Dot product of a vector with the derivative of its unit vector

In summary, the conversation discusses a problem involving the derivative of a path of class C1 and its dot product with the path itself. The question is how to show that, when the path's magnitude is always greater than zero, the dot product always equals zero. The solution involves using the fact that c(t) dotted with c'(t) equals zero when the magnitude of c(t) is a constant, and then plugging in c(t)/||c(t)|| for d(t).
  • #1
hoopsmax25
13
0

Homework Statement


Let c(t) be a path of class C1. Suppose that ||c(t)||>0 for all t.
Show that c(t) dot product with d/dt((c(t))/||c(t)||) =0 for every t.


Homework Equations


I am having trouble with the derivative of (c(t)/||c(t)||) and how to show that when its dotted with c(t) that it always equals zero.


The Attempt at a Solution


I know that c(t) dotted with c'(t) =0 when ||c(t)||= a constant but don't know how to implement this fact for this problem.
 
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  • #2
If d(t)=c(t)/||c(t)||, what is ||d(t)||?
 
  • #3
What is the magnitude of c(t)/||c(t)||?
 
  • #4
would it be 1?
 
  • #5
hoopsmax25 said:
would it be 1?

Show why it is 1. Then you can drop the '?'.
 
  • #6
Ok i get that now. So if d(t)=1, which is a constant, then d(t) dotted with d'(t) =0. Since c(t)/||c(t)|| has the same direction as c(t), we can then plug c(t) in for d(t). Does that make sense to do?
 
  • #7
hoopsmax25 said:
Ok i get that now. So if d(t)=1, which is a constant, then d(t) dotted with d'(t) =0. Since c(t)/||c(t)|| has the same direction as c(t), we can then plug c(t) in for d(t). Does that make sense to do?

Makes sense to me.
 
  • #8
Awesome. Thanks for the help
 

FAQ: Dot product of a vector with the derivative of its unit vector

1. What is the dot product of a vector with the derivative of its unit vector?

The dot product of a vector with the derivative of its unit vector is a scalar value that represents the rate of change of the vector in the direction of its unit vector. It is calculated by multiplying the magnitude of the vector with the cosine of the angle between the vector and its derivative of the unit vector.

2. How is the dot product of a vector with the derivative of its unit vector useful in physics?

In physics, the dot product of a vector with the derivative of its unit vector is useful in calculating the work done by a force on an object. It is also used in calculating the flux of a vector field through a surface and in determining the angle between two vectors.

3. Can the dot product of a vector with the derivative of its unit vector be negative?

Yes, the dot product of a vector with the derivative of its unit vector can be negative. This occurs when the angle between the vector and its derivative of the unit vector is greater than 90 degrees. In this case, the cosine of the angle is negative, resulting in a negative dot product.

4. How does the dot product of a vector with the derivative of its unit vector relate to the concept of orthogonality?

If the dot product of a vector with the derivative of its unit vector is zero, then the two vectors are said to be orthogonal or perpendicular to each other. This means that the angle between the vector and its derivative of the unit vector is 90 degrees and there is no component of the vector in the direction of its derivative of the unit vector.

5. Is the dot product of a vector with the derivative of its unit vector affected by the length of the vector?

No, the dot product of a vector with the derivative of its unit vector is not affected by the length of the vector. This is because the dot product is calculated by multiplying the magnitude of the vector with the cosine of the angle between the vector and its derivative of the unit vector, which does not depend on the length of the vector.

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