CBSE Class 12 Maths: Applications of Derivative - Exercise 6.3 (Tangents and Normals)

Understanding Tangents and Normals with Derivatives

At its heart, the derivative of a function, \( \frac{dy}{dx} \), at a specific point on its curve, represents the slope of the tangent line to the curve at that very point. This geometric interpretation is incredibly powerful and forms the basis of Exercise 6.3. If you have a curve defined by an equation \( y = f(x) \), then to find the slope of the tangent at any point \( (x_1, y_1) \) on the curve, you first calculate \( \frac{dy}{dx} \) and then substitute \( x = x_1 \) and \( y = y_1 \) into the derivative. This gives you the numerical value of the slope, let's call it \( m_T \).

The normal to a curve at a point is a line perpendicular to the tangent at that same point. We know from coordinate geometry that if two lines are perpendicular, the product of their slopes is -1. So, if \( m_T \) is the slope of the tangent, and \( m_N \) is the slope of the normal, then \( m_T \cdot m_N = -1 \). This means the slope of the normal, \( m_N = -\frac{1}{m_T} \), provided \( m_T \neq 0 \). If \( m_T = 0 \) (tangent is horizontal), the normal is vertical (slope undefined). If \( m_T \) is undefined (tangent is vertical), the normal is horizontal (slope is 0).

Once you have the slope (either \( m_T \) or \( m_N \)) and the point \( (x_1, y_1) \), you can use the point-slope form of a straight line to find its equation: \( y - y_1 = m(x - x_1) \).

• Equation of Tangent: \( y - y_1 = m_T (x - x_1) \), where \( m_T = \left( \frac{dy}{dx} \right)_{(x_1, y_1)} \).
• Equation of Normal: \( y - y_1 = m_N (x - x_1) \), where \( m_N = -\frac{1}{m_T} = -\frac{1}{\left( \frac{dy}{dx} \right)_{(x_1, y_1)}} \) (if \( m_T \neq 0 \)).

These formulas are your toolkit for solving the problems in this exercise. Remember to differentiate carefully and substitute the point coordinates correctly!

Key Terms Defined

Tangent Line

A straight line that touches a curve at a single point without crossing it at that specific point. Its slope at the point of contact is given by the derivative of the curve's function at that point.

Normal Line

A straight line that is perpendicular to the tangent line at the point of tangency on a curve. Its slope is the negative reciprocal of the tangent's slope.

Slope of Tangent (m_T)

The value of \( \frac{dy}{dx} \) evaluated at the point of tangency \( (x_1, y_1) \) on the curve.

Slope of Normal (m_N)

The negative reciprocal of the slope of the tangent, i.e., \( -\frac{1}{m_T} \), assuming \( m_T \neq 0 \).

Point of Tangency

The specific point \( (x_1, y_1) \) on the curve where the tangent and normal lines intersect the curve.

Step-by-Step Guide to Finding Tangent and Normal Equations

1. Step 1: Identify the Curve and Point — Clearly identify the equation of the curve \( y = f(x) \) and the point \( (x_1, y_1) \) at which you need to find the tangent and normal. Sometimes, only the x-coordinate is given, so you'll need to find the corresponding y-coordinate by substituting x into the curve's equation.
2. Step 2: Differentiate the Curve's Equation — Find \( \frac{dy}{dx} \) of the given curve with respect to x. Use appropriate differentiation rules (power rule, product rule, quotient rule, chain rule, etc.). If the equation is implicit, use implicit differentiation.
3. Step 3: Calculate the Slope of the Tangent — Substitute the coordinates \( (x_1, y_1) \) of the given point into the derivative \( \frac{dy}{dx} \) to get the numerical value of the slope of the tangent, \( m_T \).
4. Step 4: Calculate the Slope of the Normal — If \( m_T \neq 0 \), the slope of the normal is \( m_N = -\frac{1}{m_T} \). If \( m_T = 0 \), the normal is a vertical line \( x = x_1 \). If \( m_T \) is undefined (i.e., \( \frac{dx}{dy} = 0 \)), the normal is a horizontal line \( y = y_1 \).
5. Step 5: Write the Equation of the Tangent — Using the point-slope form \( y - y_1 = m_T (x - x_1) \), substitute \( x_1, y_1 \), and \( m_T \) to obtain the equation of the tangent. Simplify it into a standard form (e.g., \( Ax + By + C = 0 \) or \( y = mx + c \)).
6. Step 6: Write the Equation of the Normal — Using the point-slope form \( y - y_1 = m_N (x - x_1) \), substitute \( x_1, y_1 \), and \( m_N \) to obtain the equation of the normal. Simplify it.

