Problem 9 - Calculus for Biology and Medicine

Question

In Problems 9-11, find the lines that are (a) tangential and (b) normal to each curve at the given point. $$ x^{2}+y^{2}=25,(4,-3) \text { (circle) } $$

Step-by-Step Solution

Answer

Tangent line: \(y = \frac{4}{3}x - \frac{25}{3}\); Normal line: \(y = -\frac{3}{4}x\).

Step 1: Understand the Given Problem

We are given the equation of a circle, \(x^2 + y^2 = 25\), and a specific point, \((4, -3)\), on this circle. The task is to find both the tangent and normal lines to the circle at this point.

Step 2: Differentiate the Circle's Equation

To find the slope of the tangent line, differentiate the circle's equation implicitly with respect to \(x\):\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(25) \]This gives us:\[ 2x + 2y\frac{dy}{dx} = 0 \]Solving for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = -\frac{x}{y} \]

Step 3: Substitute the Point into the Derivative

Using the coordinates of the point \((4, -3)\), substitute into the derivative to find the slope of the tangent line:\[ \frac{dy}{dx} = -\frac{4}{-3} = \frac{4}{3} \]So, the slope of the tangent line at \((4, -3)\) is \(\frac{4}{3}\).

Step 4: Write the Equation of the Tangent Line

Using the point-slope form, the equation of the tangent line is given by:\[ y - y_1 = m(x - x_1) \]Substitute \(m = \frac{4}{3}, x_1 = 4, y_1 = -3\):\[ y + 3 = \frac{4}{3}(x - 4) \]Simplifying:\[ y = \frac{4}{3}x - \frac{16}{3} - 3 \]\[ y = \frac{4}{3}x - \frac{25}{3} \]

Step 5: Determine the Slope of the Normal Line

The normal line is perpendicular to the tangent line. Thus, its slope is the negative reciprocal of the tangent line's slope:\[ m_{normal} = -\frac{3}{4} \]

Step 6: Write the Equation of the Normal Line

Using the point-slope form for the normal line and \(m_{normal} = -\frac{3}{4}\), the equation is:\[ y - (-3) = -\frac{3}{4}(x - 4) \]Simplifying, we get:\[ y + 3 = -\frac{3}{4}x + 3 \]\[ y = -\frac{3}{4}x + 0 \]

Step 7: Review Final Equations

The equation of the tangent line is \(y = \frac{4}{3}x - \frac{25}{3}\), and the equation of the normal line is \(y = -\frac{3}{4}x\).

Key Concepts

Implicit Differentiation

When dealing with equations that aren't easily solved for one variable, implicit differentiation becomes very handy. In the case of the circle equation given by \(x^2 + y^2 = 25\), solving directly for \(y\) can be complicated. Thus, we differentiate the equation in relation to \(x\) without isolating \(y\).
This process helps us find how \(y\) changes with \(x\), which will further allow us to find the slopes of tangent and normal lines.
When differentiating implicitly:

• Apply the derivative to each term separately.
• For terms involving \(y\), apply the chain rule, which includes the derivative \(\frac{dy}{dx}\).
• Combine and solve the resulting equation for \(\frac{dy}{dx}\) to express the slope of the tangent line.

This approach is excellent in equations that contain a mixture of variables and would be complex to solve explicitly for one.

Slope Calculation

A critical step in finding lines that are tangent or normal to a curve at a given point involves calculating the slope. The slope describes the steepness and direction of a line, and here we are concerned with the tangent and normal lines of a circle.
Once we've used implicit differentiation on the circle's equation, \(x^2 + y^2 = 25\), to get \[ \frac{dy}{dx} = -\frac{x}{y} \], we substitute the specific point on the circle, \((4, -3)\).
The calculation \(\frac{dy}{dx} = \frac{4}{3}\) gives us the slope of the tangent line, which informs us how steep the line is and in which direction it moves as it passes through the point \((4, -3)\).

• The tangent line shares the slope of the curve at a specific point, giving direct contact without crossing.
• The normal line’s slope is the negative reciprocal of the tangent slope, offering a perpendicular intersection with the tangent.

The contrasts between tangent and normal slopes are pivotal, especially in circular geometry.

Equation of a Line

Once the slope of the line is known, writing the equation involves a straightforward formula. The equation of a line in point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope.
For a tangent line, with slope \(\frac{4}{3}\) and point \((4, -3)\), substituting gives us:

• \(y + 3 = \frac{4}{3}(x - 4)\)
• Simplifying this leads to \(y = \frac{4}{3}x - \frac{25}{3}\)

A normal line has slope \(-\frac{3}{4}\). When using the same point in the equation form, it results in:

• \(y + 3 = -\frac{3}{4}(x - 4)\)
• This simplifies to \(y = -\frac{3}{4}x\)

Through these calculations, you can precisely frame how each line traverses through given points on the circle, with unique slopes defining their distinct paths.

Analytical Geometry

This branch of mathematics deals intricately with points, lines, and surfaces using an algebraic approach. In this context, the problem involves understanding a circle, lines tangent to the circle, and those normal to it.
Analytical geometry enables the blending of algebra with geometry to solve problems like these smoothly. It explains the relationship between the circle's algebraic expression and its geometric properties.
Analyzing the circle, \(x^2 + y^2 = 25\), through differentiation and slope calculations involves observing how:

• Tangent lines interact with the circle at only one point, maintaining a single-direction spacing along the curve's edge.
• Normal lines intersect tangents perpendicularly, indicative of forces like tension or physics concepts of vectors.

Analytical geometry offers a structured pathway from algebraic manipulation to geometric visualizations, quintessential for understanding modern calculus and further mathematical applications.