Recent content by whoareyou

So then the statement "all separable equations are exact" is not true?

Homework Statement Identify the following differential equation as linear, separable, exact, or a combination of the three. $$1 + \frac{1+x}{y}\frac{dy}{dx} = 0$$ Homework Equations Start with ##F(x,y)=C## ##\displaystyle \frac{d}{dx}(F(x,y)) = \frac{d}{dx} (C)## ##\displaystyle...

Homework Statement I don't want to post the actual question because I want to understand the situation in a general case. Basically, there is a bullet that moves south along the surface of the Earth as in this diagram: http://abyss.uoregon.edu/~js/images/coriolis_effect.gif. You have to find...

Homework Statement A quasi-monochromatic beam of light illuminates Young's double-slit setup, generating a fringe pattern having 5.6-mm separation between consecutive dark bands. The distance between the place containing the apertures and the plane of observation is 7 m, and the two slits are...

But ##\dot{\theta}## is a magnitude. How can this magnitude be negative?

It's just because the formula 1/s + 1/s' = 1/f used with the standard sign convention requires that the light rays must come from the left. Also I think your ray diagram is wrong (if you were drawing it to represent the the final image in the original question)

Also, there are rules for drawing the ray diagrams when the light rays come from the left to the right. How are you drawing the ray diagram when the light rays are coming from the right?

So if the light rays come from the opposite side (i.e. from right to left instead of left to right), you need to reverse the sign convention? http://faculty.spokanefalls.edu/InetShare/AutoWebs/AsaB/Phys103/MirrorsThinLens.pdf

Homework Statement A diverging thin lens and a concave mirror have focal lengths of equal magnitude. An object is placed (3/2)f from the diverging lens and the mirror is placed a distance 3f on the other side of the lens. Using Gaussian optics, determine the final image of the system after two...

##v^2=k^2\dot{\theta^2}[\sin^2(\theta)+1+2cos(\theta)+\cos^2(\theta)]=k^2\dot{\theta^2}[2+2\cos(\theta)]=2k\dot{\theta^2}r##. I did all of this before we did the problem (this is how I got the expression for ##\dot{\theta}## in the original post).

Then, $$\vec{v}(\theta)=\dot{\theta}[-k\sin(\theta)\hat{r}+k(1+\cos(\theta))\hat{\theta}]$$

Shouldn't that be positive? $$\frac{d\hat{r}}{dθ}=\hat{\theta}$$

$$\frac{d[r(θ)\vec{i}_r(θ)]}{dt} = \frac{dr(\theta)}{d\theta}\frac{d\theta}{dt}\vec{i}_r(θ) + r(\theta)\frac{d\vec{i}_r(θ)}{d\theta}\frac{d\theta}{dt}$$

It's a constant speed ##v## along the curve. ##r## isn't constant. The graph of ##r## is a cardioid.

Homework Statement I didn't know if this was considered "advanced" physics, but it's an intermediate classical mechanics course so I'll just post my question here. Basically, if you have a cardioid ##r(\theta)=k(1+\cos(\theta))##, you can show that the ##\dot{\theta}=\frac{v}{\sqrt{2kr}}##...