What is the velocity vector of (rsin(phi), rcos(phi), 1) ?

[physicss]

Homework Statement

Hello,

Given is the following location vector: (of a string-pendulum)
x= (rsin(phi), rcos(phi), 1)
My task was to determine the velocity vector of the location vector.
Quote: "Let the location of a pendulum be given by the vector x= (rsin(phi), rcos(phi), 1), where r is the length of the filament and φ is the angle to the y-axis.

Calculate the velocity vector v for the case where only the angle φ is time-dependent."

Relevant Equations

x= (rsin(phi), rcos(phi),1)

I did try to solve the problem by forming the derivative and my result was: v=(rcos(phi), -rsin(phi),0). My solution is wrong, the tutor corrected the task but he didn’t give us the results. My question is what the solution is. Thanks in advance.

 

[TSny]

my result was: v=(rcos(phi), -rsin(phi),0)

Does your result have the dimensions of velocity?

Suppose you have a function $f(\phi$), where $\phi$ is a function of time $t$.
Recall the chain rule of calculus: $$\frac{d f}{dt} = \frac {df}{d \phi} \cdot \frac{d \phi}{dt}$$.

 

[Steve4Physics]

Hi @physicss. Welcome to PF!

My task was to determine the velocity vector of the location vector.

Being picky, I think you mean ".. . to determine the velocity vector of the pendulum 'bob' (the mass at the end of the string)".

Quote: "Let the location of a pendulum be given by the vector x= (rsin(phi), rcos(phi), 1), where r is the length of the filament and φ is the angle to the y-axis.

Calculate the velocity vector v for the case where only the angle φ is time-dependent."
Relevant Equations: x= (rsin(phi), rcos(phi),1)

I did try to solve the problem by forming the derivative and my result was: v=(rcos(phi), -rsin(phi),0). My solution is wrong ...

Are you familiar with the chain rule? $\vec x$ is a function of $\phi$. And $\phi$ is a function of $t$.

Aha! Beaten to it by @TSny.

 

[TSny]

Aha! Beaten to it by @TSny.

Very close

 

[jbriggs444]

I did try to solve the problem by forming the derivative and my result was: v=(rcos(phi), -rsin(phi),0). My solution is wrong, the tutor corrected the task but he didn’t give us the results. My question is what the solution is. Thanks in advance.

If I understand the question properly, your work looks good. However, we could clean things up a bit.

First, let us clean up the format of the question. Using $x$ for the name of the position vector is poor form. It invites confusion with the $x$ component of the position vector. Let me put that in $\LaTeX$ format as well:$$\vec{S} = (r \sin \phi, r \cos \phi, 1)$$Now let us turn to the calculus part of the exercise. We want the derivative of position with respect to time:$$\vec{v} = \frac{d\vec{S}}{dt}$$If we take the derivative of $r \sin \phi$, you say that we get $r \cos \phi$. But that is not the derivative with respect to time ($t$). That is the derivative with respect to phi ($\phi$):$$\vec{v} \ne \frac{d\vec{S}}{d\phi}$$Drat -- beaten to it by both of you.