Problem 51 - Calculus for Biology and Medicine

Question

Find the normal line, in standard form, to \(y=f(x)\) at the indicated point. $$ y=\sqrt{3} x^{4}-2 \sqrt{3} x^{2}, \text { at } x=-\sqrt{3} $$

Step-by-Step Solution

Answer

The normal line is \( x - 24\sqrt{3}y = -216 \).

Step 1: Find the derivative

To find the normal line, we first need to determine the slope of the tangent line at the given point. This requires finding the derivative of the function. Given \( y = \sqrt{3}x^4 - 2\sqrt{3}x^2 \), differentiate with respect to \( x \) to get \( y' = \frac{d}{dx}(\sqrt{3}x^4 - 2\sqrt{3}x^2) = 4\sqrt{3}x^3 - 4\sqrt{3}x \).

Step 2: Evaluate the derivative at the given point

Now calculate the derivative at \( x = -\sqrt{3} \). Substitute \( x = -\sqrt{3} \) into \( y' = 4\sqrt{3}x^3 - 4\sqrt{3}x \) to find the slope of the tangent line (\( m_t \)). This gives \( m_t = 4\sqrt{3}(-\sqrt{3})^3 - 4\sqrt{3}(-\sqrt{3}) = -36\sqrt{3} + 12\sqrt{3} = -24\sqrt{3} \).

Step 3: Determine the normal slope

The slope of the normal line (\( m_n \)) is the negative reciprocal of the tangent slope. Therefore, \( m_n = -\frac{1}{m_t} = -\frac{1}{-24\sqrt{3}} = \frac{1}{24\sqrt{3}} \).

Step 4: Find the point on the curve

Substitute \( x = -\sqrt{3} \) into the original function to find the corresponding \( y \)-coordinate. This gives \( y = \sqrt{3}(-\sqrt{3})^4 - 2\sqrt{3}(-\sqrt{3})^2 = 27\sqrt{3} - 18\sqrt{3} = 9\sqrt{3} \). Thus, the point is \((-\sqrt{3}, 9\sqrt{3})\).

Step 5: Write the equation of the normal line in point-slope form

Using the normal slope \( m_n = \frac{1}{24\sqrt{3}} \) and the point \( (-\sqrt{3}, 9\sqrt{3}) \), the point-slope form is \( y - 9\sqrt{3} = \frac{1}{24\sqrt{3}}(x + \sqrt{3}) \).

Step 6: Convert the equation to standard form

To convert the point-slope form to standard form \( Ax + By = C \), start by eliminating the fraction by multiplying through by \( 24\sqrt{3} \), giving \( 24\sqrt{3}(y - 9\sqrt{3}) = x + \sqrt{3} \). Simplifying and rearranging gives \( x - 24\sqrt{3}y = -216 \). This is the standard form of the normal line equation.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus. It represents the rate at which a function is changing at any given point. In simpler terms, consider it as the slope of the tangent line touching the curve at that specific point.

• To find the derivative of a function like \( y = \sqrt{3}x^4 - 2\sqrt{3}x^2 \), you apply differentiation rules to each term.
• This involves using a simple rule: \( \frac{d}{dx}(ax^n) = n \times ax^{n-1} \).

For our function, differentiating gives \( y' = 4 \sqrt{3}x^3 - 4 \sqrt{3}x \). These terms represent the instantaneous rate of change of \( y \) with respect to \( x \).
The derivative is crucial for finding the slope of the tangent line, which leads us to our next section.

Tangent Line

A tangent line is a straight line that just "grazes" a curve at a certain point, but does not cross it. This line has the same slope as the curve at that point.

• The slope of the tangent line is simply the value of the derivative evaluated at the specific point.
• In our exercise, the derivative at \( x = -\sqrt{3} \) is \( -24\sqrt{3} \).

Once you've calculated the derivative, the tangent line's slope at that exact point becomes clear. It describes how steeply (upward or downward) the curve is heading as it passes the specific point of interest.
Knowing the slope helps in constructing tangent and normal lines, aiding in deeper analysis of the function's behavior.

Normal Line

The normal line to a curve at a given point is a line perpendicular to the tangent line at that specific point. While the tangent line skims along the curve, the normal line dives right in, forming a perfect right angle.

• To find the slope of the normal line, take the negative reciprocal of the tangent line's slope.
• For example, if the tangent slope is \( m_t = -24\sqrt{3} \), then the normal slope \( m_n \) is \( \frac{1}{24\sqrt{3}} \).

This operation flips and changes the sign of the tangent slope, ensuring that the normal slope results in a perpendicular interaction.
The normal line is instrumental in applications such as assessing angles between curves and determining orthogonal properties in geometry.

Point-Slope Form

The point-slope form is an equation format helpful for writing the equation of a line when you know its slope and a point on the line. It is written as:

• \( y - y_1 = m(x - x_1) \)
• Here, \((x_1, y_1)\) is a known point, and \(m\) is the slope.

In our exercise, we use the point \((-\sqrt{3}, 9\sqrt{3})\) and the normal slope \(\frac{1}{24\sqrt{3}}\) to formulate: \( y - 9\sqrt{3} = \frac{1}{24\sqrt{3}}(x + \sqrt{3}) \).
This format serves as the initial setup before further simplifying or converting into other forms like the standard line equation. It's a straightforward way to get a line's equation, particularly when dealing with calculus-derived slopes.