AOS3 Topic 3: Tangents and Normal using derivatives

Understanding Tangents and Normals

Tangent: A tangent to a curve at a particular point is a straight line that just "touches" the curve at that point. It represents the direction in which the curve is heading at that specific point. The tangent line has the same slope as the curve at the point of contact.

Normal: The normal to a curve at a given point is the line that is perpendicular to the tangent at that point. This line intersects the curve at the same point where the tangent line touches the curve.

Finding the Tangent Line

To find the equation of the tangent line to a curve at a given point using derivatives, follow these steps:

  1. Find the Derivative: Compute the derivative of the function that represents the curve. The derivative gives you the slope of the tangent line at any point on the curve.
  2. Evaluate the Derivative at the Point: Substitute the x-coordinate of the point of interest into the derivative to find the slope of the tangent line at that point.
  3. Use the Point-Slope Form: Apply the point-slope form of the line equation to find the equation of the tangent line.

Point-Slope Form: y - y1 = m(x - x1)

Where m is the slope of the tangent line and (x1, y1) is the point of contact on the curve.

Example:

Find the tangent to the curve y = x2 at the point (2, 4).

- The derivative of y = x2 is dy/dx = 2x.

- At x = 2, the slope of the tangent is 2 × 2 = 4.

- The tangent line equation is y - 4 = 4(x - 2), which simplifies to y = 4x - 4.

Finding the Normal Line

To find the equation of the normal line to a curve at a given point using derivatives, follow these steps:

  1. Find the Derivative: As with the tangent line, compute the derivative of the function to get the slope of the tangent line.
  2. Compute the Slope of the Normal: The slope of the normal line is the negative reciprocal of the slope of the tangent line. If the slope of the tangent line is m, then the slope of the normal line is -1/m.
  3. Use the Point-Slope Form: Apply the point-slope form of the line equation to find the equation of the normal line.

Point-Slope Form: y - y1 = mnormal(x - x1)

Where mnormal is the slope of the normal line and (x1, y1) is the point of contact.

Example:

Find the normal to the curve y = x2 at the point (2, 4).

- The slope of the tangent is 4, so the slope of the normal is -1/4.

- The normal line equation is y - 4 = -1/4(x - 2), which simplifies to y = -1/4x + 9/2.

General Formulas

For different types of curves, the general approach remains the same, but the specific formulas for tangents and normals vary:

  • Circle: For a circle with the equation x2 + y2 = r2, the tangent at (x1, y1) is x x1 + y y1 = r2.
  • Parabola: For a parabola y2 = 4ax, the tangent at (x1, y1) is y y1 = 2a(x + x1).
  • Ellipse: For an ellipse x2/a2 + y2/b2 = 1, the tangent at (x1, y1) is x x1/a2 + y y1/b2 = 1.
  • Hyperbola: For a hyperbola x2/a2 - y2/b2 = 1, the tangent at (x1, y1) is x x1/a2 - y y1/b2 = 1.

Summary

  • The tangent line touches the curve at a specific point and has a slope equal to the derivative of the curve at that point.
  • The normal line is perpendicular to the tangent line at the same point and its slope is the negative reciprocal of the tangent’s slope.
  • Both lines are essential for understanding the geometric properties of curves and are widely used in various applications in physics, engineering, and optimization problems.

Example 1

Find the equation of tangent and normal to the circle \( x^2 + y^2 = 5 \), at the point (2, 3).

Solution:
The given equation of the circle is \( x^2 + y^2 = 5 \).
The slope of the tangent is obtained by taking the derivative of the above expression with respect to \( x \):

\[ 2x + 2y \frac{dy}{dx} = 0 \]

\[ \frac{dy}{dx} = \frac{-2x}{2y} \]

\[ \frac{dy}{dx} = \frac{-x}{y} \]

Let us substitute the point (2, 3) into the above differentiation to obtain the slope of the tangent:

\[ \text{Slope of tangent} = m = \frac{dy}{dx} = \frac{-2}{3} \]

The equation of the tangent can be computed using the point-slope form of the equation of a line:

\[ (y - y_1) = m(x - x_1) \]

Substitute \( m = \frac{-2}{3} \), \( x_1 = 2 \), and \( y_1 = 3 \):

\[ (y - 3) = \frac{-2}{3}(x - 2) \]

\[ 3(y - 3) = -2(x - 2) \]

\[ 3y - 9 = -2x + 4 \]

\[ 2x + 3y - 9 - 4 = 0 \]

\[ 2x + 3y - 13 = 0 \]

Therefore, the equation of the tangent is:

\[ 2x + 3y = 13 \]

Equation of the Normal:
The slope of the normal is:

\[ m = -\frac{dx}{dy} = -\left(\frac{-y}{x}\right) = \frac{y}{x} = \frac{3}{2} \]

Using the point (2, 3), the equation of the normal is:

\[ (y - y_1) = m(x - x_1) \]

Substitute \( m = \frac{3}{2} \), \( x_1 = 2 \), and \( y_1 = 3 \):

\[ (y - 3) = \frac{3}{2}(x - 2) \]

\[ 2(y - 3) = 3(x - 2) \]

\[ 2y - 6 = 3x - 6 \]

\[ 3x - 2y = 0 \]

Therefore, the equation of the normal is:

\[ 3x - 2y = 0 \]

Example 2

Find the equation of tangent and normal to the curve \( y = (x - 5)(x - 2)(x - 3) \), where the curve cuts the x-axis.

