Tangents & Normals to Exponential Curves
Learn tangents and normals to exponential curves for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-04). The derivative gives the slope of a curve such as \(y = e^{2x}\) at any point, and that slope lets you write the line that just touches the curve there.
You will learn to find the gradient at a point, write the tangent using point-gradient form, find the normal as the perpendicular line through the same point, and solve problems on where these lines meet the axes — combining exponential derivatives with coordinate geometry.
Theory
The tangent to a curve matches its gradient at a point; the normal is perpendicular. This Year 12 Mathematics Advanced topic (MAV-12-04) finds tangents and normals to exponential curves.
At \((x_0,y_0)\) on \(y=f(x)\), the tangent has gradient \(m=f'(x_0)\) and equation \(y-y_0=m(x-x_0)\). The normal is perpendicular, with gradient \(-\dfrac1m\).
For exponentials the gradient comes from \(\dfrac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}\).
Method
- Find the point: substitute \(x_0\) into \(f\) for \(y_0\).
- Find the gradient: \(m=f'(x_0)\).
- Write the tangent \(y-y_0=m(x-x_0)\) (or use \(-\tfrac1m\) for the normal).
| \(m\) | \(=\) | \(e^{0}=1\) |
| \(y\) | \(=\) | \(x+1\) |
| \(m\) | \(=\) | \(2e^{0}=2\) |
| \(y\) | \(=\) | \(2x+1\) |
| \(\text{gradient}\) | \(=\) | \(-1\) |
| \(y\) | \(=\) | \(-x+1\) |
| \(\text{point}\) | \((1,e)\) | |
| \(m\) | \(=\) | \(e\) |
| \(y\) | \(=\) | \(ex\) |
Common pitfalls
Frequently asked questions
How do you find the equation of a tangent to a curve?
Find the point by substituting into the function, find the gradient m by substituting into the derivative, then use y minus y nought equals m times (x minus x nought).
What is the gradient of the tangent to y = e to the x at x = 0?
It is e to the 0, which is 1, so the tangent is y equals x plus 1.
What is the gradient of a normal line?
The normal is perpendicular to the tangent, so its gradient is minus 1 divided by the tangent gradient.
How do you find a tangent to an exponential curve?
Differentiate using the chain rule to get the gradient function, evaluate it at the point, then write the straight line through the point with that gradient.