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Calculus - Tangents and Normals

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The derivative dy/dx represents the gradient of the tangent to the curve y = f(x) at any given point.

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A tangent is a straight line that touches a curve at a specific point and has the same gradient as the curve at that point.

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A normal is a straight line that is perpendicular (at 90 degrees) to the tangent at the point of contact.

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The product of the gradients of two perpendicular lines is -1 (m1 * m2 = -1).

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To find the equation of a tangent or normal, you need a point (x1, y1) and the gradient (m) using the formula y - y1 = m(x - x1).

📐Formulae

Gradient of Tangent (mtm_t): mt=f′(x1)m_t = f'(x_1)

Gradient of Normal (mnm_n): mn=−1f′(x1)m_n = -\frac{1}{f'(x_1)}

Equation of Tangent: y−y1=f′(x1)(x−x1)y - y_1 = f'(x_1)(x - x_1)

Equation of Normal: y−y1=−1f′(x1)(x−x1)y - y_1 = -\frac{1}{f'(x_1)}(x - x_1)

💡Examples

Problem 1:

Find the equation of the tangent and the normal to the curve y=x3−2x2+4y = x^3 - 2x^2 + 4 at the point where x=2x = 2.

Solution:

  1. Find the y-coordinate: When x=2x = 2, y=(2)3−2(2)2+4=8−8+4=4y = (2)^3 - 2(2)^2 + 4 = 8 - 8 + 4 = 4. Point is (2, 4).
  2. Find the derivative: dydx=3x2−4x\frac{dy}{dx} = 3x^2 - 4x.
  3. Find the gradient of the tangent (mtm_t): At x=2x = 2, mt=3(2)2−4(2)=12−8=4m_t = 3(2)^2 - 4(2) = 12 - 8 = 4.
  4. Tangent Equation: y−4=4(x−2)⇒y=4x−4y - 4 = 4(x - 2) \Rightarrow y = 4x - 4.
  5. Find the gradient of the normal (mnm_n): mn=−1/mt=−1/4m_n = -1/m_t = -1/4.
  6. Normal Equation: y−4=−1/4(x−2)⇒4y−16=−x+2⇒x+4y=18y - 4 = -1/4(x - 2) \Rightarrow 4y - 16 = -x + 2 \Rightarrow x + 4y = 18.

Explanation:

First, we determine the point of contact by substituting x into the original equation. We differentiate the function to get the gradient expression. Substituting x=2 into the derivative gives the tangent's slope. The normal's slope is the negative reciprocal of the tangent's slope. Finally, we use the point-slope formula for both.

Problem 2:

Find the coordinates of the point on the curve y=xy = \sqrt{x} where the tangent is parallel to the line x−4y+10=0x - 4y + 10 = 0.

Solution:

  1. Rewrite the line in y=mx+cy = mx + c form: 4y=x+10⇒y=14x+2.54y = x + 10 \Rightarrow y = \frac{1}{4}x + 2.5. The gradient m=1/4m = 1/4.
  2. Differentiate the curve: y=x1/2⇒dydx=12x−1/2=12xy = x^{1/2} \Rightarrow \frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.
  3. Set the derivative equal to the gradient: 12x=14\frac{1}{2\sqrt{x}} = \frac{1}{4}.
  4. Solve for x: 2x=4⇒x=2⇒x=42\sqrt{x} = 4 \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4.
  5. Find y: y=4=2y = \sqrt{4} = 2.
  6. Point: (4, 2).

Explanation:

If a tangent is parallel to a line, they share the same gradient. We find the gradient of the given line, set the derivative of the curve equal to that value, and solve for x. Then, we find the corresponding y-value from the original curve equation.