Solution:
The given equation of the curve is \( y = (x - 5)(x - 2)(x - 3) \).
The point where the above curve cuts the x-axis can be obtained by substituting \( y = 0 \) in the above equation:

\[ 0 = (x - 5)(x - 2)(x - 3) \]

Thus, \( x - 5 = 0 \), which gives:

\[ x = 5 \]

Hence, the curve cuts the x-axis at the point (5, 0).
The equation of the tangent and normal can be calculated by first determining the slope of the curve at the point (5, 0). To do so, we differentiate \( y = (x - 5)(x - 2)(x - 3) \).

\[ \frac{dy}{dx} = \frac{d}{dx}\left((x^2 - 5x + 6)\right) \]

After differentiating, we get:

\[ \frac{dy}{dx} = (x^2 - 5x + 6) \cdot 1 - (x - 5)(2x - 5) \]

Now, substitute the point \( (5, 0) \) into the derivative to find the slope of the tangent:

\[ m = \frac{dy}{dx} = \frac{6}{36} = \frac{1}{6} \]

Equation of the Tangent:
Let us find the equation of the tangent with slope \( m = \frac{1}{6} \), passing through the point (5, 0):

\[ (y - 0) = \frac{1}{6}(x - 5) \]

Multiplying both sides by 6:

\[ 6y = x - 5 \]

\[ x - 6y - 5 = 0 \]

Therefore, the equation of the tangent is:

\[ x - 6y = 5 \]

Equation of the Normal:
The slope of the normal is \( m = -\frac{dx}{dy} = -6 \).
The equation of the normal with slope \( m = -6 \), passing through the point (5, 0), is:

\[ (y - 0) = -6(x - 5) \]

\[ y = -6(x - 5) \]

\[ y = -6x + 30 \]

Therefore, the equation of the normal is:

\[ 6x + y = 30 \]

Example 3

Determine the Equation of the Normal to the Parabola \( y^2 = 8x \) at the Point \( (2, 4) \).

Solution:
The parabola is \( y^2 = 8x \). Differentiate implicitly with respect to \( x \) to find the derivative:
\[ 2y \frac{dy}{dx} = 8 \]
\[ \frac{dy}{dx} = \frac{4}{y} \]
Evaluate the derivative at \( (2, 4) \):
\[ \frac{dy}{dx} \bigg|_{(2, 4)} = \frac{4}{4} = 1 \]
So, the slope \( m \) of the tangent at \( (2, 4) \) is \( 1 \).

The slope of the normal is \( -\frac{1}{m} = -1 \).
To find the normal line equation, use the point-slope form \( y - y_1 = m_{normal}(x - x_1) \):
\[ y - 4 = -1(x - 2) \]
Simplify:
\[ y - 4 = -x + 2 \]
\[ y = -x + 6 \]
Thus, the equation of the normal line is \( y = -x + 6 \).

Example 4

Find the Tangent and Normal to the Ellipse \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \) at the Point \( (3, 4) \)

Solution:
The ellipse equation is \( \frac{x^2}{9} + \frac{y^2}{16} = 1 \). Differentiate implicitly with respect to \( x \):
\[ \frac{2x}{9} + \frac{2y}{16} \frac{dy}{dx} = 0 \]
\[ \frac{dy}{dx} = -\frac{8x}{9y} \]
Evaluate the derivative at \( (3, 4) \):
\[ \frac{dy}{dx} \bigg|_{(3, 4)} = -\frac{8 \cdot 3}{9 \cdot 4} = -\frac{24}{36} = -\frac{2}{3} \]
So, the slope \( m \) of the tangent at \( (3, 4) \) is \( -\frac{2}{3} \).
To find the equation of the tangent line, use the point-slope form \( y - y_1 = m(x - x_1) \):
\[ y - 4 = -\frac{2}{3}(x - 3) \]
Simplify:
\[ y - 4 = -\frac{2}{3}x + 2 \]
\[ y = -\frac{2}{3}x + 6 \]
Thus, the equation of the tangent line is \( y = -\frac{2}{3}x + 6 \).

The slope of the normal is the negative reciprocal of the slope of the tangent. So, it is \( \frac{3}{2} \).
To find the equation of the normal line, use the point-slope form \( y - y_1 = m_{normal}(x - x_1) \):
\[ y - 4 = \frac{3}{2}(x - 3) \]
Simplify:
\[ y - 4 = \frac{3}{2}x - \frac{9}{2} \]
\[ y = \frac{3}{2}x - \frac{1}{2} \]
Thus, the equation of the normal line is \( y = \frac{3}{2}x - \frac{1}{2} \).

Example 5

Given the Hyperbola \( \frac{x^2}{25} - \frac{y^2}{16} = 1 \), Find the Tangent and Normal at the Point \( (5, 0) \).

Solution:
The hyperbola equation is \( \frac{x^2}{25} - \frac{y^2}{16} = 1 \). Differentiate implicitly with respect to \( x \):
\[ \frac{2x}{25} - \frac{2y}{16} \frac{dy}{dx} = 0 \]
\[ \frac{dy}{dx} = \frac{8x}{25y} \]
Evaluate the derivative at \( (5, 0) \):
\[ \frac{dy}{dx} \bigg|_{(5, 0)} = \text{undefined} \]
The derivative is undefined because \( y = 0 \). The slope of the tangent at \( (5, 0) \) is vertical, so the tangent is a vertical line \( x = 5 \).

To find the equation of the normal line, observe that since the slope of the tangent is vertical, the normal line must be horizontal. Hence, the normal line at \( (5, 0) \) is a horizontal line:
\[ y = 0 \]

Exercise 1

Exercise 2

Exercise 3