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Calculus - Tangent line – Normal line

Grade 11IB_AA

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

🔑Concepts

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The derivative of a function f(x)f(x), denoted as f′(x)f'(x) or dydx\frac{dy}{dx}, represents the gradient (slope) of the tangent line to the curve at any point xx.

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A tangent line to a curve at a point P(a,f(a))P(a, f(a)) is a straight line that just touches the curve at that point and has the same gradient as the curve at PP.

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A normal line is a straight line that is perpendicular to the tangent line at the point of contact P(a,f(a))P(a, f(a)).

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If the gradient of the tangent is mTm_T, then the gradient of the normal mNm_N is the negative reciprocal: mN=−1mTm_N = -\frac{1}{m_T}, provided mT≠0m_T \neq 0.

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To find the equation of either line, we typically use the point-gradient form: y−y1=m(x−x1)y - y_1 = m(x - x_1), where (x1,y1)(x_1, y_1) is the point of tangency.

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Horizontal tangents occur where f′(x)=0f'(x) = 0. Vertical tangents occur where the derivative is undefined (approaches infinity).

📐Formulae

mT=f′(a)m_T = f'(a) balances

mN=−1f′(a)m_N = -\frac{1}{f'(a)}

Equation of Tangent: y−f(a)=f′(a)(x−a)\text{Equation of Tangent: } y - f(a) = f'(a)(x - a)

Equation of Normal: y−f(a)=−1f′(a)(x−a)\text{Equation of Normal: } y - f(a) = -\frac{1}{f'(a)}(x - a)

mT×mN=−1m_T \times m_N = -1

💡Examples

Problem 1:

Find the equation of the tangent to the curve f(x)=x3−2x2+4f(x) = x^3 - 2x^2 + 4 at the point where x=2x = 2.

Solution:

  1. Find the yy-coordinate: f(2)=23−2(22)+4=8−8+4=4f(2) = 2^3 - 2(2^2) + 4 = 8 - 8 + 4 = 4. The point is (2,4)(2, 4).
  2. Find the derivative: f′(x)=3x2−4xf'(x) = 3x^2 - 4x.
  3. Calculate the gradient at x=2x = 2: mT=f′(2)=3(22)−4(2)=12−8=4m_T = f'(2) = 3(2^2) - 4(2) = 12 - 8 = 4.
  4. Use the point-slope formula: y−4=4(x−2)y - 4 = 4(x - 2).
  5. Simplify: y−4=4x−8⇒y=4x−4y - 4 = 4x - 8 \Rightarrow y = 4x - 4.

Explanation:

First, evaluate the function to find the point of tangency. Then, differentiate the function to find the gradient function. Substitute the xx-value into the derivative to get the specific gradient mm. Finally, substitute the point and gradient into the line equation.

Problem 2:

Given the function g(x)=xg(x) = \sqrt{x}, find the equation of the normal to the curve at x=9x = 9.

Solution:

  1. Find the yy-coordinate: g(9)=9=3g(9) = \sqrt{9} = 3. The point is (9,3)(9, 3).
  2. Find the derivative: g(x)=x1/2⇒g′(x)=12x−1/2=12xg(x) = x^{1/2} \Rightarrow g'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}.
  3. Gradient of tangent: mT=g′(9)=129=16m_T = g'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}.
  4. Gradient of normal: mN=−1mT=−6m_N = -\frac{1}{m_T} = -6.
  5. Equation of normal: y−3=−6(x−9)y - 3 = -6(x - 9).
  6. Simplify: y−3=−6x+54⇒y=−6x+57y - 3 = -6x + 54 \Rightarrow y = -6x + 57.

Explanation:

The normal is perpendicular to the tangent. After finding the tangent's gradient using the derivative, take the negative reciprocal to find the normal's gradient, then use the point-slope form.

Problem 3:

Find the coordinates of the point(s) on the curve y=x2−6x+5y = x^2 - 6x + 5 where the tangent is horizontal.

Solution:

  1. Find the derivative: dydx=2x−6\frac{dy}{dx} = 2x - 6.
  2. A horizontal tangent has a gradient of 00. Set dydx=0\frac{dy}{dx} = 0:
    2x−6=0⇒2x=6⇒x=32x - 6 = 0 \Rightarrow 2x = 6 \Rightarrow x = 3
  3. Find the corresponding yy-value: y=(3)2−6(3)+5=9−18+5=−4y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4.
  4. The point is (3,−4)(3, -4).

Explanation:

Horizontal lines have a slope of zero. By setting the derivative equal to zero, we find the xx-coordinates where the curve 'flattens out', which usually corresponds to local maxima or minima.