A Particle Moves Along The { Y $}$-axis With Position At Time { T $}$ Given By { Y(t) = E^{-t} \sin T $}$ For { 0 \leq T \leq 2\pi $}$.(a) Find The Time { T $}$ At Which The Particle Is Farthest
Introduction
In this problem, we are given the position of a particle at time t as y(t) = e^(-t) sin t for 0 ≤ t ≤ 2π. Our goal is to find the time t at which the particle is farthest from the origin. This problem involves analyzing the given function and using mathematical techniques to determine the maximum value of the function within the specified interval.
Understanding the Function
The given function y(t) = e^(-t) sin t is a product of two functions: e^(-t) and sin t. The function e^(-t) is an exponential function that decreases as t increases, while the function sin t is a trigonometric function that oscillates between -1 and 1. The product of these two functions will result in a function that oscillates between -e^(-t) and e^(-t).
Finding the Maximum Value
To find the maximum value of the function y(t) = e^(-t) sin t, we need to analyze the behavior of the function within the specified interval 0 ≤ t ≤ 2π. We can start by finding the critical points of the function, which are the points where the derivative of the function is equal to zero.
Derivative of the Function
To find the derivative of the function y(t) = e^(-t) sin t, we can use the product rule of differentiation. The derivative of the function is given by:
y'(t) = e^(-t) cos t - e^(-t) sin t
Critical Points
To find the critical points of the function, we need to set the derivative equal to zero and solve for t:
e^(-t) cos t - e^(-t) sin t = 0
We can factor out the common term e^(-t) from the equation:
e^(-t) (cos t - sin t) = 0
Since e^(-t) is never equal to zero, we can set the term in parentheses equal to zero:
cos t - sin t = 0
We can rewrite the equation as:
cos t = sin t
Using the identity cos^2 t + sin^2 t = 1, we can rewrite the equation as:
cos^2 t = sin^2 t
Taking the square root of both sides, we get:
cos t = ± sin t
We can solve for t by using the inverse tangent function:
t = arctan (± sin t / cos t)
Since cos t and sin t are both positive in the first quadrant, we can take the positive square root:
t = arctan (sin t / cos t)
Solving for t
To solve for t, we need to find the values of t that satisfy the equation t = arctan (sin t / cos t). We can use a calculator or a computer program to find the values of t.
Numerical Solution
Using a numerical method, we can find the values of t that satisfy the equation t = arctan (sin t / cos t). The values of t are approximately:
t ≈ 0.7854, 2.3562, 4.7124, 6.2832
Analyzing the Results
The values of t that satisfy the equation t = arctan (sin t / cos t) are approximately 0.7854, 2.3562, 4.7124, and 6.2832. These values correspond to the points where the function y(t) = e^(-t) sin t has a maximum value.
Conclusion
In this problem, we found the time t at which the particle is farthest from the origin by analyzing the given function y(t) = e^(-t) sin t. We used mathematical techniques to determine the maximum value of the function within the specified interval 0 ≤ t ≤ 2π. The values of t that satisfy the equation t = arctan (sin t / cos t) are approximately 0.7854, 2.3562, 4.7124, and 6.2832.
Further Discussion
This problem can be extended to find the time t at which the particle is farthest from the origin for a more general function y(t) = f(t) sin t, where f(t) is a continuous function. The solution would involve finding the critical points of the function and analyzing the behavior of the function within the specified interval.
References
- [1] Calculus, 3rd edition, by Michael Spivak
- [2] Differential Equations and Dynamical Systems, 3rd edition, by Lawrence Perko
Appendix
The following is a list of the mathematical symbols used in this problem:
- y(t) = e^(-t) sin t
- y'(t) = e^(-t) cos t - e^(-t) sin t
- t = arctan (± sin t / cos t)
Q: What is the position of the particle at time t?
A: The position of the particle at time t is given by the function y(t) = e^(-t) sin t.
Q: What is the interval of time for which the position of the particle is given?
A: The position of the particle is given for the interval 0 ≤ t ≤ 2π.
Q: How do we find the time t at which the particle is farthest from the origin?
A: To find the time t at which the particle is farthest from the origin, we need to analyze the behavior of the function y(t) = e^(-t) sin t within the specified interval 0 ≤ t ≤ 2π. We can start by finding the critical points of the function, which are the points where the derivative of the function is equal to zero.
Q: What is the derivative of the function y(t) = e^(-t) sin t?
A: The derivative of the function y(t) = e^(-t) sin t is given by y'(t) = e^(-t) cos t - e^(-t) sin t.
Q: How do we find the critical points of the function?
A: To find the critical points of the function, we need to set the derivative equal to zero and solve for t. We can factor out the common term e^(-t) from the equation and set the term in parentheses equal to zero.
Q: What is the equation that we need to solve to find the critical points?
A: The equation that we need to solve to find the critical points is cos t - sin t = 0.
Q: How do we solve the equation cos t - sin t = 0?
A: We can solve the equation cos t - sin t = 0 by using the identity cos^2 t + sin^2 t = 1. We can rewrite the equation as cos^2 t = sin^2 t and take the square root of both sides to get cos t = ± sin t.
Q: What is the next step in solving the equation cos t - sin t = 0?
A: The next step in solving the equation cos t - sin t = 0 is to use the inverse tangent function to solve for t. We can rewrite the equation as t = arctan (± sin t / cos t).
Q: How do we find the values of t that satisfy the equation t = arctan (± sin t / cos t)?
A: We can use a calculator or a computer program to find the values of t that satisfy the equation t = arctan (± sin t / cos t).
Q: What are the values of t that satisfy the equation t = arctan (± sin t / cos t)?
A: The values of t that satisfy the equation t = arctan (± sin t / cos t) are approximately 0.7854, 2.3562, 4.7124, and 6.2832.
Q: What do these values of t represent?
A: These values of t represent the times at which the particle is farthest from the origin.
Q: How do we determine the maximum value of the function y(t) = e^(-t) sin t?
A: We can determine the maximum value of the function y(t) = e^(-t) sin t by analyzing the behavior of the function within the specified interval 0 ≤ t ≤ 2π. We can use the critical points that we found earlier to determine the maximum value of the function.
Q: What is the maximum value of the function y(t) = e^(-t) sin t?
A: The maximum value of the function y(t) = e^(-t) sin t is approximately 0.7071.
Q: What is the significance of the maximum value of the function y(t) = e^(-t) sin t?
A: The maximum value of the function y(t) = e^(-t) sin t represents the farthest position of the particle from the origin.
Q: How does this problem relate to real-world applications?
A: This problem relates to real-world applications in physics and engineering, where the motion of particles and objects is often described using mathematical functions.
Q: What are some potential extensions of this problem?
A: Some potential extensions of this problem include finding the time t at which the particle is farthest from the origin for a more general function y(t) = f(t) sin t, where f(t) is a continuous function.
Q: What are some potential applications of this problem?
A: Some potential applications of this problem include modeling the motion of particles and objects in physics and engineering, and analyzing the behavior of complex systems in fields such as biology and economics.