A tangent line is a straight line that touches a curve at a point and shares the same instantaneous slope as the curve at that point. Unlike a secant line, which intersects a curve at two or more points, a tangent line represents the direction of the curve at a single point.
Tangent Line: Touches the curve at a point and has the same slope as the curve at that point.
Secant Line: Intersects the curve at two distinct points.
The point where the tangent line touches the curve is called the point of tangency.
The equation of the line with slope(m) and passing through the point (x₀, y₀) is given by the point-slope form: y - y₀ = m(x - x₀).
Now, let's consider the tangent line to a curve y = f(x) at the point (x₀, y₀).
Slope of the tangent line, m = f'(x₀, y₀)
Hence, the equation of the tangent line is:
y – y₀ = f '(x₀, y₀)(x – x₀)
Steps to Find the Equation of the Tangent Line
Find the equation of the tangent line to the curve y = f(x) at the point (x₀, y₀) or at x = x₀.
Step 1: If it's not given, that is, the question only states the tangent is at x = x₀, find the y-coordinate by plugging x₀ into the function y = f(x), i.e., y₀ = f(x₀).
Step 2: Find the derivative of the function y = f(x) and label this f'(x).
Step 3: Replace the point (x₀, y₀) into the derivative f'(x) to obtain the slope of the tangent (m).
Step 4: Use the slope-point form: y - y₀ = m(x - x₀), to write the equation of the tangent line.
Geometric Interpretation
The derivative tells us how steep a curve is at a particular point.
- Positive Slope : If f′(x) > 0, the tangent line rises from left to right.
- Negative Slope: If f′(x) < 0, the tangent line falls from left to right.
- Zero Slope: If f′(x)=0, the tangent line is horizontal.
- Undefined Slope: If the derivative is undefined, the tangent may be vertical.
Understanding the slope helps visualize the behavior of the function near the point of tangency.
Solved Examples
Example 1: Find the equation of the tangent line of the curve y = 3x2 - 4x at x = -1. Also, verify it.
The point at which the tangent is drawn is, (x0, y0) = (-1, f(-1)) = (-1, 3(-1)2 - 4(-1)) = (-1, 7).
The given curve is, f(x) = 3x2 - 4x.
Its derivative is f'(x) = 6x - 4.
The slope of the tangent is, m = f'(-1) = 6(-1) - 4 = -10.
The equation of the tangent line is,
y - y0 = m (x - x0)
⇒ y - 7 = -10 (x - (-1))
⇒ y - 7 = -10 (x + 1)
⇒ y - 7 = -10x - 10
⇒ y = -10x - 3
Example 2: What is the tangent line equation of the curve x = cos t and y = sin t at t = π/2?
The point at which the tangent is drawn is, (x0, y0) = (cos π/2, sin π/2) = (0, 1). (by using unit circle)
The curve is defined by parametric equations.
So dy/dx = (dy/dt) / (dx/dt) = (cos t) / (- sin t).
The slope of the tangent is, m = (dy/dx)ₜ ₌ π/₂ = (cos π/2) / (-sin π/2) = 0/(-1) = 0.
The equation of the tangent line is,
y - y0 = m (x - x0)
⇒ y - 1 = 0 (x - 0)
⇒ y - 1 = 0
⇒ y = 1
Example 3: Find the parametric equations of the tangent line drawn to the curve given by the equations x = 2t2, y = 3t, z = t3 at t = 2.
The point at which the tangent line is drawn is,
(x0, y0, z0) = (x(2), y(2), z(2)) = (2(2)2, 3(2), 23) = (8, 6, 8).
Compute the derivatives
x'(t) = 4t, y'(t) = 3, z'(t) = 3t2
The direction ratios of the tangent line are,
<a, b, c> = <x'(2), y'(2), z'(2)> = <4(2), 3, 3(2)2> = <8, 3, 12>
The parametric equations of the tangent line are:
x = x0 + at, y = y0 + bt, z = z0 + ct
⇒ x = 8 + 8t, y = 6 + 3t, z = 8 + 12t
Practice Problems
- Find the equation of the tangent line to y = x2 at x = 1.
- Find the equation of the tangent line to y = sinx at x = π/4.
- Find the equation of the tangent line to y = ex at x = 1.
- Find the equation of the tangent line to y = lnx at x = 1.
- Determine whether the function y = x4 has a horizontal tangent at x = 0.
