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Application of Derivatives - Tangents and normals

Grade 12CBSE

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

🔑Concepts

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Geometrical Interpretation of the Derivative: The derivative of a function y=f(x)y = f(x) at a point P(x1,y1)P(x_1, y_1), denoted as dydx\frac{dy}{dx} evaluated at (x1,y1)(x_1, y_1), represents the slope of the tangent line to the curve at that specific point. Visually, if you imagine a curve on a Cartesian plane, the tangent is a straight line that 'just touches' the curve at point PP without crossing through it immediately at that spot, representing the curve's instantaneous direction.

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The Tangent Line: A tangent to a curve at a given point is a line that best approximates the curve locally near that point. On a graph, as you zoom in closer and closer to the point of contact, the curve and its tangent line become indistinguishable. The slope of this line is given by the first derivative of the curve's equation.

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The Normal Line: The normal to a curve at a point P(x1,y1)P(x_1, y_1) is a straight line that is perpendicular to the tangent line at the same point of contact. Visually, the tangent and normal lines intersect at the point PP to form a 90∘90^{\circ} angle, resembling a 'cross' or a 'T-junction' situated exactly on the curve.

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Relationship between Slopes: Because the tangent and normal are perpendicular to each other, the product of their slopes is always −1-1 (provided neither is vertical). If the slope of the tangent is mm, the slope of the normal is −1m-\frac{1}{m}. This inverse relationship is a fundamental geometric property of perpendicular lines on a coordinate plane.

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Horizontal and Vertical Tangents: A tangent is horizontal (parallel to the x-axis) when the derivative dydx=0\frac{dy}{dx} = 0. Visually, these occur at the 'peaks' (maxima) or 'valleys' (minima) of a smooth curve. A tangent is vertical (parallel to the y-axis) when the derivative dydx\frac{dy}{dx} is undefined (tends to infinity), often occurring where the curve has a sharp vertical 'climb'.

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Point-Slope Form Application: The equations for both tangents and normals are derived using the point-slope form of a linear equation, y−y1=m(x−x1)y - y_1 = m(x - x_1). To find the equation, you need the coordinates of the point of contact (x1,y1)(x_1, y_1) and the slope calculated from the derivative.

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Orthogonal Curves: Two curves are said to be orthogonal if they intersect at right angles. This happens if, at every point of intersection, the tangent to the first curve is perpendicular to the tangent of the second curve. Visually, the two curves cross each other like a perfectly aligned '+' sign.

📐Formulae

Slope of the tangent at (x1,y1)(x_1, y_1): m=[dydx](x1,y1)m = \left[ \frac{dy}{dx} \right]_{(x_1, y_1)}

Equation of the tangent: y−y1=m(x−x1)y - y_1 = m(x - x_1)

Slope of the normal at (x1,y1)(x_1, y_1): mn=−1m=−1[dydx](x1,y1)m_n = -\frac{1}{m} = -\frac{1}{\left[ \frac{dy}{dx} \right]_{(x_1, y_1)}}

Equation of the normal: y−y1=−1m(x−x1)y - y_1 = -\frac{1}{m}(x - x_1)

Condition for horizontal tangent: dydx=0\frac{dy}{dx} = 0

Condition for vertical tangent: dxdy=0\frac{dx}{dy} = 0 or dydx→∞\frac{dy}{dx} \to \infty

Condition for orthogonality of two curves with tangent slopes m1m_1 and m2m_2: m1⋅m2=−1m_1 \cdot m_2 = -1

💡Examples

Problem 1:

Find the equations of the tangent and normal to the curve y=x4−6x3+13x2−10x+5y = x^4 - 6x^3 + 13x^2 - 10x + 5 at the point (1,3)(1, 3).

Solution:

  1. Find the derivative of the curve: dydx=4x3−18x2+26x−10\frac{dy}{dx} = 4x^3 - 18x^2 + 26x - 10
  2. Calculate the slope of the tangent (mm) at x=1x = 1: $m = 4(1)^3 - 18(1)^2 + 26(1) - 10 = 4 - 18 + 26 - 10 = 2$$
  3. Equation of the tangent: Using y−y1=m(x−x1)y - y_1 = m(x - x_1), we get y−3=2(x−1)y - 3 = 2(x - 1) y−3=2x−2⇒2x−y+1=0y - 3 = 2x - 2 \Rightarrow 2x - y + 1 = 0
  4. Slope of the normal (mnm_n): mn=−1m=−12m_n = -\frac{1}{m} = -\frac{1}{2}
  5. Equation of the normal: Using y−y1=mn(x−x1)y - y_1 = m_n(x - x_1), we get y−3=−12(x−1)y - 3 = -\frac{1}{2}(x - 1) 2y−6=−x+1⇒x+2y−7=02y - 6 = -x + 1 \Rightarrow x + 2y - 7 = 0

Explanation:

We first compute the derivative to find the general slope function. By substituting the x-coordinate of the given point into the derivative, we find the specific slope of the tangent. We then use the point-slope formula for the tangent and the negative reciprocal slope for the normal.

Problem 2:

Find the points on the curve y=x3−3x2−9x+7y = x^3 - 3x^2 - 9x + 7 where the tangents are parallel to the x-axis.

Solution:

  1. Differentiate the function: dydx=3x2−6x−9\frac{dy}{dx} = 3x^2 - 6x - 9
  2. For tangents to be parallel to the x-axis, the slope must be zero: 3x2−6x−9=03x^2 - 6x - 9 = 0
  3. Solve for xx: x^2 - 2x - 3 = 0 \Rightarrow (x - 3)(x + 1) = 0$$ So, x = 3ororx = -1$
  4. Find corresponding yy values: For x=3x = 3: y=(3)3−3(3)2−9(3)+7=27−27−27+7=−20y = (3)^3 - 3(3)^2 - 9(3) + 7 = 27 - 27 - 27 + 7 = -20 For x=−1x = -1: y=(−1)3−3(−1)2−9(−1)+7=−1−3+9+7=12y = (-1)^3 - 3(-1)^2 - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12
  5. The points are (3,−20)(3, -20) and (−1,12)(-1, 12).

Explanation:

Tangents parallel to the x-axis have a slope of 00. By setting the derivative to zero and solving the resulting quadratic equation, we find the x-coordinates of the points. Substituting these back into the original curve equation provides the corresponding y-coordinates.