Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.1.10

Chapter 3, Problem 3.1.10

In Exercises 5–10, find an equation for the tangent line to the curve at the given point. Then sketch the curve and tangent line together.

y = (1 / x³), (−2, −1/8)

Verified step by step guidance

First, identify the function given: \( y = \frac{1}{x^3} \). We need to find the derivative of this function to determine the slope of the tangent line at the given point.

To find the derivative, use the power rule. Rewrite the function as \( y = x^{-3} \) and differentiate: \( \frac{dy}{dx} = -3x^{-4} \).

Evaluate the derivative at the given point \( x = -2 \) to find the slope of the tangent line. Substitute \( x = -2 \) into the derivative: \( \frac{dy}{dx} = -3(-2)^{-4} \).

With the slope calculated, use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point \((-2, -\frac{1}{8})\).

Substitute the slope and the point into the point-slope form to get the equation of the tangent line. Simplify the equation to express it in the form \( y = mx + b \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function at a point provides the slope of the tangent line to the curve at that point. For the function y = (1 / x³), the derivative can be found using the power rule, which helps in determining the rate of change of the function with respect to x. This slope is crucial for writing the equation of the tangent line.

Point-Slope Form of a Line

The point-slope form is a method for writing the equation of a line when you know a point on the line and its slope. It is expressed as y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the given point. This form is used to find the equation of the tangent line once the slope is determined from the derivative.

Graphing Functions and Tangent Lines

Graphing involves plotting the function and its tangent line to visualize their relationship. For y = (1 / x³), sketching the curve helps in understanding its behavior, while the tangent line at the point (-2, -1/8) shows the instantaneous rate of change. This visual representation aids in comprehending how the tangent line approximates the curve locally.

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Slopes of Tangent Lines

Related Practice

Textbook Question

In Exercises 41–58, find dy/dt.

y = ((3t − 4) / (5t + 2))⁻⁵

Textbook Question

Theory and Examples

The equations in Exercises 49 and 50 give the position s = f(t) of a body moving on a coordinate line (s in meters, t in seconds). Find the body’s velocity, speed, acceleration, and jerk at time t = π/4 sec.

s = 2 − 2 sin t

Textbook Question

Find the derivatives of the functions in Exercises 1–42.

𝔂 = x⁵ - 0.125x² + 0.25x

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Textbook Question

Find the derivatives of the functions in Exercises 1–42.

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𝓻 = sin √ 2θ

Textbook Question

Graphs

Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d).