Ch. 4 - Applications of Derivatives

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

Hass 15th Edition

Ch. 4 - Applications of Derivatives

Problem 45

Chapter 4, Problem 45

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

Verified step by step guidance

Start by integrating the acceleration function a(t) = 32 with respect to time t to find the velocity function v(t). This involves finding the antiderivative of 32.

The antiderivative of a constant 32 is 32t plus a constant of integration, C1. So, v(t) = 32t + C1.

Use the initial condition v(0) = 20 to solve for C1. Substitute t = 0 and v(0) = 20 into the velocity equation: 20 = 32(0) + C1, which gives C1 = 20.

Now, integrate the velocity function v(t) = 32t + 20 with respect to time t to find the position function s(t). This involves finding the antiderivative of 32t + 20.

The antiderivative of 32t is 16t² and the antiderivative of 20 is 20t. So, s(t) = 16t² + 20t + C2. Use the initial condition s(0) = 5 to solve for C2 by substituting t = 0 and s(0) = 5 into the position equation: 5 = 16(0)² + 20(0) + C2, which gives C2 = 5. Thus, the position function is s(t) = 16t² + 20t + 5.

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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.

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) = 20 and initial position s(0) = 5 are used to find the constants of integration when solving for velocity and position functions.

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. Using the initial conditions, we can solve for any constants and determine the exact position function for the object.

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Relations and Functions

Related Practice

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' = 𝓍⁴ ― 2𝓍²

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Textbook Question

The range R of a projectile fired from the origin over horizontal ground is the distance from the origin to the point of impact. If the projectile is fired with an initial velocity at an angle with the horizontal, then in Chapter 13 we find that R-v_0^2/g(sin 2α) where g is the downward acceleration due to gravity. Find the angle α for which the range R is the largest possible.

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Textbook Question

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

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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

y' = 𝓍² ― 𝓍―6

Graphs and Graphing

Graph the curves in Exercises 33–42.

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y = 𝓍√4 ― 𝓍²