Tangent and normal lines
Tangent and Normal Lines
Tangent and Normal Lines is the opening topic of Applications of Differentiation in the California Calculus Standards. It is aligned to Standard 4.1, which requires students to understand the derivative as the slope of the tangent line to a curve at a point.
The tangent line touches a curve at a point with the same slope \(f'(a)\), while the normal line is perpendicular to it, with slope \(-\dfrac{1}{f'(a)}\). Both are found with the point-slope form once the derivative is known.
Theory
The tangent line touches a curve at a point with the same slope as the curve, \(f'(a)\); the normal line is perpendicular to it. Both are found with the point-slope form once you know the derivative.
The tangent line at \(\big(a,f(a)\big)\) has slope \(f'(a)\) and equation
The normal line is perpendicular to the tangent at the same point, so its slope is the negative reciprocal \(-\dfrac{1}{f'(a)}\).
A horizontal tangent occurs where \(f'(a)=0\) — the slope, and hence the tangent line, is flat.
The two lines through the point of tangency:
How to find a tangent or normal line
- Differentiate to get \(f'(x)\).
- Evaluate \(f'(a)\) for the tangent slope (or \(-\dfrac{1}{f'(a)}\) for the normal).
- Find the point \((a,f(a))\) and substitute into point-slope form.
Slope is the derivative; the point is \((2,5)\).
| \(f'(x)\) | \(=\) | \(2x\) |
| \(f'(2)\) | \(=\) | \(4\) |
| \(y-5\) | \(=\) | \(4(x-2)\) |
The tangent is \(y=4x-3\).
Tangent slope \(f'(1)=2\); the normal slope is \(-\dfrac{1}{2}\), and the point is \((1,1)\).
| \(y-1\) | \(=\) | \(-\dfrac{1}{2}(x-1)\) |
| \(y\) | \(=\) | \(-\dfrac{1}{2}x+\dfrac{3}{2}\) |
A horizontal tangent means \(f'(x)=0\).
| \(f'(x)\) | \(=\) | \(2x-4\) |
| \(2x-4\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(2\) |
With \(y=x^{1/2}\), \(f'(x)=\dfrac{1}{2\sqrt{x}}\); the point is \((4,2)\).
| \(f'(4)\) | \(=\) | \(\dfrac{1}{4}\) |
| \(y-2\) | \(=\) | \(\dfrac{1}{4}(x-4)\) |
The tangent is \(y=\dfrac{1}{4}x+1\).
Common pitfalls
Frequently asked questions
How do you find the equation of a tangent line?
Differentiate for the slope \(f'(a)\), find the point \((a,f(a))\), then use point-slope form \(y-f(a)=f'(a)(x-a)\).
What is a normal line?
The line perpendicular to the tangent at the point of contact. Its slope is the negative reciprocal of the tangent slope, \(-\dfrac{1}{f'(a)}\).
Where is the tangent line horizontal?
Where \(f'(x)=0\). Solve that equation to find the x-values where the curve has a flat tangent.
What is the slope of the tangent line?
The derivative evaluated at the point, \(f'(a)\). That single number is the instantaneous slope of the curve there.