Path of particle following gradient of a function of three variables

Question

A particle moves in 3D-space in such way that its direction of motion at any point is perpendicular to the level surface of $f(x, y, z) = 4 − x^2 − 2y^2 + 3z^2$ through that point. If the path of the particle passes through the point $(1, 1, 8)$, show that it also passes through $(2, 4, 1)$. Does it pass through $(3, 7, 0)$?

Clearly the gradient of the function is some scalar multiple of the velocity of the particle, and from this follows:

$dx/dt = -2k(x), dy/dt = -4k(y)$, and $dz/dt = 6k(z)$ where k is some scalar.

Then, $-1/2(dx/x) = -1/4(dy/y) = 1/6(dz/z)$.

How, then, can the path of the particle be parameterized in terms of t, if at all?

Answer 1

How about this:

$dx/dt = -2 k(t) x$;
$dy/dt = - 4k(t) y$;
$dz/dt = 6k(t) z$.

Allowed for a $k(t)$.

Separate variables :

$\int dx/x = -2 \int k(t)dt$ ; $\int dy/y = -4 \int k(t)dt$ ; $\int dz/z = 6 \int k(t)dt;$

Integrate :

$ln(x) = -2 \int k(t)dt + E;$

$ln(y) = -4 \int k(t)dt + F$;

$ln(z) = 6 \int k(t)dt + G;$

Let $\int k(t)dt =: p(t);$

We get :

$x = A exp(-2p(t)),$ with $A = exp(E);$

$y = B exp(-4p(t)),$ with $B = exp(F);$

$z = C exp(6p(t)),$ with $C = exp (G),$

$A,B,C$ are constants to be determined by the initial condition, path passes through$ (1,1,8).$ Let $t= 0$ at this point.

$1 = A exp (-2p(0))$ ;

$1 = B exp(-4p(0))$;

$8 = C exp(6p(0))$.

If follows:

$A = exp (2p(0))$; $B = exp(4p(0))$; $C = 8 exp(-6p(0))$.

Note : $B = A^2$ and $C = 8 A^{-3}$.

Does the particle pass through $(2,4,1)$?

$2 = exp(2p(0)) exp (-2p(t))$;

$4 = exp(4p(0)) exp(-4p(t))$;

$1 = 8 exp(-6p(0)) exp(6p(t))$;

Square the first equation to get the second, and rearrange the first eq.

$ 2 exp(-2p(0)) exp(2p(t)) = 1$ and cube it

to get the third, so all is well.

$(3,7,0)$??

Does not pass through.

Comments welcome.

Answer 2

If $r(t)$ is position of the particle, then the hypothesis can be written: $$\dot{r}(t)=k(t)\nabla f(r(t))$$ It's important to note that $k$ depends on time and not on the position of the particle. This equation defines three differential equations. This equations are of the form $\dot{u}(t)=ak(t)u(t)$, where $a$ is a constant and $u$ is the unknown. It can be solved: $\ln(u(t))=\int \frac{\dot{u}(\tau)}{u(\tau)}d\tau=a\int k(\tau)d\tau$. Although $\int k(\tau)d\tau$ depends on time, it does not affect the direction of the particle. I left to you the details.