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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.2.47
Chapter 4, Problem 4.2.47

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 = 9.8, v(0) = −3, s(0) = 0

Verified step by step guidance
1
Start by integrating the acceleration function a(t) = 9.8 with respect to time t to find the velocity function v(t). This involves finding the antiderivative of a constant function.
The antiderivative of a constant 9.8 is 9.8t. Therefore, v(t) = 9.8t + C, where C is the constant of integration.
Use the initial condition v(0) = -3 to solve for the constant C. Substitute t = 0 and v(0) = -3 into the velocity equation: -3 = 9.8(0) + C.
Solve for C to find that C = -3. Thus, the velocity function is v(t) = 9.8t - 3.
Integrate the velocity function v(t) = 9.8t - 3 with respect to time t to find the position function s(t). Use the initial condition s(0) = 0 to solve for the constant of integration in the position function.

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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 find the velocity function from the acceleration function by integrating acceleration with respect to time. This step is crucial for determining the velocity at any given time.
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Introduction to Indefinite Integrals

Initial Conditions

Initial conditions are values given at the start of a problem that help determine the specific solution to a differential equation. Here, the initial velocity v(0) = -3 and initial position s(0) = 0 are used to find the constants of integration when solving for velocity and position functions.
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Initial Value Problems

Position Function

The position function s(t) describes the location of an object at any time t. It is found by integrating the velocity function, which itself is derived from the acceleration function. The position function is essential for understanding the object's movement over time and is determined using integration and initial conditions.
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Relations and Functions
Related Practice
Textbook Question

Each of Exercises 67–88 gives the first derivative of a continuous function y=f(x). Find y'' and then use Steps 2–4 of the graphing procedure described in this section to sketch the general shape of the graph of f.

69. y' = x(x - 3)²

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Textbook Question

6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.

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Textbook Question

Initial Value Problems


Solve the initial value problems in Exercises 71–90.


dy/dx = 2x − 7, y(2) = 0

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Textbook Question

Theory and Examples


Determine the values of constants a and b so that f(x) = ax² + bx has an absolute maximum at the point (1,2).

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Textbook Question

30. Find a positive number for which the sum of its reciprocal and four times its square is the smallest possible.

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Textbook Question

Finding Functions from Derivatives


Suppose that f(0) = 5 and that f'(x) = 2 for all x. Must f(x) = 2x + 5 for all x? Give reasons for your answer.

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