Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 43

Chapter 4, Problem 43

Finding Position from Velocity or Acceleration

Exercises 41–44 give the velocity v = ds/dt and initial position of an object moving along a coordinate line. Find the object’s position at time t.

v = sin πt, s(0) = 0

Verified step by step guidance

To find the position function s(t), we need to integrate the velocity function v(t) = sin(πt) with respect to time t. This is because velocity is the derivative of position, so integrating velocity will give us the position.

Set up the integral: ∫sin(πt) dt. This integral will give us the position function s(t) up to a constant of integration, which we will determine using the initial condition.

Perform the integration: The integral of sin(πt) with respect to t is -1/π * cos(πt) + C, where C is the constant of integration.

Use the initial condition s(0) = 0 to find the constant C. Substitute t = 0 into the position function: s(0) = -1/π * cos(0) + C = 0. Since cos(0) = 1, this simplifies to -1/π + C = 0, which means C = 1/π.

Substitute the constant C back into the position function to get the final expression for s(t): s(t) = -1/π * cos(πt) + 1/π.

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Key Concepts

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

Integration

Integration is the process of finding the antiderivative or the area under a curve. In this context, it is used to determine the position function from the velocity function. By integrating the velocity function v(t) = sin(πt), we can find the position function s(t), which describes the object's position over time.

Initial Conditions

Initial conditions are values that specify the state of a system at a particular time, often used to solve differential equations. Here, the initial condition s(0) = 0 indicates the object's position at time t = 0. This information is crucial for determining the constant of integration when finding the position function from the velocity function.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, showing that they are inverse processes. It states that if a function is continuous over an interval, the integral of its derivative over that interval equals the change in the function's values. This theorem is applied to find the position function from the velocity function by integrating and using the initial condition to solve for any constants.

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Related Practice

Textbook Question

Finding Functions from Derivatives

In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.

r'(t) = sec t tan t − 1, P(0, 0)

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Textbook Question

In Exercises 9–66, graph the function using appropriate methods from the graphing procedures presented just before Example 9, identifying the coordinates of any local extreme points and inflection points. Then find coordinates of absolute extreme points, if any.

46. y = cos(x) + √3 * sin(x), 0 ≤ x ≤ 2π

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Textbook Question

Each of Exercises 43–48 gives the first derivative of a function y = ƒ(𝓍). (a) At what points, if any, does the graph of ƒ have a local maximum, local minimum, or inflection point? (b) Sketch the general shape of the graph.

y' = 𝓍² ― 𝓍―6

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Textbook Question

Finding Position from Velocity or Acceleration

Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.

a = 32, v(0) = 20, s(0) = 5

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Textbook Question

Graphs and Graphing

Graph the curves in Exercises 33–42.

______

y = 𝓍√4 ― 𝓍²

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Textbook Question

Finding Position from Velocity or Acceleration