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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.9.95
Chapter 4, Problem 4.9.95

Velocity to position Given the following velocity functions of an object moving along a line, find the position function with the given initial position.


v(t) = 6t² + 4t - 10; s(0) = 0

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Step 1: Understand the problem. The velocity function v(t) = 6t² + 4t - 10 represents the rate of change of the position function s(t). To find the position function, we need to integrate the velocity function with respect to time t.
Step 2: Set up the integral. The position function s(t) is obtained by integrating v(t): s(t) = ∫(6t² + 4t - 10) dt.
Step 3: Perform the integration term by term. Use the power rule for integration: ∫tⁿ dt = (tⁿ⁺¹)/(n+1), and for constants, ∫C dt = Ct. Applying this to each term: ∫6t² dt = 2t³, ∫4t dt = 2t², and ∫-10 dt = -10t.
Step 4: Combine the results of the integration. The position function becomes s(t) = 2t³ + 2t² - 10t + C, where C is the constant of integration.
Step 5: Use the initial condition s(0) = 0 to solve for C. Substitute t = 0 into s(t): s(0) = 2(0)³ + 2(0)² - 10(0) + C = 0. This simplifies to C = 0. Therefore, the position function is s(t) = 2t³ + 2t² - 10t.

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Key Concepts

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

Velocity and Position Functions

Velocity is the rate of change of position with respect to time, represented mathematically as the derivative of the position function. To find the position function from a given velocity function, one must integrate the velocity function. This process reverses differentiation, allowing us to recover the original position function from its rate of change.
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Integration

Integration is a fundamental concept in calculus that involves finding the antiderivative of a function. In this context, integrating the velocity function v(t) = 6t² + 4t - 10 will yield the position function s(t). The result of integration includes a constant of integration, which can be determined using initial conditions, such as the initial position s(0) = 0.
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Initial Conditions

Initial conditions are specific values that help determine the constants of integration when solving differential equations. In this problem, the initial position s(0) = 0 provides a boundary condition that allows us to solve for the constant after integrating the velocity function. This ensures that the position function accurately reflects the object's position at the start of the observation.
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Related Practice
Textbook Question

{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.


f(x) = sin x + x - 1; x₀ = 0.5

Textbook Question

Use the following graphs to identify the points (if any) on the interval [a, b] at which the function has an absolute maximum or an absolute minimum value <IMAGE>

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Textbook Question

Particular antiderivatives For the following functions f, find the antiderivative F that satisfies the given condition.

f(x) = (3y + 5)/y; F(1) = 3. y > 0

Textbook Question

17–83. Limits Evaluate the following limits. Use l’Hôpital’s Rule when it is convenient and applicable.


lim_Θ→π/2⁻ (tan Θ)ᶜᵒˢ ᶿ

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Textbook Question

Increasing and decreasing functions. Find the intervals on which f is increasing and the intervals on which it is decreasing.


f(x) = -2x⁴ + x² + 10

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Textbook Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = x³ - 6x² + 9x