Problems Based on Normals to a Curve

In calculus, the concept of a normal to a curve at a given point is closely related to the tangent to the curve at that point. While the tangent line touches the curve at a single point and has the same direction as the curve at that point, the normal line is perpendicular to the tangent line and therefore to the curve at the point of tangency.

Understanding Normals

A normal to a curve at a given point is a straight line that is perpendicular to the tangent of the curve at that point. The slope of the normal is the negative reciprocal of the slope of the tangent if the slope of the tangent is not zero. If the tangent is vertical, the normal is horizontal and vice versa.

Slope of the Normal

If the slope of the tangent to the curve at a point (x, y) is m_t, then the slope of the normal m_n is given by:

$$ m_n = -\frac{1}{m_t} $$

Equation of the Normal

Given a curve defined by a function y = f(x), and a point (x_1, y_1) on the curve, the equation of the normal can be found using the point-slope form of a line:

$$ y - y_1 = m_n (x - x_1) $$

where m_n is the slope of the normal.

Table of Differences and Important Points

Aspect	Tangent	Normal
Definition	A line that touches the curve at one point and has the same direction as the curve at that point.	A line that is perpendicular to the tangent at the point of contact.
Slope	The derivative of the function at the point of tangency (m_t = f'(x_1)).	The negative reciprocal of the slope of the tangent (m_n = -1/m_t).
Equation	y - y_1 = m_t (x - x_1)	y - y_1 = m_n (x - x_1)
Relationship	The tangent is parallel to the curve at the point of tangency.	The normal is perpendicular to both the curve and the tangent at the point of tangency.

Formulas

For a curve y = f(x), the slope of the tangent at a point (x_1, y_1) is given by the derivative of f at x_1:

$$ m_t = f'(x_1) $$

The slope of the normal is then:

$$ m_n = -\frac{1}{m_t} = -\frac{1}{f'(x_1)} $$

The equation of the normal at (x_1, y_1) is:

$$ y - y_1 = -\frac{1}{f'(x_1)} (x - x_1) $$

Examples

Example 1: Find the equation of the normal

Consider the curve y = x^2. Find the equation of the normal to the curve at the point (1, 1).

Solution:

First, find the slope of the tangent (m_t) by taking the derivative of y with respect to x:

$$ f'(x) = 2x $$

At the point (1, 1), the slope of the tangent is:

$$ m_t = f'(1) = 2(1) = 2 $$

The slope of the normal (m_n) is the negative reciprocal of m_t:

$$ m_n = -\frac{1}{m_t} = -\frac{1}{2} $$

Now, use the point-slope form to find the equation of the normal:

$$ y - 1 = -\frac{1}{2} (x - 1) $$

Simplifying, we get:

$$ y = -\frac{1}{2}x + \frac{3}{2} $$

This is the equation of the normal to the curve y = x^2 at the point (1, 1).

Example 2: Normal to a Circle

Find the equation of the normal to the circle x^2 + y^2 = r^2 at the point (r/\sqrt{2}, r/\sqrt{2}).

Solution:

The slope of the tangent to the circle at any point (x_1, y_1) is given by:

$$ m_t = -\frac{x_1}{y_1} $$

At the point (r/\sqrt{2}, r/\sqrt{2}), the slope of the tangent is:

$$ m_t = -\frac{r/\sqrt{2}}{r/\sqrt{2}} = -1 $$

The slope of the normal is the negative reciprocal of m_t:

$$ m_n = -\frac{1}{m_t} = 1 $$

Using the point-slope form, the equation of the normal is:

$$ y - \frac{r}{\sqrt{2}} = 1 \left(x - \frac{r}{\sqrt{2}}\right) $$

Simplifying, we get:

$$ y = x $$

This is the equation of the normal to the circle at the point (r/\sqrt{2}, r/\sqrt{2}).

Understanding how to find normals to a curve is essential for solving many problems in calculus, especially those involving optimization and geometric applications. It is also a fundamental concept in differential geometry and plays a crucial role in the study of curves and surfaces.