Problem 5 - Thomas Calculus

Question

Find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together. \(y=4-x^{2}, \quad(-1,3)\)

Step-by-Step Solution

Answer

The tangent line at \((-1, 3)\) is \(y = 2x + 5\).

Step 1: Finding the Derivative

The first step in finding the equation of the tangent line is to find the derivative of the function. Given the function \(y = 4 - x^2\), the derivative \(y'\) will be \(-2x\). This derivative represents the slope of the tangent line at any point \(x\) on the curve.

Step 2: Evaluating the Derivative at Given Point

Next, substitute \(x = -1\) into the derivative to find the slope of the tangent line at the point \((-1, 3)\). The slope \(m\) is \(-2(-1) = 2\).

Step 3: Using Point-Slope Form

With the slope and the point, use the point-slope form of a line: \(y - y_1 = m(x - x_1)\). Using the point \((-1, 3)\) and slope \(m = 2\), the equation becomes \(y - 3 = 2(x + 1)\).

Step 4: Simplifying the Equation

Expand and simplify the equation from the previous step: \(y - 3 = 2x + 2\), which simplifies to \(y = 2x + 5\). This is the equation of the tangent line.

Key Concepts

Derivative

The derivative is a critical concept in calculus that measures how a function changes as its input changes. In simpler terms, it's a way of finding the slope of a function at any given point. For the exercise provided, the function is given by the equation:

• \(y = 4 - x^2\)

To find the slope of the curve at any point, we compute its derivative. Derivatives provide a powerful tool for understanding how functions behave.
In this case, the derivative is computed as:

• \(y' = -2x\)

This derivative gives us the slope of the tangent line at each point on the curve. By substituting the specific x-value of the point of interest into the derivative, we can find the exact slope at that point.

Point-Slope Form

The point-slope form is a straightforward way to write the equation of a line when you know the slope and a point on the line. It is expressed as:

• \(y - y_1 = m(x - x_1)\)

Here, \(m\) represents the slope, and \((x_1, y_1)\) is a specific point on the line.
This formula is particularly helpful in our exercise, where the slope provided by the derivative is used along with the given point

• \((-1,3)\)

Using the point-slope form, we use the calculated slope (\(m = 2\)) and substitute this along with the coordinates

• \(x_1 = -1, y_1 = 3\)

Thus, the equation of the tangent line is

• \(y - 3 = 2(x + 1)\)

This equation can then be expanded to give us a clearer view of the tangent line in slope-intercept form.

Slope of Tangent Line

The slope of the tangent line is central to understanding how a curve behaves at a given point. It tells us whether the curve is increasing or decreasing and how steeply. In the derivative section, we found that the slope of the tangent line at

• \((-1, 3)\)

was calculated by evaluating

• \(y' = -2x\)

The slope, therefore, is

• \(-2(-1) = 2\)

This indicates that at the point

• \((-1, 3)\)

The curve is increasing with a slope of 2.
Understanding the slope helps in making predictions about the curve's behavior near that point. A positive slope indicates that as you move to the right on the coordinate plane, the curve moves upwards.

Curve Sketching

Curve sketching involves drawing both the function's graph and the tangent line at the given point. Let's break it down:
First, we consider the curve

• \(y = 4 - x^2\)

This is a downward-opening parabola with a vertex at

• \((0, 4)\)

Now, the tangent line at point

• \((-1, 3)\)

is given by

• \(y = 2x + 5\)

When sketching both the curve and the tangent line, notice how the tangent line just "touches" the curve at the provided point and reflects the curve's instantaneous rate of change at that point.
This technique not only helps in visualizing the relationship between the curve and the tangent line but also in understanding how calculus can be used to analyze real-world problems with visual representations.