/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Position from velocity Consider ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Position from velocity Consider an object moving along a line with the given velocity v and initial position. a. Determine the position function, for \(t \geq 0,\) using the antiderivative method b. Determine the position function, for \(t \geq 0,\) using the Fundamental Theorem of Calculus (Theorem 6.1 ). Check for agreement with the answer to part (a). \(v(t)=\sin t\) on \([0,2 \pi] ; s(0)=1\).

Short Answer

Expert verified
Question: Given an object is moving along a line with a velocity function \(v(t) = \sin t\), and an initial position \(s(0) = 1\), find the position function \(s(t)\) for \(t \geq 0\). Answer: The position function of the object moving along the line is \(s(t) = -\cos t + 2\) for \(t \geq 0\).

Step by step solution

01

Finding the position function using the antiderivative method

Since we know that \(v(t) = \frac{ds}{dt}\), we can find the position function \(s(t)\) by taking the antiderivative of the velocity function. The antiderivative of \(\sin t\) is \(-\cos t + C\), where C is the constant of integration. We can use the initial condition \(s(0) = 1\) to find C. \(1 = -\cos(0) + C \Rightarrow C = 2\) Therefore, the position function using the antiderivative method is \(s(t) = -\cos t + 2\).
02

Finding the position function using the Fundamental Theorem of Calculus

Next, we'll use the Fundamental Theorem of Calculus to find the position function \(s(t)\). According to the theorem, if \(v(t)\) is continuous on \([0, 2\pi]\), then: \(s(t) = s(0) + \int_0^t v(\tau)d\tau\) We are given \(s(0) = 1\) and \(v(t) = \sin t\), so we can set up the integral as follows: \(s(t) = 1 + \int_0^t \sin \tau d\tau\) To solve the integral, we recall that the antiderivative of \(\sin \tau\) is \(-\cos \tau\). Therefore, \(s(t) = 1 - [\cos \tau]_0^t = 1 - (\cos t - \cos 0) = -\cos t + 2\)
03

Comparing the results

We found the position function using both the antiderivative method and the Fundamental Theorem of Calculus. In both cases, we arrived at the same position function: \(s(t) = -\cos t + 2\). Since the results agree, we can conclude that the position function of the object moving along the line with velocity \(v(t) = \sin t\) and initial position \(s(0) = 1\) is: \(s(t) = -\cos t + 2\) for \(t \geq 0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Antiderivative Method
The antiderivative method is an essential concept in calculus for finding a function given its derivative. This technique is particularly useful in problems that involve motion, where velocity is known and position needs to be determined. In essence, the antiderivative provides the original function before differentiation occurred. When we have a velocity function, such as v(t) = sin(t), the antiderivative is the position function s(t).

Here's how we apply the method: To find s(t), we take the antiderivative of v(t). Since the antiderivative of sin(t) is -cos(t) + C, where C is a constant, we determine C by using the initial condition — the position of the object at time zero. For instance, with s(0) = 1, we find C by plugging t = 0 into the equation, leading to 1 = -cos(0) + C. After solving for C, we get the position function s(t) = -cos(t) + C.

The initial condition is crucial because it gives us a specific point on our function, which helps us determine the constant of integration. Without this information, we would have an infinite number of possible antiderivatives.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a groundbreaking discovery that links the concept of differentiation to that of integration. It essentially states that if a function f(t) is continuous over an interval, then the definite integral of f(t) from a to b is related to any antiderivative F(t) of f(t) by the equation F(b) - F(a). In the realm of motion, this theorem allows us to find the position of an object at any given time if we know its velocity.

To use this theorem, you start by recognizing that the integral of a velocity function v(t) with respect to time from 0 to t will give us the change in position over that time interval. Then, by adding the initial position, we can find the current position. In the example provided, the position function calculated using the Fundamental Theorem agrees with that found by the antiderivative method, confirming the validity of the result.

