Application of Derivative in Geometry

Let $y=f(x)$ be a continuous function of $x$ representing a curve in the $XY$-plane, and let $P(x_1,y_1)$ be any point on the curve. The derivative of $y$ with respect to $x$, evaluated at $P$, is written $\left.\dfrac{dy}{dx}\right|_{(x_1,y_1)} = [f'(x)]_{(x_1,y_1)}$, and this quantity represents the slope (also called the gradient) of the tangent to the curve at $P$.

The normal to the curve at $P$ is defined as the line through $P$ that is perpendicular to the tangent there. Since two perpendicular lines have slopes that multiply to $-1$, if $m$ is the tangent's slope and $m'$ is the normal's slope, then $m'=-\dfrac{1}{m}$, valid whenever $m\ne0$.

Equation of the tangent at $P(x_1,y_1)$: using the point-slope form of a line, $y-y_1=m(x-x_1)$, i.e. $y-y_1=\left.\dfrac{dy}{dx}\right|_{(x_1,y_1)}(x-x_1)$.

Equation of the normal at $P(x_1,y_1)$: $y-y_1=m'(x-x_1)$, where $m'=-\dfrac{1}{\left.dy/dx\right|_{(x_1,y_1)}}$ (provided $\left.dy/dx\right|_{(x_1,y_1)}\ne0$).

When the curve is given by an equation that mixes $x$ and $y$ together (an implicit equation) rather than solved explicitly as $y=f(x)$, the slope $dy/dx$ is still found by differentiating both sides of the equation with respect to $x$, treating $y$ as an (unknown) function of $x$ — every term containing $y$ picks up a factor of $dy/dx$ by the chain rule, and every product of $x$ and $y$ needs the product rule. Once every term is differentiated, the equation is solved algebraically for $dy/dx$, which can then be evaluated at the specific point of interest.

When the curve is given parametrically, with both $x$ and $y$ expressed in terms of a third variable (a parameter, commonly $t$ or $\theta$), the slope is found using $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ (or with $\theta$ in place of $t$) — each of $x$ and $y$ is differentiated separately with respect to the parameter, and the two derivatives are divided.

Worked Example 1 — Tangent and normal at a point, three different curve forms

(i) For $y=2x^3-x^2+2$ at the point $\left(\dfrac12,2\right)$: differentiating gives $\dfrac{dy}{dx}=6x^2-2x$. At $x=\dfrac12$: slope $m=6\left(\dfrac14\right)-2\left(\dfrac12\right)=\dfrac32-1=\dfrac12$. Tangent: $y-2=\dfrac12\left(x-\dfrac12\right)$, which simplifies to $2x-4y+7=0$. The normal's slope is $m'=-2$, giving $y-2=-2\left(x-\dfrac12\right)$, which simplifies to $2x+y-3=0$.

(ii) For the implicit curve $x^3+2x^2y-9xy=-2$ at $(2,1)$: differentiating every term with respect to $x$ (using the product rule on $2x^2y$ and on $9xy$) and collecting the $dy/dx$ terms gives $\dfrac{dy}{dx}=\dfrac{9y-4xy-3x^2}{2x^2-9x}$. At $(2,1)$: numerator $=9-8-12=-11$, denominator $=8-18=-10$, so $m=\dfrac{-11}{-10}=11$. Tangent: $y-1=11(x-2)$, i.e. $11x-y-21=0$. Normal slope $=-\dfrac{1}{11}$, giving $x+11y-13=0$. …