Recent content by zvjaginsevfan

I finally solved it, thank you very much!

It's -1/x^2 but you said you should do it wrt to t, am I missing something?

So applying d/dt to both sides of v = 1/x 1- dv/dt = d/dt * x/x^2 2- a = dx/dt * 1/x ^2 3- a = v / x^2 4- a = 1/x^3 for x = 4, a is found 1/256 m/s^2 here. But shouldn't accelaration be negative, did I make a mistake in these steps?

Four message above, there is an attached photo. I used chain rule and I used the equation you wrote but it turns out accelaration is a positive number in my case, which is illogical as 1/x is a decreasing function.

I stated it above, I could not do it due to problem in retrieving calculus knowledge. It felt like 1/x is gone and there is 0 at the right side of the equation but probably it is not.

Sorry for flood, but I also found something with chain rule, but it turns out accelaration is a positive number in my case, which is illogical as 1/x is a decreasing function.

1- dv/dx = dv/dt * dt / dx 2- v = 1/x 3-differentiating w.r.t x, dv/dx = -1/x^2 *dx 4- as dv/dt = a, I found -1/x^2*dx = a*dt/dx -1/x^2*dx = a*dt/dx is where I stuck. I could not go further.

I want to differentiate it w.r.t t but I think my calculus skills are rusty after holiday, so I went with this way: 1- dv/dx = dv/dt * dt / dx 2- v = 1/x 3-differentiating w.r.t x, dv/dx = -1/x^2 *dx 4- as dv/dt = a, I found -1/x^2*dx = a*dt/dx -1/x^2*dx = a*dt/dx is where I stuck. I could not...

I used that formula. At the end I got dv = axdx, and as a also depends on x, I could not figure out how to find the a from here. Can you explain?

This is a homework question from my friend, I found the time but a tough differential equation occurred when I was trying to find accelaration, is there a simple solution for this?