Prove that the derivative of the position vector equals the velocity vector

[winnie_d_poop]

Homework Statement

Given a constant direction, take the time derivative of both sides of the position vector and show that they are equal

Relevant Equations

$\dot{\vec{r}}=\dot{x}\hat{i}+\dot{y}\hat{j}+\dot{z}\hat{k}$

$\vec{r}=r\hat{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$r=\sqrt{x^2+y^2+z^2}$

$\hat{r}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}$

$\dot{r}=\frac{\dot{x}x+\dot{y}y+\dot{z}z}{\sqrt{x^2+y^2+z^2}}$

1.)$\dot{\vec{r}}=\dot{x}\hat{i}+\dot{y}\hat{j}+\dot{z}\hat{k}=\dot{r}\hat{r}$ since the unit vector is constant

2.) $\dot{r}\hat{r}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}\frac{\dot{x}x+\dot{y}y+\dot{z}z}{\sqrt{x^2+y^2+z^2}}$

3.)$\dot{x}\hat{i}+\dot{y}\hat{j}+\dot{z}\hat{k}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}\frac{\dot{x}x+\dot{y}y+\dot{z}z}{\sqrt{x^2+y^2+z^2}}$

This is where I'm stuck. How do I get both sides to be equal?

EDIT: I made it through the problem but did I do it right?

 

[haruspex]

1.)$\dot{\vec{r}}=\dot{x}\hat{i}+\dot{y}\hat{j}+\dot{z}\hat{k}=\dot{r}\hat{r}$ since the unit vector is constant

Constant in magnitude but not direction. The coordinate unit vectors, $\hat{i}$ etc., are constant.

 

[PeroK]

Homework Statement:: Given a constant direction, take the time derivative of both sides of the position vector and show that they are equal

If two functions (of time) are equal, then their time derivatives must be equal.
$$\vec a(t) = \vec b(t) \ \Rightarrow \ \vec a'(t) = \vec b'(t)$$
If you start with an equation and differentiate it, you still have an equation. That's generally and trivially true.

Am I misunderstanding the question?

 

[Delta2]

If two functions (of time) are equal, then their time derivatives must be equal.
$$\vec a(t) = \vec b(t) \ \Rightarrow \ \vec a'(t) = \vec b'(t)$$
If you start with an equation and differentiate it, you still have an equation. That's generally and trivially true.

Am I misunderstanding the question?

I think the problem is poorly stated in the OP. I think the problem wants us to verify that taking the time derivative of the position vector in the cartesian coordinate system equals the time derivative of the position vector in the polar coordinate system. Of course we know that they are equal as you note and of course this holds in any case and not only if the direction is constant.

 

[PeroK]

I think the problem is poorly stated in the OP. I think the problem wants us to verify that taking the time derivative of the position vector in the cartesian coordinate system equals the time derivative of the position vector in the polar coordinate system. Of course we know that they are equal as you note and of course this holds in any case and not only if the direction is constant.

Is the question to find an expression for velocity in polar coordinates? That would make sense.

And, then, an expression for acceleration in polar coordinates (which is, of course, quite complicated).

Although, in that case we would normally have only two dimensions.

 

[Delta2]

Homework Statement:: Given a constant direction, take the time derivative of both sides of the position vector and show that they are equal
Relevant Equations:: $\dot{\vec{r}}=\dot{x}\hat{i}+\dot{y}\hat{j}+\dot{z}\hat{k}$

$\vec{r}=r\hat{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$r=\sqrt{x^2+y^2+z^2}$

$\hat{r}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}$

$\dot{r}=\frac{\dot{x}x+\dot{y}y+\dot{z}z}{\sqrt{x^2+y^2+z^2}}$

This is where I'm stuck. How do I get both sides to be equal?

You need to use that when the direction is constant then $\hat v=\hat r$.

Also that $$x\dot x+y\dot y+z\dot z=\vec{r}\cdot\vec{v}=|\vec{r}||\vec{v}|$$. The first equality is from the expression of dot product in cartesian coordinates. The second equality is from the expression of dot product as $|\vec{r}||\vec{v}|\cos\theta$ where the $\theta$ angle is zero, again because $\hat v=\hat r$.

 

[Delta2]

Constant in magnitude but not direction. The coordinate unit vectors, $\hat{i}$ etc., are constant.

It is given that the direction (of motion) is constant. I think the purpose of this problem is to show that the condition of constant direction implies many different things, check post #6.

 

[PeroK]

You need to use that when the direction is constant then $\hat v=\hat r$.

That's true for radial motion. In which case, we have constant $\theta$ and $\phi$. And only $r(t)$ is a function of time. Then, we can express $x(t), y(t), z(t)$ in terms of $r(t), \theta, \phi$.

In any case, you need a relationship between $x, y, z$ here.

 

[Delta2]

That's true for radial motion. In which case, we have constant $\theta$ and $\phi$. And only $r(t)$ is a function of time. Then, we can express $x(t), y(t), z(t)$ in terms of $r(t), \theta, \phi$.

In any case, you need a relationship between $x, y, z$ here.

Well, if the direction is constant, radial motion (constant $\theta$ and $\phi$) is the only choice. Or are there other choices? I suppose it can be done by using the expressions of x,y,z in spherical coordinates and using the fact that the derivatives of $\theta$ and $\phi$ are zero.
I don't know if it can be done using the relationship that $x(t),y(t),z(t)$are in the same line for all t.

 

[PeroK]

Well, if the direction is constant, radial motion (constant $\theta$ and $\phi$) is the only choice. Or are there other choices?

I initially assumed we were talking about the direction of motion. Which clearly doesn't have to be radial. Probably we are talking about constant $\hat r$, as you've assumed.

 

[winnie_d_poop]

Sorry for the confusion everybody, I was just trying to show that one side of 3.) is equal to the other side.

3.)$\dot{x}\hat{i}+\dot{y}\hat{j}+\dot{z}\hat{k}=\frac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{x^2+y^2+z^2}}\frac{\dot{x}x+\dot{y}y+\dot{z}z}{\sqrt{x^2+y^2+z^2}}$

My solution (attached to the original problem) used the definition of the dot product and the fact that the angle between r and v is 0 (both pointed out by delta!) I just want to make sure I didn't do anything wrong and get the right answer by mistake.

Thanks!

 

[PeroK]

Sorry for the confusion everybody, I was just trying to show that one side of 3.) is equal to the other side.
My solution (attached to the original problem) used the definition of the dot product and the fact that the angle between r and v is 0 (both pointed out by delta!) I just want to make sure I didn't do anything wrong and get the right answer by mistake.

Thanks!

What are you actually trying to do here?