Recall that the gradient function tells us the gradient of a curve at any point, given the $x$ value (or any other required inputs).
The gradient at a point on the curve is the same as the gradient of the tangent to the curve at that point. Therefore the gradient function is used to find the gradient of the tangent at any given point on a curve.
Example
Find the equation of the tangent to the curve $y=x^2+5x+6$ at the point $(1,12)$.
Solution The gradient of the tangent is given by the value of $dy/dx$: $$ \begin{equation}\begin{aligned} y&=x^2+5x+6\\ \frac{dy}{dx}&=2x+5\\ \end{aligned}\end{equation} $$
At the point $(1,12)$, the value of the gradient function is $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=2(1)+5\\ \end{aligned}\end{equation} $$
Thus with the gradient of the tangent as $m_T=1$ and the point $(1,12)$ we can find the equation of the tangent: $$ \begin{equation}\begin{aligned} y-y_1&=m_T(x-x_1)\\ y-12&=7(x-1)\\ y&=7x-7+12\\ y&=7x+5\\ \end{aligned}\end{equation} $$
Equation of tangent practice
Calculate the equations of the tangents to the following curves at the given points:
- $y=3x^4$ at $(1,3)$
- $y=-x^3+3x$ at $x=0$
- $y=\sin{x}$ at $x=90\degree$
- $y=\cos{2x}$ at $x=30\degree$
- $y=12$ at $x=4$
Equation of the normal line at a point
A normal line is simply a line which is perpendicular to the tangent. We know that when two lines are perpendicular, the product of their gradients is $-1$: $$ \begin{equation}\begin{aligned} m_T\times m_N &=-1\\ \therefore m_N&=\frac{-1}{m_T}\\ &=\frac{-1}{\frac{dy}{dx}}\\ \end{aligned}\end{equation} $$
Example
Find the equation of the normal to the curve $y=x^3-7x$ at the point $(2,-6)$.
Solution The gradient of the tangent is the value of $dy/dx$: $$ \begin{equation}\begin{aligned} y&=x^3-7x\\ \therefore \frac{dy}{dx}&=3x^2-7\\ \end{aligned}\end{equation} $$
At $x=2$: $$ \begin{equation}\begin{aligned} \frac{dy}{dx}&=3(2)^2-7\\ &=5\\ \end{aligned}\end{equation} $$
This is the gradient of the tangent. The gradient of the normal will be the negative reciprocal of the gradient of the tangent: $$ \begin{equation}\begin{aligned} m_N&=\frac{-1}{m_T}\\ &=\frac{-1}{5}\\ \end{aligned}\end{equation} $$
The equation of the normal will be $$ \begin{equation}\begin{aligned} y-y_1&=m(x-x_1)\\ y-(-6)&=\frac{-1}{5}(x-2)\\ y+6&=\frac{-1}{5}x+\frac{2}{5}\\ y&=\frac{-1}{5}x+\frac{2}{5}-6\\ y&=\frac{-1}{5}x+\frac{2-30}{5}\\ y&=\frac{-1}{5}x-\frac{28}{5}\\ \end{aligned}\end{equation} $$
Equation of normal practice
Determine the equations of the normals to the following curves at the given points:
- $y=\cos{2x}$ at $x=0\degree$
- $y=5x^2+3x$ at $x=1$
- $y=\sqrt{x}$ at $x=4$
- $y=1-x^4$ at $x=-1$
- $y=\frac{1}{4x^2}$ at $x=-2$