A normal curve is a straight line passing through the point where the tangent touches the curve and is perpendicular (at right angles) to the tangent at that point.
The gradient of the tangent to a curve is $m$, then the gradient of the normal is $\displaystyle -\dfrac{1}{m}$, as the product of the gradients of $2$ perpendicular lines equals to $-1$.
The gradient of the tangent at $x=a$ is $f^{\prime}(a)$. Therefore the gradient of the normal is $\displaystyle -\dfrac{1}{f^{\prime}(a)}$. The equation of the normal is:
$$y-f(a) = -\dfrac{1}{f^{\prime}(a)}(x-a)$$
Example 1
Find the gradient of the normal to the curve $f(x)=2x^3-x^2+1$ at $x=1$.
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
f^{\prime}(x) &= 6x^2-2x \\
f^{\prime}(1) &= 6 \times 1^2-2 \times 1 &\color{red} \text{gradient of tangent} \\
&= 4 \\
\end{align} \)
Therefore, the normal gradient is $\displaystyle -\dfrac{1}{4}$.
Example 2
Find the equation of the normal to $f(x)=x^3-2x+3$ at the point $(1,2)$.
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
f^{\prime}(x) &= 3x^2-2 \\
f^{\prime}(1) &= 3 \times 1^2-2 &\color{red} \text{gradient of tangent} \\
&= 1
\end{align} \)
Therefore the gradient of the normal is $-1$.
The equation of the normal is:
\( \begin{align} \displaystyle \require{color}
y-2 &= -1(x-1) \\
\therefore y &= -x+3
\end{align} \)
Example 3
Find the equation of the normal to $f(x)=3x^2-1$ at $x=1$.
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
f(1) &= 3 \times 1^2-1 \\
&= 2 &\color{red} y\text{-coordinate of the point} \\
f^{\prime}(x) &= 6x \\
f^{\prime}(1) &= 6 \times 1 &\color{red} \text{gradient of tangent} \\
&= 6
\end{align} \)
Therefore, the normal gradient is $\displaystyle -\dfrac{1}{6}$.
The equation of the normal is:
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
y-2 &= -\dfrac{1}{6}(x-1) &\color{red} y-f(1) = -\dfrac{1}{f^{\prime}(1)}(x-1) \\
6y-12 &= -x+1 \\
\therefore x+6y-13 &= 0
\end{align} \)
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