Tangents & Normals to Exponential Curves
Tangents and normals to exponential curves use the gradient \(m=f'(x_0)\) to write the tangent \(y-y_0=m(x-x_0)\), with the normal gradient \(-\dfrac1m\).
Part of the NSW Year 12 Mathematics Advanced course, in the Calculus area of study (Differential calculus focus area) of the 2024 syllabus. Work through practice questions with fully worked solutions and video lessons, or scroll down for the theory summary and worked examples.
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.