Equation of Tangent Line to a Curve Using Derivative

Sahabat Smaridasa - Mathematical Concepts

One of the applications of derivatives in mathematics is determining the slope of the tangent line to a curve at a given point. In this article, we will study the Equation of the Tangent Line to a Curve Using Derivatives. To better understand this material, it is recommended to also read the topics on the definition of derivatives, derivatives of algebraic functions, and derivatives of trigonometric functions.

Finding the Slope of the Tangent Line

The slope of the tangent line at point \(P(x, f(x))\) on the curve \(y = f(x)\) is given by:

\[ m = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x) \]

Thus, at point \(A(a, f(a))\), the slope is \(m = f'(a)\).

Steps to find the slope at point \(A(a, f(a))\):

1. Find the derivative \(f'(x)\).
2. Substitute \(x = a\) to get \(m = f'(a)\).

Gambar 1: Ilustrasi garis secan \(PQ\) dan garis singgung di titik \(P(x, f(x))\) untuk menentukan gradien

Equation of the Tangent Line

The equation of the tangent line at point \((x_1, y_1)\) on the curve \(y = f(x)\) is:

\[ y - y_1 = m(x - x_1) \]

where \(m = f'(x_1)\).

Gambar 2: Ilustrasi garis singgung pada kurva \(y = f(x)\) di titik \(P\)

📘 Example 1 : Tangent Line at a Given Point

Find the equation of the tangent line to the curve \(y = x^3 - 3x + 4\) at the point \((2, 6)\).

Solution:

\(f'(x) = 3x^2 - 3\).

At \(x = 2\), \(m = 3(2)^2 - 3 = 12 - 3 = 9\).

Using point-slope form: \(y - 6 = 9(x - 2)\) → \(y - 6 = 9x - 18\) → \(y = 9x - 12\).

Thus, the tangent line equation is \(y = 9x - 12\).

Gambar 3: Kurva \(y = x^3 - 3x + 4\) dan garis singgung \(y = 9x - 12\) di titik \((2, 6)\)

📘 Example 2 : Tangent Line at a Given Absissa

Find the equation of the tangent line to the curve \(y = x^2 - x + 2\) at the point where \(x = 1\), and determine its intersection points with the x-axis and y-axis.

Solution:

At \(x = 1\), \(y = 1^2 - 1 + 2 = 2\). So the point is \((1, 2)\).

\(f'(x) = 2x - 1\). At \(x = 1\), \(m = 2(1) - 1 = 1\).

Tangent line: \(y - 2 = 1(x - 1)\) → \(y = x + 1\).

Intersection with x-axis (\(y = 0\)): \(0 = x + 1\) → \(x = -1\) → point \((-1, 0)\).

Intersection with y-axis (\(x = 0\)): \(y = 0 + 1 = 1\) → point \((0, 1)\).

📘 Example 3 : Intersection of Two Tangent Lines

The line \(y = x + 1\) intersects the parabola \(y = x^2 + 2x + 1\) at points A and B. Find the equations of the tangent lines to the parabola at A and B. If the intersection point of these two tangent lines is \((a, b)\), find \(a + b\).

Solution:

Find intersection points: \(x^2 + 2x + 1 = x + 1\) → \(x^2 + x = 0\) → \(x(x+1) = 0\) → \(x = 0\) or \(x = -1\).

For \(x = 0\), \(y = 1\) → point \(A(0, 1)\). For \(x = -1\), \(y = 0\) → point \(B(-1, 0)\).

\(f'(x) = 2x + 2\).

At \(A(0,1)\): \(m = 2(0) + 2 = 2\) → tangent: \(y - 1 = 2(x - 0)\) → \(y = 2x + 1\).

At \(B(-1,0)\): \(m = 2(-1) + 2 = 0\) → tangent: \(y - 0 = 0(x + 1)\) → \(y = 0\).

Intersection of \(y = 0\) and \(y = 2x + 1\): \(0 = 2x + 1\) → \(x = -\frac{1}{2}\). So intersection is \((-\frac{1}{2}, 0)\).

Thus \(a = -\frac{1}{2}\), \(b = 0\), and \(a + b = -\frac{1}{2}\).

Tangent Line Given the Slope

If the slope \(m\) is known, we find the point of tangency using \(m = f'(x_1)\), then substitute to find \(y_1\).

Relationships with other lines:

• Parallel lines: \(m_1 = m_2\)
• Perpendicular lines: \(m_1 \cdot m_2 = -1\)

📘 Example 4 : Tangent Line with Given Slope

Find the equation of the tangent line to the curve \(y = x^2 - 2x + 3\) with slope 2.

Solution:

\(f'(x) = 2x - 2\). Set \(2x - 2 = 2\) → \(2x = 4\) → \(x = 2\).

At \(x = 2\), \(y = 2^2 - 2(2) + 3 = 4 - 4 + 3 = 3\). So point \((2, 3)\).

Tangent line: \(y - 3 = 2(x - 2)\) → \(y - 3 = 2x - 4\) → \(y = 2x - 1\).

📘 Example 5 : Tangent Line Parallel to a Given Line

Find the equation of the tangent line to the curve \(y = x^2 + x - 1\) that is parallel to the line \(y = 7x + 4\).

Solution:

The slope of the given line is \(m_1 = 7\). Since parallel lines have equal slopes, \(m = 7\).

\(f'(x) = 2x + 1\). Set \(2x + 1 = 7\) → \(2x = 6\) → \(x = 3\).

At \(x = 3\), \(y = 3^2 + 3 - 1 = 9 + 3 - 1 = 11\). Point \((3, 11)\).

Tangent line: \(y - 11 = 7(x - 3)\) → \(y - 11 = 7x - 21\) → \(y = 7x - 10\).

📘 Example 6 : Tangent Line Perpendicular to a Given Line

Find the equation of the tangent line to the curve \(y = \sqrt{x - 3}\) that is perpendicular to the line \(6x + 3y - 4 = 0\).

Solution:

Rewrite \(6x + 3y - 4 = 0\) as \(3y = -6x + 4\) → \(y = -2x + \frac{4}{3}\). Slope \(m_1 = -2\).

Perpendicular lines satisfy \(m \cdot m_1 = -1\) → \(m \cdot (-2) = -1\) → \(m = \frac{1}{2}\).

\(f'(x) = \frac{1}{2\sqrt{x-3}}\). Set \(\frac{1}{2\sqrt{x-3}} = \frac{1}{2}\) → \(2\sqrt{x-3} = 2\) → \(\sqrt{x-3} = 1\) → \(x - 3 = 1\) → \(x = 4\).

At \(x = 4\), \(y = \sqrt{4-3} = \sqrt{1} = 1\). Point \((4, 1)\).

Tangent line: \(y - 1 = \frac{1}{2}(x - 4)\) → multiply by 2: \(2y - 2 = x - 4\) → \(x - 2y = 2\).

Note: The derivative gives the slope of the tangent line at any point on the curve. This concept is fundamental in many applications, including optimization, physics (velocity), and geometry.