Help with a multivariable calculus question.

Question

Assume that the distribution of pressure in space is given by $p(x, y, z) = F(x^2 + y^2 + z^2)$, where $F$ is some differentiable function of one variable.

a) By computing $\nabla p$, show that a particle in space evolving according to the ordinary differential equation $$\mathbf{r}'(t) = −\nabla p(\mathbf{r}(t))$$ either remains stationary, or moves in a radial direction, i.e. directly towards or away from the origin $(x, y, z) = (0, 0, 0)$.

b) Find the trajectory of the particle in the particular case where $F(u) = u$ and at time $0$, the particle is at $\mathbf{r}(0) = \mathbf{R} = (R_1, R_2, R_3) \in \mathbb{R}^3$.

The main part of the question I don't understand is where to begin. I also don't fully understand how to take the gradient of a parametrised curve. The abstract nature of the function $F$ also doesn't help either. I just need some guidance and further understanding of what the question is asking us to derive.

Answer 1

$\vec r'(t)=(x'(t),y'(t),z'(t))=$

$-(2xF'(u),2yF'(u),2zF'(u))$;

$\vec r'(t)=-2F'(u)\vec r,$ where $u:=x^2+y^2+z^2.$

Trajectory: $F'(u)=1$;

$\vec r'(t)=-(2 \cdot 1)\vec r$;

$\dfrac{d}{dt}(\vec r \cdot \vec r)=2\vec r \cdot \vec r'(t)=-4r^2$;

$\dfrac{d}{dt}(r^2)=-4r^2$

$\dfrac{d(r^2)}{r^2}=-4dt;$

Integrating :

$\log (r^2)-\log (r_0^2 )=-4(t-t_0).$

Recall: $\nabla =(\frac{\partial }{\partial x},\frac{\partial }{\partial y}, \frac{ \partial }{\partial z})$.

First component:

$\frac{\partial F(u)}{\partial x}= F'(u)\frac{\partial (x^2+y^2+z^2)}{\partial x}=$

$2xF'(u)$.

Answer 2

Let $f(u)=\dfrac{dF(u)}{du}$. Then by the chain rule

$$\nabla F(\mathbf{r}^2(t))=2\mathbf{r}(t)\,f(\mathbf{r}^2(t)),$$

which is a radial vector. Hence by the differential equation on time, $\mathbf{r}(t)$ is radial.

The trajectory is a radial line from the point $\mathbf{r}(0)$, and considering only the modulus $r(t)$,

$$\dot r(t)=-2r(t)\cdot1$$ has the solution

$$\log(r(t))-\log(r(0))=-2(t-0).$$