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Chapter 24: Problem 6

Given that the \(x\) - and \(y\) -coordinates of a moving par. ticle are given by the indicated parametric equations, find the magnitude and direction of the velocity for the specific value of \(t .\) Sketch the curves and show the velocity and its components. $$x=\sqrt{1+2 t}, y=t-t^{2}, t=4$$

Short Answer

Expert verified
Magnitude of velocity is approximately 7.0179. Direction is \(272.73^\circ\) from the positive x-axis.

Step by step solution

01

Find the Derivatives

To determine the velocity vector, find the derivatives of the parametric equations with respect to time \(t\). For \(x\): \(\frac{dx}{dt} = \frac{1}{2\sqrt{1+2t}} \cdot 2 = \frac{1}{\sqrt{1+2t}}\).For \(y\): \(\frac{dy}{dt} = 1 - 2t\).
02

Evaluate the Derivatives at \(t=4\)

Substitute \(t=4\) into the derivatives to find the components of the velocity vector.\(\frac{dx}{dt}\Big|_{t=4} = \frac{1}{\sqrt{1+2 \times 4}} = \frac{1}{\sqrt{9}} = \frac{1}{3}\).\(\frac{dy}{dt}\Big|_{t=4} = 1 - 2 \times 4 = 1 - 8 = -7\).
03

Calculate the Magnitude of the Velocity

Use the Pythagorean theorem to find the magnitude of the velocity vector:\[ v = \sqrt{\left(\frac{1}{3}\right)^2 + (-7)^2} = \sqrt{\frac{1}{9} + 49} = \sqrt{\frac{1}{9} + \frac{441}{9}} = \sqrt{\frac{442}{9}} = \sqrt{49.1111}\approx 7.0179 \]
04

Determine the Direction of the Velocity

The direction is calculated using the tangent of the angle \(\theta\) which the vector makes with the x-axis.\( \tan \theta = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-7}{\frac{1}{3}} = -21 \).Thus, \( \theta = \tan^{-1}(-21) \). Calculate \(\theta = -87.27^\circ \) or \(272.73^\circ\) based on the quadrant.
05

Sketch the Curves and Velocity

On the graph, plot the parametric curve obtained from \(x = \sqrt{1 + 2t}\) and \(y = t - t^2\). At point \(t=4\), mark the position, plot the velocity vector and its components. The components are: horizontal \(\frac{1}{3}\), vertical \(-7\). Highlight the angle \(\theta\) with the x-axis.

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Key Concepts

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

Velocity Vector
When studying motion in physics and calculus, the velocity vector is crucial. It's derived from parametric equations that describe a particle's position over time. In such a scenario:
  • The velocity vector gives the speed and direction of the particle's movement.
  • It is composed of two main components: the rate of change of the x-coordinate (horizontal movement) and the y-coordinate (vertical movement).
By finding the derivatives of the parametric equations, we can express these components. For the given equations, the velocity vector at a specific time \(t\) is represented as \((\frac{dx}{dt}, \frac{dy}{dt})\). This will help in visualizing the particle's movement along its path.
Magnitude of Velocity
Determining the magnitude of the velocity is essential to understanding how fast a particle is moving at a specific point in time. This can be calculated using the Pythagorean theorem on the velocity vector:
  • The formula used is: \(v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\).
  • For the values found at \(t=4\), substituting the derivatives, the exact expression becomes \( \sqrt{\left(\frac{1}{3}\right)^2 + (-7)^2} \).
  • Simplifying gives \(\sqrt{49.1111} \approx 7.0179\).
This magnitude reflects the actual speed at which the particle travels without considering the direction.
Direction of Velocity
Understanding the direction of velocity is fundamental to predicting a particle's path. The direction is determined by the angle \(\theta\) that the velocity vector forms with the positive x-axis.
  • The tangent of this angle is given by \(\tan \theta = \frac{\text{vertical component}}{\text{horizontal component}} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
  • Substituting our specific values found at \(t=4\), we have \(\tan \theta = \frac{-7}{\frac{1}{3}} = -21\).
  • This yields \(\theta = \tan^{-1}(-21)\), approximately \(-87.27^\circ\) or, in standard position terms, \(272.73^\circ\).
The calculated direction helps in plotting the velocity vector on a graph accurately.
Derivatives
Derivatives play a key role in identifying how a function changes over time. They express rates of change, which are paramount in forming velocity vectors.
  • For parametric equations, the derivative of \(x\) with respect to time \(t\), denoted as \(\frac{dx}{dt}\), denotes horizontal speed.
  • The derivative of \(y\), \(\frac{dy}{dt}\), represents vertical speed.
At \(t=4\), these derivatives become analytical tools to predict how the particle's position changes. Derivatives enable the calculation of both the velocity vector components and the overall speed (magnitude of velocity) in calculus.
Calculus
Calculus allows us to analyze and predict a particle's behavior over time. It encompasses several key concepts:
  • With calculus, we can tackle the dynamic nature of motion through the concepts of differentiation and integration.
  • Differentiation, the primary focus here, leads to the calculation of rates of change and is instrumental in forming velocity expressions.
  • This systematic approach offers a step-by-step breakdown of complex problems into manageable tasks, such as finding derivatives and evaluating them.
In our exercise, calculus facilitated the transition from understanding a position to figuring out speed and direction, demonstrating its utility across disciplines.

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