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Calculus - Tangents and normals

Grade 11IB_AI

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

🔑Concepts

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The gradient of a curve y=f(x)y = f(x) at a specific point (x1,y1)(x_1, y_1) is given by the value of the derivative at that point, denoted as f′(x1)f'(x_1).

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A tangent is a straight line that touches a curve at a single 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 to the tangent at the point of contact. The product of the gradients of two perpendicular lines is −1-1.

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To find the equation of a tangent or normal, you typically need a point (x1,y1)(x_1, y_1) and a gradient mm, then use the point-gradient formula: y−y1=m(x−x1)y - y_1 = m(x - x_1).

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If the tangent is horizontal, its gradient is 00 and its equation is y=y1y = y_1. If the tangent is vertical, the normal is horizontal.

📐Formulae

f′(x)=dydxf'(x) = \frac{dy}{dx}

mtangent=f′(a)m_{tangent} = f'(a)

mnormal=−1f′(a)m_{normal} = -\frac{1}{f'(a)}

y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a) (Equation of the Tangent)

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

m1×m2=−1m_1 \times m_2 = -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)=(2)3−2(2)2+4=8−8+4=4f(2) = (2)^3 - 2(2)^2 + 4 = 8 - 8 + 4 = 4. So the point is (2,4)(2, 4).

  2. Find the derivative: f′(x)=3x2−4xf'(x) = 3x^2 - 4x.

  3. Find the gradient of the tangent at x=2x = 2: m=f′(2)=3(2)2−4(2)=12−8=4m = f'(2) = 3(2)^2 - 4(2) = 12 - 8 = 4.

  4. Use the point-gradient formula: y−4=4(x−2)y - 4 = 4(x - 2) y−4=4x−8y - 4 = 4x - 8 y=4x−4y = 4x - 4.

Explanation:

First, evaluate the original function to find the coordinates of the point. Then, differentiate the function to find the gradient function. Substitute the xx-value into the derivative to find the specific gradient, and finally use the linear equation formula.

Problem 2:

Find the equation of the normal to the curve y=xy = \sqrt{x} at the point (9,3)(9, 3).

Solution:

  1. Rewrite the function for differentiation: y=x12y = x^{\frac{1}{2}}.

  2. Find the derivative: dydx=12x−12=12x\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.

  3. Find the gradient of the tangent at x=9x = 9: mtangent=129=12×3=16m_{tangent} = \frac{1}{2\sqrt{9}} = \frac{1}{2 \times 3} = \frac{1}{6}.

  4. Find the gradient of the normal: mnormal=−11/6=−6m_{normal} = -\frac{1}{1/6} = -6.

  5. Equation of the normal: y−3=−6(x−9)y - 3 = -6(x - 9) y−3=−6x+54y - 3 = -6x + 54 y=−6x+57y = -6x + 57.

Explanation:

The normal is perpendicular to the tangent. After finding the tangent's gradient (16\frac{1}{6}), we take the negative reciprocal to find the normal's gradient (−6-6) and then apply the point-gradient formula.