Solved Examples for Practice

• Example 1: Find the slope of the tangent to the curve \( y = 3x^4 - 4x \) at \( x = 4 \). Step 1: The curve is \( y = 3x^4 - 4x \). The x-coordinate is given as \( x = 4 \). Step 2: Differentiate \( y \) with respect to \( x \). \( \frac{dy}{dx} = \frac{d}{dx}(3x^4 - 4x) = 12x^3 - 4 \) Step 3: Substitute \( x = 4 \) into the derivative to find the slope of the tangent. \( m_T = (12(4)^3 - 4) = 12(64) - 4 = 768 - 4 = 764 \) Final answer: The slope of the tangent to the curve at \( x = 4 \) is 764.
• Example 2: Find the equations of the tangent and normal to the curve \( y = x^3 - 3x^2 - 9x + 7 \) at \( x = 0 \). Step 1: The curve is \( y = x^3 - 3x^2 - 9x + 7 \). The x-coordinate is \( x = 0 \). Find the y-coordinate at \( x = 0 \): \( y = (0)^3 - 3(0)^2 - 9(0) + 7 = 7 \) So, the point of tangency is \( (0, 7) \). Step 2: Differentiate \( y \) with respect to \( x \). \( \frac{dy}{dx} = \frac{d}{dx}(x^3 - 3x^2 - 9x + 7) = 3x^2 - 6x - 9 \) Step 3: Calculate the slope of the tangent at \( (0, 7) \). \( m_T = (3(0)^2 - 6(0) - 9) = -9 \) Step 4: Calculate the slope of the normal. \( m_N = -\frac{1}{m_T} = -\frac{1}{-9} = \frac{1}{9} \) Step 5: Write the equation of the tangent using \( (0, 7) \) and \( m_T = -9 \). \( y - 7 = -9(x - 0) \) \( y - 7 = -9x \) \( 9x + y - 7 = 0 \) Step 6: Write the equation of the normal using \( (0, 7) \) and \( m_N = \frac{1}{9} \). \( y - 7 = \frac{1}{9}(x - 0) \) \( 9(y - 7) = x \) \( 9y - 63 = x \) \( x - 9y + 63 = 0 \) Final answer: Equation of tangent is \( 9x + y - 7 = 0 \). Equation of normal is \( x - 9y + 63 = 0 \).
• Example 3: Find the points on the curve \( y = x^3 - 11x + 5 \) at which the tangent is \( y = x - 11 \). Step 1: The curve is \( y = x^3 - 11x + 5 \). The tangent line is \( y = x - 11 \). Step 2: Find the slope of the given tangent line. From \( y = x - 11 \), the slope is \( m = 1 \). Step 3: Differentiate the curve's equation to find the general slope of the tangent. \( \frac{dy}{dx} = 3x^2 - 11 \) Step 4: Equate the general slope of the tangent to the slope of the given tangent line to find the x-coordinates of the points. \( 3x^2 - 11 = 1 \) \( 3x^2 = 12 \) \( x^2 = 4 \) \( x = \pm 2 \) Step 5: Find the corresponding y-coordinates for \( x = 2 \) and \( x = -2 \) using the curve's equation. For \( x = 2 \): \( y = (2)^3 - 11(2) + 5 = 8 - 22 + 5 = -9 \). Point is \( (2, -9) \). For \( x = -2 \): \( y = (-2)^3 - 11(-2) + 5 = -8 + 22 + 5 = 19 \). Point is \( (-2, 19) \). Step 6: Verify if these points lie on the given tangent line \( y = x - 11 \). For \( (2, -9) \): \( -9 = 2 - 11 \) (True). So, \( (2, -9) \) is a valid point. For \( (-2, 19) \): \( 19 = -2 - 11 \) (False, \( 19 \neq -13 \)). So, \( (-2, 19) \) is not the point where the given line is tangent. Final answer: The point on the curve at which the tangent is \( y = x - 11 \) is \( (2, -9) \).

