
In calculus, the tangent line to a curve at a given point is the line that “just touches” the curve at that point.
The normal line is the line perpendicular to the tangent line at the same point.
For a function y=f(x):
The slope of the tangent line at x=a is the derivative:
$$m_{tangent} = f^\prime(a) $$
The point of tangency is:
(a , f(a))
The equation of the tangent line is:
$$y – f(a) = m_{tangent}(x – a) = f^\prime(a)(x – a)$$
$$y = f^\prime(a)(x – a) + f(a)$$
The normal line is perpendicular to the tangent line.
So its slope is the negative reciprocal of the tangent’s slope:
$$m_{normal} = -\frac{1}{f^\prime(a)}$$
Then, the equation of the normal line is:
$$y – f(a) = m_{normal}(x – a) = -\frac{1}{f^\prime(a)}(x – a)$$
$$y = -\frac{1}{f^\prime(a)}(x – a) + f(a)$$
Let’s find the tangent and normal lines to:
$$y = x^2+2x$$ at $$x=1$$
Step 1. Find derivative function$$f^\prime(x)$$
$$f^\prime(x) = 2x +2$$
Step 2. Find slope and tangency point:
$$slope = m_{tangent} = f^\prime(1) = 2(1) + 2 =4$$
$$f(1) = 1^2+2(1) = 3$$ tangency point (1 , 3)
Step 3. Tangent line:
$$y – f(1) = f^\prime(1) (x -1) \longrightarrow y = 4x -1$$
Step 4. Normal line:
$$y -f(1) = -\frac{1}{f^\prime(1)}(x -1) \longrightarrow y=-\frac{1}{4}x+\frac{13}{4}$$
The tangent line gives instantaneous rate of change.
The normal line is useful in geometry and physics (e.g., reflections, perpendicular forces).
If f′(a)=0, then the tangent is horizontal and the normal is vertical.
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