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.
Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
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 guidance1
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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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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Related Practice
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
dy/dx = 2x − 7, y(2) = 0
Textbook Question
Theory and Examples
Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.
Textbook Question
Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(cscθ cotθ) / 2 dθ
Textbook Question
[Technology Exercises] When solving Exercises 14–30, you may need to use appropriate technology (such as a calculator or a computer).
26. Factoring a quartic Find the approximate values of r_1 through r_4 in the factorization
8x^4-14x^3-9x^2+11x-1=8(x-r_1)(x-r_2)(x-r_3)(x-r_4)
Textbook Question
Identifying Extrema
In Exercises 15–18:
a. Find the open intervals on which the function is increasing and those on which it is decreasing.
b. Identify the function’s local and absolute extreme values, if any, saying where they occur.