Time Derivatives of Unit Vectors

In summary, the Hyperbolic coordinate system is given by u=2xy and v=x^2 - y^2. To find the unit vectors u and v in terms of u,v,x(hat),y(hat), we use the formula u(hat) = (du/dx)i + (du/dy)j and v(hat) = (dv/dx)i + (dv/dy)j, and substitute in for x and y to get equations in terms of v and u. The correct magnitude for the unit vectors is 1, and the final equations are u(hat) = (2yi + 2xj)/(sqrt(4x^2 + 4y^2
  • #1
Jshroomer
2
0

Homework Statement



The Hyperbolic coordinate system is given by: u=2xy and v=x^2 - y^2

a.) Find the unit vectors u and v in terms of u,v,x(hat),y(hat)
b.)Find the time derivatives of u(hat) and v(hat), your answers will have du/dt and dv/dt in them

Homework Equations



None really


The Attempt at a Solution



I found the unit vectors by

u(hat) = (du/dx)i + (du/dy)j
v(hat) = (dv/dx)i + (dv/dy)j

then I substituted in for x and y to give equations in terms of v and u

u(hat) = (sqrt(2))sqrt(sqrt(u^2 + v^2) - v))x(hat) + (sqrt(2))sqrt(sqrt(u^2 + v^2) +v))y(hat)

v(hat) = (sqrt(2))sqrt(sqrt(u^2 + v^2) + v))x(hat) + (-sqrt(2))sqrt(sqrt(u^2 + v^2) - v))y(hat)

I just don't understand how to take the time derivatives of these unit vectors, any help would be greatly appreciated.

Thanks
 
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  • #2
Jshroomer said:

Homework Statement



The Hyperbolic coordinate system is given by: u=2xy and v=x^2 - y^2

a.) Find the unit vectors u and v in terms of u,v,x(hat),y(hat)
b.)Find the time derivatives of u(hat) and v(hat), your answers will have du/dt and dv/dt in them

Homework Equations



None really

The Attempt at a Solution



I found the unit vectors by

u(hat) = (du/dx)i + (du/dy)j
v(hat) = (dv/dx)i + (dv/dy)j

then I substituted in for x and y to give equations in terms of v and u

u(hat) = (sqrt(2))sqrt(sqrt(u^2 + v^2) - v))x(hat) + (sqrt(2))sqrt(sqrt(u^2 + v^2) +v))y(hat)

v(hat) = (sqrt(2))sqrt(sqrt(u^2 + v^2) + v))x(hat) + (-sqrt(2))sqrt(sqrt(u^2 + v^2) - v))y(hat)

I just don't understand how to take the time derivatives of these unit vectors, any help would be greatly appreciated.

Thanks
Your unit vectors do not appear to have the correct magnitude, which should be 1.

For [itex]\hat{u}[/itex] I get: [itex]\displaystyle\hat{u}=\frac{\displaystyle
\frac{\partial u}{\partial x}\hat{i}+\frac{\partial u}{\partial y}\hat{j}}{\displaystyle\sqrt{\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2}}=\frac{y\hat{i}+x\hat{j}}{\sqrt{x^2+y^2}}[/itex]
 
  • #3
Thank you for the correction sammy, I'm not very good at vector calculus, and my book is hard to get useful information from.

so, I have

u(hat) = (2yi + 2xj)/(sqrt(4x^2 + 4y^2))

and

v(hat) = (2xi -2yj)/(sqrt(4x^2 + 4y^2))

I still do not understand how to take the time derivatives though, thanks
 
Last edited:

1. What are unit vectors?

Unit vectors are vectors that have a magnitude of 1 and are used to define a specific direction in space. They are important in physics and mathematics as they provide a basis for describing and analyzing other vectors.

2. What is the significance of time derivatives of unit vectors?

The time derivatives of unit vectors are used to describe the change in direction of a vector over time. They are crucial in understanding how a vector is changing and can help in predicting its future direction and magnitude.

3. How are time derivatives of unit vectors calculated?

The time derivatives of unit vectors are calculated by taking the derivative of each component of the unit vector with respect to time. This can be done using calculus and is often used in physics and engineering to analyze the motion of objects.

4. What is the relationship between time derivatives of unit vectors and velocity?

The time derivatives of unit vectors are directly related to velocity. In fact, the time derivative of a position vector is equal to the velocity vector. This means that the time derivatives of unit vectors can be used to calculate the velocity of an object in a given direction at a specific time.

5. Can time derivatives of unit vectors be negative?

Yes, time derivatives of unit vectors can be negative. This indicates that the vector is changing in the opposite direction of its current direction. For example, a negative time derivative of a unit vector in the x-direction means that the vector is decreasing in the x-direction over time.

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