Question

Normal lines \(A\) line perpendicular to another line or to a tangent line is often called a normal line. Find an equation of the line perpendicular to the line that is tangent to the following curves at the given point \(P\) $$y=\sqrt{x} ; P(4,2)$$

Short answer

Answer: The equation of the normal line to the curve \(y = \sqrt{x}\) at point \(P(4, 2)\) is \(y - 2 = -4(x - 4)\).

Step-by-step solution

Find the derivative of the curve

Given the curve \(\displaystyle y=\sqrt{x}\), first, find its derivative with respect to \(x\). As \(\displaystyle y=x^{\frac{1}{2}}\), we can use the power rule to find the derivative. $$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}}$$

Evaluate the derivative at point P

Evaluate the derivative at point \(\displaystyle P( 4,2)\): $$\frac{dy}{dx}|_{x=4} = \frac{1}{2}(4)^{-\frac{1}{2}} = \frac{1}{4}$$ The slope of the tangent line at point \(\displaystyle P( 4,2)\) is \(\displaystyle \frac{1}{4}\).

Find the equation of the tangent line

Using the point-slope form of a line, we can find the equation of the tangent line: $$y - y_{1} = m(x - x_{1})$$ Inserting point \(\displaystyle P\displaystyle \left( 4,2\right) \) and the slope \(\displaystyle \frac{1}{4}\), we get: $$y - 2 = \frac{1}{4}(x - 4)$$

Determine the slope of the normal line

Since the normal line is perpendicular to the tangent line, its slope is the negative inverse of the tangent's slope: $$m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{\frac{1}{4}} = -4$$

Find the equation of the normal line

Using the point-slope form of a line again with point \(\displaystyle P( 4,2)\) and slope \(\displaystyle -4\), we can find the equation of the normal line: $$y - y_{1} = m(x - x_{1})$$ Inserting the point and the slope, we get: $$y - 2 = -4(x - 4)$$ That is the equation of the normal line.

Key concepts

Tangent Lines

A tangent line to a curve at a given point is a straight line that just touches the curve at that point, without crossing it. It represents the immediate direction the curve is heading at that exact point. Imagine a car driving on a curved road, the tangent line shows the path of the car at that moment.
To find a tangent line, first, understand:

• Tangent lines have the same slope as the curve's derivative at the point of tangency.
• They are key in finding normal lines since normal lines are perpendicular to tangent lines.

For the curve given by the equation \(y = \sqrt{x}\), and at point \(P(4,2)\), the tangent line equation uses the point-slope formula with the derivative's value as the slope.

Derivative

Derivatives are fundamental in calculus and measure how a function changes as its input changes. They can be aptly described as the rate of change or the function's instantaneous slope at any given point.
For functions, derivatives can be computed using different rules, with one of the simplest being the power rule.

• The power rule: If \(y = x^n\), then \( \frac{dy}{dx} = n \cdot x^{n-1} \)

In this problem, the derivative of \(y = \sqrt{x}\) or \(y = x^{1/2}\) is calculated as \( \frac{dy}{dx} = \frac{1}{2} x^{-1/2} \).
This derivative tells us the slope of the tangent line at any point on the curve \(y = \sqrt{x}\). By evaluating it at \(x = 4\), we find it is \( \frac{1}{4} \). This shows how steep the curve is at point \(P\).

Slope

The slope of a line in mathematics is a measure of its steepness or the angle at which it inclines from the horizontal. It's a vital concept when determining the characteristics of a line in algebra.
Slopes:

• Positive slope: Line goes upwards as you move right.
• Negative slope: Line goes downwards as you move right.
• Zero slope: Line is horizontal.
• Undefined slope: Line is vertical.

In the context of tangent and normal lines:

• The slope of the tangent line at \(P(4,2)\) is \(1/4\), as computed from the derivative.
• For normal lines, which are perpendicular to tangent lines, the slope is the negative reciprocal. Thus, the normal line at \(P\) has a slope of \(-4\).

Point-Slope Form

The point-slope form of a linear equation is an algebraic way to express lines on a coordinate plane. It is given by the formula:

• \(y - y_1 = m(x - x_1)\)

Where:

• \(m\) represents the slope.
• \((x_1, y_1)\) is a specific point on the line.

In this particular exercise:

• The tangent line at \(P(4,2)\) uses \(m = \frac{1}{4}\) and rearranges into the formula \(y - 2 = \frac{1}{4}(x - 4)\).
• The normal line uses a slope of \(-4\) and the same point to form \(y - 2 = -4(x - 4)\).

Understanding this form helps in quickly writing the equation of a line when given a point and its slope, making it an essential tool in geometry and calculus.