Problem 10 - University Calculus: Early Transcendentals

Question

Find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$y=\frac{1}{x^{3}}, \quad\left(-2,-\frac{1}{8}\right)$$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = -\frac{3}{16}x - \frac{1}{2} \).

Step 1: Find the Derivative of the Function

The first step is to find the derivative of the function since the derivative represents the slope of the tangent line at any given point. We start with the function \( y = \frac{1}{x^3} \). Rewrites the function as \( y = x^{-3} \) for easier differentiation. Using the power rule, the derivative is: \[ \frac{dy}{dx} = -3x^{-4} = -\frac{3}{x^4} \].

Step 2: Evaluate the Derivative at the Given Point

Now that we have the derivative, we find the slope of the tangent at the point \((-2, -\frac{1}{8})\). Substitute \(x = -2\) into the derivative: \[ \frac{dy}{dx} \bigg|_{x = -2} = -\frac{3}{(-2)^4} = -\frac{3}{16} \]. This is the slope of the tangent line.

Step 3: Use the Point-Slope Form to Find the Equation of the Tangent Line

With the slope \(-\frac{3}{16}\) and the point \((-2, -\frac{1}{8})\), use the point-slope form formula: \( y - y_1 = m(x - x_1) \), where \(m\) is the slope and \((x_1, y_1)\) is the point on the line. Substitute the values: \[ y + \frac{1}{8} = -\frac{3}{16}(x + 2) \].

Step 4: Simplify the Equation of the Tangent Line

Expand and simplify the equation: \[ y + \frac{1}{8} = -\frac{3}{16}x - \frac{3}{8} \]. Subtract \(\frac{1}{8}\) from both sides to solve for \(y\): \[ y = -\frac{3}{16}x - \frac{3}{8} - \frac{1}{8} \]. Combine the constants: \[ y = -\frac{3}{16}x - \frac{4}{8} = -\frac{3}{16}x - \frac{1}{2} \].

Step 5: Sketch the Curve and Tangent

Draw the curve \( y = \frac{1}{x^3} \), which is a typical reciprocal power function with a decreasing curve in quadrants I and III. Next, add the tangent line \( y = -\frac{3}{16}x - \frac{1}{2} \), passing through the point \((-2, -\frac{1}{8})\), with a slope of \(-\frac{3}{16}\). The tangent line should just touch the curve at this point without crossing it.

Key Concepts

Tangent Line

A tangent line is a straight line that "just touches" a curve at a given point. This line has the same slope as the curve at that specific point, meaning it represents the direction in which the curve is proceeding as closely as possible at that point.

• For the function given, the tangent line at the point of interest will provide a snapshot of the curve's slope there.
• This is crucial for understanding the behavior of the function locally around the point.

When you lay tangent lines and curves together, it gives a visual understanding and confirms analytical work. The tangent won't cross the curve at the tangent point—showing it only "touches."

Derivative

Finding the derivative of a function is critical because it represents the slope of that function at any given point. It's a mathematic tool used in calculus to uncover functions' rates of change.

• The derivative function tells us how quickly or slowly the original function's value is changing.
• Using the power rule simplifies the process when dealing with polynomial expressions.
In our example, the original function was transformed into an easily differentiable form, so finding the derivative was straightforward. The derivative discovered gives us a formula that tells us the slope of the tangent line for any x value in the function.

Slope of a Tangent

The slope of a tangent line is essentially what you get when you plug a specific value into the derivative of a function. This slope dictates the angle at which the tangent meets the curve at the given point. When describing the slope, we think of the incline or decline steepness of the tangent line.

• It's important to properly substitute into the derivative to ensure accurate slope calculations.
• A correct slope will confirm the tangent line perfectly aligns with the curve's behavior at that point.

In the example provided, after evaluating the derivative at our specific x-value, we found a slope of \(-\frac{3}{16}\). This slope is essential for drawing the tangent and further calculations.

Point-Slope Form

The point-slope form is a mathematical equation used to describe a line in algebra, which is particularly useful when dealing with problems of finding tangents to curves.

• The general form is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is a known point on the line.
• This form eases the transition from having a "slope" to constructing a usable "linear equation."
By using the slope from the derivative, and the specific point on the curve, the equation for the tangent line was formulated. Simplifying leads to a comprehensible equation ready for graphing or further analysis.