The Vector R(t) = 3cos T3i + 3sin T +t²k Gives The Position Of A Moving Body At Any Time T. Find The Speed And Direction Of The Body At T=2
Introduction
In mathematics, the study of vector motion is crucial in understanding the position, speed, and direction of moving objects. Given a vector function r(t) that represents the position of a moving body at any time t, we can use this information to determine the speed and direction of the body at a specific time. In this article, we will explore how to find the speed and direction of a moving body using the vector function r(t) = 3cos t3i + 3sin t +t²k.
Understanding the Vector Function
The vector function r(t) = 3cos t3i + 3sin t +t²k represents the position of a moving body at any time t. Here, i and k are unit vectors in the x and z directions, respectively. The function 3cos t represents the x-coordinate of the body, while 3sin t represents the y-coordinate. The function t² represents the z-coordinate of the body.
Finding the Speed of the Body
To find the speed of the body at a specific time t, we need to find the magnitude of the velocity vector. The velocity vector is the derivative of the position vector with respect to time. In this case, we have:
r'(t) = -9sin t3i + 3cos t + 2tk
The speed of the body is the magnitude of the velocity vector, which is given by:
| r'(t) | = √((-9sin t)² + (3cos t)² + 2²) = √(81sin² t + 9cos² t + 4) = √(81sin² t + 9(1 - sin² t) + 4) = √(81sin² t + 9 - 9sin² t + 4) = √(13 - 18sin² t)
To find the speed of the body at t = 2, we substitute t = 2 into the expression for the speed:
| r'(2) | = √(13 - 18sin² 2) = √(13 - 18(0.9093)) = √(13 - 16.368) = √(-3.368) = 1.835 (approximately)
Finding the Direction of the Body
To find the direction of the body at a specific time t, we need to find the unit vector in the direction of the velocity vector. The unit vector is given by:
u(t) = r'(t) / | r'(t) |
Substituting the expression for the velocity vector, we get:
u(t) = (-9sin t3i + 3cos t + 2tk) / √(13 - 18sin² t)
To find the direction of the body at t = 2, we substitute t = 2 into the expression for the unit vector:
u(2) = (-9sin 2)3i + 3cos 2 + 2k) / √(13 - 18sin² 2) = (-9(0.9093))3i + 3(0.3420) + 2k) / √(13 - 18(0.9093)) = (-8.1823)3i + 1.0260 + 2k) / √(-3.368) = (-8.1823)3i + 1.0260 + 2k) / -1.835 = (4.5911)3i - 0.5611 - 1.094k)
Conclusion
In this article, we have explored how to find the speed and direction of a moving body using the vector function r(t) = 3cos t3i + 3sin t +t²k. We have found the speed of the body at t = 2 to be approximately 1.835 units per second, and the direction of the body at t = 2 to be approximately (4.5911)3i - 0.5611 - 1.094k).
References
- [1] "Vector Calculus" by Michael Corral
- [2] "Calculus: Early Transcendentals" by James Stewart
Mathematical Formulas
- r(t) = 3cos t3i + 3sin t +t²k
- r'(t) = -9sin t3i + 3cos t + 2tk
- | r'(t) | = √((-9sin t)² + (3cos t)² + 2²)
- u(t) = r'(t) / | r'(t) |
The Vector Motion of a Moving Body: Q&A =============================================
Introduction
In our previous article, we explored how to find the speed and direction of a moving body using the vector function r(t) = 3cos t3i + 3sin t +t²k. In this article, we will answer some frequently asked questions related to the vector motion of a moving body.
Q: What is the significance of the vector function r(t) = 3cos t3i + 3sin t +t²k?
A: The vector function r(t) = 3cos t3i + 3sin t +t²k represents the position of a moving body at any time t. The function 3cos t represents the x-coordinate of the body, while 3sin t represents the y-coordinate. The function t² represents the z-coordinate of the body.
Q: How do I find the speed of the body at a specific time t?
A: To find the speed of the body at a specific time t, you need to find the magnitude of the velocity vector. The velocity vector is the derivative of the position vector with respect to time. In this case, we have:
r'(t) = -9sin t3i + 3cos t + 2tk
The speed of the body is the magnitude of the velocity vector, which is given by:
| r'(t) | = √((-9sin t)² + (3cos t)² + 2²) = √(81sin² t + 9cos² t + 4) = √(81sin² t + 9(1 - sin² t) + 4) = √(81sin² t + 9 - 9sin² t + 4) = √(13 - 18sin² t)
Q: How do I find the direction of the body at a specific time t?
A: To find the direction of the body at a specific time t, you need to find the unit vector in the direction of the velocity vector. The unit vector is given by:
u(t) = r'(t) / | r'(t) |
Substituting the expression for the velocity vector, we get:
u(t) = (-9sin t3i + 3cos t + 2tk) / √(13 - 18sin² t)
Q: What is the difference between the speed and velocity of the body?
A: The speed of the body is the magnitude of the velocity vector, while the velocity is the direction of the body. In other words, the speed tells us how fast the body is moving, while the velocity tells us in which direction the body is moving.
Q: Can I use the vector function r(t) = 3cos t3i + 3sin t +t²k to find the acceleration of the body?
A: Yes, you can use the vector function r(t) = 3cos t3i + 3sin t +t²k to find the acceleration of the body. The acceleration is the derivative of the velocity vector with respect to time. In this case, we have:
r''(t) = -9cos t3i - 9sin t + 2k
Q: How do I use the vector function r(t) = 3cos t3i + 3sin t +t²k to find the position of the body at a specific time t?
A: To find the position of the body at a specific time t, you can substitute the value of t into the vector function r(t) = 3cos t3i + 3sin t +t²k. For example, to find the position of the body at t = 2, you can substitute t = 2 into the vector function:
r(2) = 3cos 2)3i + 3sin 2) + 2²k = 3(0.4161)3i + 3(0.9093) + 4k = 1.2483)3i + 2.7279 + 4k
Conclusion
In this article, we have answered some frequently asked questions related to the vector motion of a moving body. We have discussed how to find the speed and direction of the body using the vector function r(t) = 3cos t3i + 3sin t +t²k. We have also discussed how to find the acceleration of the body and how to use the vector function to find the position of the body at a specific time t.
References
- [1] "Vector Calculus" by Michael Corral
- [2] "Calculus: Early Transcendentals" by James Stewart
Mathematical Formulas
- r(t) = 3cos t3i + 3sin t +t²k
- r'(t) = -9sin t3i + 3cos t + 2tk
- | r'(t) | = √((-9sin t)² + (3cos t)² + 2²)
- u(t) = r'(t) / | r'(t) |
- r''(t) = -9cos t3i - 9sin t + 2k