Tangents & Normals to Logarithmic Curves
Learn tangents and normals to logarithmic curves for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-04). Since the derivative of \(\ln x\) is \(\dfrac{1}{x}\), the slope of a log curve at any point is easy to find, and from it you can write the touching line.
You will learn to find the gradient at a given point, write the tangent and the perpendicular normal, and answer questions such as where a tangent passes through the origin — practice that ties logarithmic derivatives to the geometry of straight lines.
Theory
Tangents to a log curve use the gradient \(\dfrac{d}{dx}\ln x=\dfrac1x\). This Year 12 Mathematics Advanced topic (MAV-12-04) finds tangents and normals to logarithmic curves.
At \((x_0,y_0)\) on \(y=f(x)\): the tangent has gradient \(m=f'(x_0)\), equation \(y-y_0=m(x-x_0)\); the normal has gradient \(-\dfrac1m\).
For \(y=\ln x\) the gradient at \(x_0\) is \(\dfrac{1}{x_0}\), and the curve passes through \((1,0)\).
Method
- Find the point \((x_0,\ln x_0)\).
- Find the gradient \(m=\dfrac{1}{x_0}\).
- Write the tangent (or normal with \(-\tfrac1m\)).
| \(m\) | \(=\) | \(\dfrac11=1\) |
| \(y\) | \(=\) | \(x-1\) |
| \(\text{point}\) | \((e,1)\) | |
| \(m\) | \(=\) | \(\dfrac1e\) |
| \(y\) | \(=\) | \(\dfrac{x}{e}\) |
| \(\text{gradient}\) | \(=\) | \(-1\) |
| \(y\) | \(=\) | \(-x+1\) |
| \(m\) | \(=\) | \(\dfrac12\) |
Common pitfalls
Frequently asked questions
How do you find the tangent to y = ln x?
Find the point, compute the gradient as 1 over the x-value, then write y minus y nought equals m times (x minus x nought).
What is the gradient of y = ln x at x = 1?
It is 1 over 1, which is 1, so the tangent there is y equals x minus 1.
Where does y = ln x cross the x-axis?
At the point (1, 0), because ln 1 equals 0.
What is the normal to a logarithmic curve?
It is the line perpendicular to the tangent at the point, with gradient minus 1 over the tangent gradient.