# Determining Motion from a Derivative

• friendbobbiny
In summary, the given position function r(t) and its derivative r'(t) = c X r(t), where c is a constant vector, indicate that the particle's path is perpendicular to both c and r(t). This means that r'(t) lies on a plane that c is normal to. By using Newton's method, we can approximate the particle's next position by multiplying the cross product of c and r(t) by the time interval t. It may be helpful to choose a coordinate system where c is aligned with the z-axis for easier analysis. Additionally, by dotting both sides with r(t) and solving, we can determine the magnitude of r(t).

## Homework Statement

Given position function r(t) and r'(t) = c X r(t), where c is some constant vector, describe the path of the particle. In other words, describe r(t).

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## The Attempt at a Solution

a) r'(t) points in the direction of motion. If we can understand how r'(t) changes direction, we can understand how r(t) moves.
b) r'(t) is the cross product of c and r(t). Therefore, the function is perpendicular to both c and r(t).
c) If r'(t) is perpendicular to c, r'(t) lies on a plane that c is normal to.

Consider iterating a few times with Newton's method to see what happens.
For starters, assume that C = Identity and evaluate the cross product.

what does C = identity mean, sorry? a constant function?

if c is given by <C1,C2,C3> and I cross that by r(t) <R1,R2,R3>, I get a cross product.:
<C2(R3) - C3(R2), C1(R3) - C3(R1), C1(R2) - C2(R1)>

I then multiply that cross product by time interval t. This gives the position function's next approximate position.

\$<t[C2(R3) - C3(R2)], t[C1(R3) - C3(R1)], t[C1(R2) - C2(R1)]t>

Subsequent iterations become very messy. Are you sure about this? No conceptual way of understanding the problem?

Dot both sides with r(t) and solve. What does this tell you about |r(t)|?

You might find it helpful to choose a coordinate system to make the analysis easier. Orient the coordinate system so that ##\vec{c}## points along the z-axis. That'll give you a much simpler expression for ##\vec{c}\times\vec{r}##.

## 1. What is a derivative and how is it used to determine motion?

A derivative is a mathematical concept that represents the rate of change of a function. In the context of motion, the derivative is used to calculate the velocity and acceleration of an object at a specific point in time. This is done by taking the derivative of the position function with respect to time.

## 2. What is the difference between velocity and acceleration?

Velocity is a measure of an object's displacement over time, while acceleration is a measure of an object's change in velocity over time. In other words, velocity tells us how fast an object is moving and in what direction, while acceleration tells us how quickly the object's velocity is changing.

## 3. How do you use a derivative to determine the direction of motion?

The direction of motion can be determined by looking at the sign of the derivative. If the derivative is positive, the object is moving in the positive direction, and if the derivative is negative, the object is moving in the negative direction. If the derivative is zero, the object is at rest.

## 4. Can a derivative be used to calculate the position of an object?

Yes, a derivative can be integrated to find the original function. In the context of motion, this means that the position of an object can be calculated by integrating the velocity function with respect to time. However, this only works if the initial position of the object is known.

## 5. How is the concept of derivatives applied in real-life situations?

The concept of derivatives is widely used in various fields such as physics, engineering, economics, and biology. In physics, derivatives are used to analyze the motion of objects and in engineering, they are used to design and optimize systems. In economics, derivatives are used to model changes in variables such as stock prices and interest rates. In biology, derivatives are used to study the growth and development of organisms.