acceleration vectors are $$ \begin{align*} \\ \mathbf{q}'(x) &= \mathbf{i} + f'(x)\,\mathbf{j} \\ \mathbf{q}''(x) &= f''(x)\,\mathbf{j} \\ \\ \mathbf{r}'(x) &= \mathbf{i} + g'(x)\,\mathbf{j} \\ \mathbf{r}''(x) &= g''(x)\,\mathbf{j} \end{align*} $$ Now, recall from Exercise 6 (b) that for the above-defined Cartesian equations, their respective curvatures are $$ \begin{align*} \\ \kappa_q(x) &= \frac{|f''(x)|}{\left\{1 + \left[f'(x)\right]^2\right\}^{3/2}} \\ \\ \kappa_r(x) &= \frac{|g''(x)|}{\left\{1 + \left[g'(x)\right]^2\right\}^{3/2}} \end{align*} $$ Since these two values are equal at $x = a,$ we see that $$ \begin{align*} |f''(a)|\left\{1 + \left[g'(a)\right]^2\right\}^{3/2}&= |g''(a)|\left\{1 + \left[f'(a)\right]^2\right\}^{3/2} \tag{8.1} \end{align*} $$ But since $f$ and $g$ have the same tangent line at $x = a,$ this means that $f'(a) = g'(a)$ and by extension, $$ \begin{align*} \left\{1 + \left[g'(a)\right]^2\right\}^{3/2}&= \left\{1 + \left[f'(a)\right]^2\right\}^{3/2} \end{align*} $$ Dividing both sides of (8.1) by $\left\{1 + \left[g'(a)\right]^2\right\}^{3/2}$ (since this value can never be zero for any real-valued g'(a)), we arrive at the conclusion $$ \begin{align*} \\ |f''(a)| &= |g''(a)| \quad \blacksquare \\ \\ \end{align*} $$