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Are $\frac{d(f(r(t),\theta(t)))}{dt}$ and $\frac{\bar{v}}{|\bar{v}|}\bar{\nabla}f$ same in spherical polar coordinates?

Where $\bar{v}$ defines the vector given by the derivative w.r.t t of the parametric curve $(r(t),\theta(t))$.

In words, I want to ask if the directional derivative is same as the derivative $\frac{d(f(r(t),\theta(t)))}{dt}$ in plane polar coordinates?

I am confused as I am not able to find them to be equal.

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Note that from the chain rule we have

$$\frac{d f}{d t}=\frac{\partial f}{\partial r}\frac{dr}{dt}+\frac{\partial f}{\partial \theta}\frac{d\theta}{dt}+\frac{\partial f}{\partial \phi}\frac{d\phi}{dt}\tag1$$

With $\vec r=\hat r(t)r(t)$ and $\vec v=\frac{d\vec r}{dt}$ we can write

$$\vec v=\hat r\frac{d r}{dt}+\hat \theta r \frac{d\theta}{dt}+\hat \phi r\sin(\theta)\frac{d\phi}{dt}\tag2$$

In spherical coordinates $\nabla f(r,\theta)$ is

$$\nabla f=\hat r\frac{\partial f}{\partial r}+\hat \theta \frac1r \frac{\partial f}{\partial \theta}+\hat \phi\frac1{r\sin(\theta)}\frac{\partial f}{\partial \phi}\tag3$$

Using $(2)$ and $(3)$, we see that

$$\vec v\cdot \nabla f=\frac{\partial f}{\partial r}\frac{dr}{dt}+\frac{\partial f}{\partial \theta}\frac{d \theta}{d t}+\frac{\partial f}{\partial \phi}\frac{d\phi}{dt}\tag4$$

Comparing $(1)$ and $(4)$ reveals

$$\vec v\cdot \nabla f=\frac{d f}{d t}$$

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  • $\begingroup$ I have been confused, because, we always use {x(t),y(t)} for parametric plot, but if I have {r(t),theta(t)} then feel we can define a circle as {a,t} and the vector tangent to any point at t is (0,1) where this is in polar coordinates. Am I correct? $\endgroup$ – Chetan Waghela Mar 9 at 4:40
  • $\begingroup$ I am unsure as to the meaning of your question. The position vector $\vec r(t)$ as a function of $t$ traces out a path. That path is not a circle in general. For a circle of radius $a$, the position vector is $\vec r(t)=a \hat r(t)$ with $\vec v(t) =a\hat \phi(t)$ with $\hat r\cdot\hat \phi=0$. $\endgroup$ – Mark Viola Mar 9 at 17:19
  • $\begingroup$ I am trying to involve parametric curve into the picture here. The tangent vector to this parametric curve is the vector for directional derivative. Anyway, I seem to have solved my questions. Thanks. $\endgroup$ – Chetan Waghela Mar 9 at 17:48
  • $\begingroup$ You're welcome. My pleasure and pleased this was useful. $\endgroup$ – Mark Viola Mar 9 at 17:55
  • $\begingroup$ I think I made a mistake in questioning what I wanted. However, your answer seems perfectly right. It also solves my dilemma that $v^i$ is equal to $\frac{d\gamma^i(t)}{dt}$ "only" in cartesian coordinates, where $\gamma^i$ represents the curve in cartesian coordinates. For example for a closed circle with radius 'a', $\gamma^i$ is ($\gamma^1$=acos(t),$\gamma^2$=asin(t)). However, in plane polar coordinates $\gamma^i$ is ($\gamma^1$=a,$\gamma^2$=t) here $\frac{d\gamma(t)}{dt}$ is (0,1) however it should have been (0,a) as shown by second term of eq2. $\endgroup$ – Chetan Waghela Mar 14 at 6:28

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