Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• (a) Let $P$ be a point on the parabola $y = x^2.$ Let $Q$ be the point of intersection of the normal line at $P$ with the $y$-axis. What is the limiting position of $Q$ as $P$ tends to the $y$-axis? (b) Solve the same problem for the curve $y = f(x),$ where $f'(0) = 0.$
• (a) Turn the equation of the parabola into a vector $P$ parameterized by $x.$ What is the derivative of $P?$ What is the vector $N$ normal to $P?$ How can we use the line described by $P$ and $N$ to find $Q?$ (b) The approach from (a) can be used in the general case where $y = f(x)$ for some real-valued $f.$ Use L'Hôpital's rule (Theorem 7.9) for the indeterminate form if applicable.
• (a) We can describe any point $P$ on the curve with the parametric vector $P(x) = (x, x^2).$ Its derivative $P'(x)$ is then $(1, 2x).$ Thus, the line normal to the curve at point $P$ must be parallel to the vector $N = \left(1, -\frac{1}{2x}\right)$ for $x \neq 0.$ Then, by definition, the line $L,$ normal to the curve at $P = (x, x^2),$ is the following set of points: \begin{align*} \\ L &= \{P + tN\, |\, t\ \text{real}\} \end{align*} Or in other words, for all real $t,$ $L$ is the set of points $\left(x + t, x^2 - \frac{t}{2x}\right).$ Setting $t = -x$ we find the point $Q$ at which the normal line intersects the $y$-axis. That is, \begin{align*} Q = \left(0, x^2 + \frac{1}{2}\right) \end{align*} As $x \rightarrow 0,$ we find that $Q \rightarrow \left(0, \frac{1}{2}\right).\,\blacksquare$ (b) We can use the same approach as in part (a) for the general case $y = f(x).$ We first turn the point $P$ into a vector parameterized by $x,$ namely $P = \left(x, f(x)\right).$ Then, its derivative is $P' = \left(1, f'(x)\right).$ The normal line at $P$ is then parallel to $N = \left(1, -\frac{1}{f'(x)}\right)$ for $f'(x) \neq 0,$ and the normal line $L$ is defined as: \begin{align*} \\ L &= \{P + tN\, |\, t\ \text{real}\} \end{align*} In other words, $L$ is the set of points $\left(x + t, f(x) - \frac{t}{f'(x)}\right)$ for all real $t$ and where $f'(x) \neq 0.$ Then, setting $t = -x,$ we see that the $y$-intercept of the normal line is the point \begin{align*} \\ Q &= \left(0, f(x) + \frac{x}{f'(x)}\right) \end{align*} Taking the limit of $Q$ as $x \rightarrow 0$ gives us the indeterminate form $\frac{0}{0},$ since $f'(0) = 0.$ We can then use L'Hôpital's rule for the indeterminate form to give us \begin{align*} \\ \lim_{x \to\,0} Q &= \left(0, f(x)\right) + \left(0, \frac{1}{f''(x)}\right) \\ \\ &= \lim_{x \to\,0} \left(0, f(x) + \frac{1}{f''(x)}\right). \end{align*} If $f''(0) \neq 0,$ \begin{align*} \\ \lim_{x \to\,0} Q &= \left(0, f(0) + \frac{1}{f''(0)}\right) \end{align*} Otherwise, \begin{align*} \\ \lim_{x \to\,0} \left|f(0) + \frac{1}{f''(0)}\right| &= +\infty \quad \blacksquare \end{align*}