It's important to remember that this theorem requires the function v(t) to be continuous over the interval we're considering. If this condition is met, we can apply the theorem with confidence to solve problems related to cumulative effect, such as finding the total distance traveled given a velocity function over time.
Velocity Function Integration
Velocity function integration is a practical application of integral calculus, used to understand motion. When working with velocity functions, like v(t) = sin(t), we use integration to find how much the position has changed over a particular time period. Here’s what makes this process so intriguing: velocity is the rate of change of position, and by integrating the velocity function, we’re essentially summing up all these tiny changes in position to get the position function s(t).

In our exercise, we integrated v(t) from 0 to t to find the change in position, and then added the initial position to get s(t). This process showcases how integration can 'accumulate' all of the small instantaneous velocities over the time interval to provide a complete picture of the object's location at any time t. Keep in mind, to carry out the integration correctly, one needs to understand the antiderivatives for common functions and apply them accurately.
Initial Condition in Calculus
Initial conditions in calculus are vital pieces of information that allow us to determine the specific solution to a differential equation from a family of potential solutions. In the context of motion, initial conditions typically give us the object's position at time zero. For example, s(0) = 1 denotes that the object starts at position 1 when t = 0.

Applying the initial condition enables us to find the specific constant C in the antiderivative, which ensures that our position function s(t) correctly describes the object's movement based on its actual starting point. It’s like solving a mystery with a crucial clue that leads us to a single, definite answer. When combining this with velocity integration, we finalize the position equation that accurately portrays the object's journey along its path. Without such initial conditions, we would be left with many possible trajectories, each representing a different starting scenario, and unable to pinpoint the true path of the object.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Shock absorber A heavy-duty shock absorber is compressed \(2 \mathrm{cm}\) from its equilibrium position by a mass of \(500 \mathrm{kg} .\) How much work is required to compress the shock absorber \(4 \mathrm{cm}\) from its equilibrium position? (A mass of 500 kg exerts a force (in newtons) of \(\left.500 g, \text { where } g=9.8 \mathrm{m} / \mathrm{s}^{2} .\right)\)

Different axes of revolution Suppose \(R\) is the region bounded by \(y=f(x)\) and \(y=g(x)\) on the interval \([a, b],\) where \(f(x) \geq g(x) \geq 0\) a. Show that if \(R\) is revolved about the horizontal line \(y=y_{0}\) that lies below \(R,\) then by the washer method, the volume of the resulting solid is $$ V=\int_{a}^{b} \pi\left(\left(f(x)-y_{0}\right)^{2}-\left(g(x)-y_{0}\right)^{2}\right) d x $$ b. How is this formula changed if the line \(y=y_{0}\) lies above \(R ?\)

Emptying a water trough A cattle trough has a trapezoidal cross section with a height of \(1 \mathrm{m}\) and horizontal sides of length \(\frac{1}{2} \mathrm{m}\) and \(1 \mathrm{m} .\) Assume the length of the trough is \(10 \mathrm{m}\) (see figure). a. How much work is required to pump the water out of the trough (to the level of the top of the trough) when it is full? b. If the length is doubled, is the required work doubled? Explain.

Shell method Use the shell method to find the volume of the following solids. A hole of radius \(r \leq R\) is drilled symmetrically along the axis of a bullet. The bullet is formed by revolving the parabola \(y=6\left(1-\frac{x^{2}}{R^{2}}\right)\) about the \(y\) -axis, where \(0 \leq x \leq R\).

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The distance traveled by an object moving along a line is the same as the displacement of the object. b. When the velocity is positive on an interval, the displacement and the distance traveled on that interval are equal. c. Consider a tank that is filled and drained at a flow rate of $$V^{\prime}(t)=1-\frac{t^{2}}{100}(\mathrm{gal} / \mathrm{min}),$$ for $$t \geq 0,\( where \)t\( is measured in minutes. It follows that the volume of water in the tank increases for 10 min and then decreases until the tank is empty. d. A particular marginal cost function has the property that it is positive and decreasing. The cost of increasing production from \)A\( units to \)2 A\( units is greater than the cost of increasing production from \)2 A\( units to \)3 A$ units.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.