YoLearn AI's Exam Tips for Tangents and Normals

1. Careful Differentiation: The most common mistake is errors in finding \( \frac{dy}{dx} \). Double-check your differentiation, especially for complex functions (e.g., trigonometric, exponential, logarithmic, or implicit functions).
2. Point Substitution: Ensure you substitute the correct \( x_1 \) and \( y_1 \) values into \( \frac{dy}{dx} \) to get the numerical slope. If only \( x_1 \) is given, always find \( y_1 \) first using the curve's equation.
3. Slope of Normal: Remember that the slope of the normal is the negative reciprocal of the slope of the tangent. If \( m_T = 0 \), the normal is vertical (equation \( x = x_1 \)). If \( m_T \) is undefined (e.g., \( \frac{dy}{dx} = \frac{1}{0} \) ), the tangent is vertical (equation \( x = x_1 \)), and the normal is horizontal (equation \( y = y_1 \)).
4. Special Conditions: Pay close attention to conditions like "tangent is parallel to x-axis" (means \( m_T = 0 \)), "tangent is parallel to y-axis" (means \( m_T \) is undefined, or \( \frac{dx}{dy} = 0 \)), or "tangent is parallel/perpendicular to a given line". In these cases, you'll first find the slope of the reference line and then use it to find the points on the curve.
5. Simplify Equations: Always simplify your final equations of tangent and normal into a neat form, preferably \( Ax + By + C = 0 \).
6. Parametric Equations: If the curve is given in parametric form (e.g., \( x = f(t), y = g(t) \)), remember to find \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) and then substitute the value of 't' corresponding to the given point.

Practice Questions with Solutions

• Q: Find the slope of the tangent to the curve \( y = x^3 - x \) at \( x = 2 \). A: Step 1: Given curve is \( y = x^3 - x \). Step 2: Differentiate with respect to \( x \): \( \frac{dy}{dx} = 3x^2 - 1 \). Step 3: Substitute \( x = 2 \) into \( \frac{dy}{dx} \). \( m_T = 3(2)^2 - 1 = 3(4) - 1 = 12 - 1 = 11 \). Final answer: The slope of the tangent is 11.
• Q: Find the equation of the tangent and normal to the curve \( y = x^2 - 4x + 3 \) at the point \( (2, -1) \). A: Step 1: Curve is \( y = x^2 - 4x + 3 \), point is \( (2, -1) \). Step 2: Differentiate: \( \frac{dy}{dx} = 2x - 4 \). Step 3: Slope of tangent at \( (2, -1) \): \( m_T = 2(2) - 4 = 4 - 4 = 0 \). Step 4: Slope of normal: Since \( m_T = 0 \), the tangent is horizontal. The normal is vertical. Step 5: Equation of tangent: \( y - (-1) = 0(x - 2) \Rightarrow y + 1 = 0 \). Step 6: Equation of normal: Since it's a vertical line passing through \( (2, -1) \), its equation is \( x = 2 \) or \( x - 2 = 0 \). Final answer: Equation of tangent is \( y + 1 = 0 \). Equation of normal is \( x - 2 = 0 \).
• Q: Find the points on the curve \( x^2 + y^2 - 2x - 3 = 0 \) at which the tangents are parallel to the x-axis. A: Step 1: Curve is \( x^2 + y^2 - 2x - 3 = 0 \). Tangents are parallel to x-axis, so their slope is 0. Step 2: Differentiate implicitly with respect to \( x \): \( 2x + 2y \frac{dy}{dx} - 2 = 0 \) \( 2y \frac{dy}{dx} = 2 - 2x \) \( \frac{dy}{dx} = \frac{2 - 2x}{2y} = \frac{1 - x}{y} \) Step 3: Set \( \frac{dy}{dx} = 0 \) for tangents parallel to x-axis. \( \frac{1 - x}{y} = 0 \Rightarrow 1 - x = 0 \Rightarrow x = 1 \) Step 4: Substitute \( x = 1 \) back into the curve's equation to find \( y \). \( (1)^2 + y^2 - 2(1) - 3 = 0 \) \( 1 + y^2 - 2 - 3 = 0 \) \( y^2 - 4 = 0 \) \( y^2 = 4 \Rightarrow y = \pm 2 \) Final answer: The points are \( (1, 2) \) and \( (1, -2) \).
• Q: Find the equation of the normal to the curve \( y^2 = 4x \) at the point \( (1, 2) \). A: Step 1: Curve is \( y^2 = 4x \), point is \( (1, 2) \). Step 2: Differentiate implicitly with respect to \( x \): \( 2y \frac{dy}{dx} = 4 \) \( \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \) Step 3: Slope of tangent at \( (1, 2) \): \( m_T = \frac{2}{2} = 1 \). Step 4: Slope of normal: \( m_N = -\frac{1}{m_T} = -\frac{1}{1} = -1 \). Step 5: Equation of normal using \( (1, 2) \) and \( m_N = -1 \). \( y - 2 = -1(x - 1) \) \( y - 2 = -x + 1 \) \( x + y - 3 = 0 \) Final answer: The equation of the normal is \( x + y - 3 = 0 \).

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