Proving Vector A and d\vec{}A/dt Are Perpendicular With Constant Magnitude

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In summary, The problem asks for the proof that if a vector \vec{}A has constant magnitude, then it is perpendicular to its time derivative d\vec{}A/dt, given that d\vec{}A/dt is not equal to 0. This can be illustrated by imagining a circular motion with a constant radius, where the two vectors with magnitude A are displaced by an amount delta theta. Taking the limit of the vector difference delta A divided by delta theta as delta theta approaches 0 proves the perpendicularity. An example of this problem can be seen in a circular motion with constant radius.
  • #1
jawwadaziz
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please help me(a vector problem)

Homework Statement



i m very confuse about this assignment. please help me how to prove.

If [tex]\vec{}A[/tex] has constant magnitude show that [tex]\vec{}A[/tex] and d[tex]\vec{}A[/tex]/dt are perpendicular provided that d[tex]\vec{}A[/tex]/dt [tex]\neq[/tex]0

also give one physical example of this problem.

please help me in this problem.

thnx.


Homework Equations





The Attempt at a Solution

 
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  • #2


This can be thought of as a circular motion problem with constant radius. Since the radius vector has a constant magnitude (constant radius), then only the position of the vector changes with time. Draw two vectors with magnitue A, one displaced an amount delta theta from the other. Find the vector difference which will be delta A, then take the limit of delta A divided by delta theta as delta theta goes to zero.
 
  • #3


i didnt got u !

Hey Chris buddy can u pleasez solve it for me pleasez..

thnx in advance.
 
  • #4


jawwadaziz said:
i didnt got u !

Hey Chris buddy can u pleasez solve it for me pleasez..

thnx in advance.

No, we will not solve this for you. You have been given a great hint, and now the rest is up to you.

Please re-read the Rules link at the top of the page, especially the part about how you must do the bulk of the work on homework/coursework problems.

Show your work, and if you have a specific question, we can offer hints and tutorial help. We do not do your homework for you.
 
  • #5


ooops sorry !

I didnt knew the rules..
it won't happen next tym.
 
  • #6


The magnitude of A squared is A.A (dot product). Is that a help?
 

Related to Proving Vector A and d\vec{}A/dt Are Perpendicular With Constant Magnitude

1. How do you prove that Vector A and d\vec{}A/dt are perpendicular?

To prove that Vector A and d\vec{}A/dt are perpendicular, we can use the dot product method. If the dot product of two vectors is equal to zero, then the vectors are perpendicular. So, we can take the dot product of Vector A and d\vec{}A/dt and show that it equals zero.

2. What is the significance of proving that Vector A and d\vec{}A/dt are perpendicular?

The significance of proving this is that it shows that the rate of change of Vector A is always perpendicular to Vector A itself. This is an important concept in vector calculus and has many applications in physics and engineering.

3. What is the constant magnitude in this proof?

The constant magnitude in this proof refers to the fact that the magnitude of Vector A remains constant, while its direction changes. This is important in proving that Vector A and d\vec{}A/dt are always perpendicular.

4. Can this proof be applied to any vector?

Yes, this proof can be applied to any vector, as long as its magnitude remains constant and it has a rate of change vector, d\vec{}A/dt. The proof relies on the dot product method, which can be applied to any two vectors.

5. What are some real-world examples of Vector A and d\vec{}A/dt being perpendicular with constant magnitude?

One example is a car driving in a circular motion. The velocity vector of the car, which is tangent to the circular path, is perpendicular to the acceleration vector, which is pointing towards the center of the circle. Another example is a swinging pendulum, where the velocity vector at any point is perpendicular to the acceleration vector due to gravity.

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