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Hi guys, thanks for the replies! Was sure it was correct!

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Hi there, I wanted to check whether I got this one correct: Q: A bubble rises from a height [math]h[/math] from the bottom of a tall open tank of liquid that is at uniform temperature. Write an expression for the bubble's volume [math]V_{f}[/math] just before it bursts at the surface in terms of it's original volume [math]V_{i}[/math], and the uniform density of the liquid [math]\rho[/math]. A: By taking the temperature to be constant, the initial pressure to be that of the atmosphere plus the pressure at a height [math]h[/math] in the liquid and the final pressure to be simply that of the atmosphere. By using the [math]\frac{PV}{T} = constant[/math] equation I derive: [math]\frac{P_{i}V_{i}}{T} = \frac{P_{f}V_{f}}{T}\rightarrow (P_{atm} + \rho gh)V_{i} = P_{atm}V_{f}\rightarrow V_{f} = \frac{P_{atm} + \rho gh}{P_{atm}}V_{i}[/math] Is this correct? Thanks

Sorry, mistake(s)! [math]\lim_{x \to -2} (x^2 - 1) = 3[/math] if [math]0 < |x - (-2)| < \delta[/math] then [math]|(x^2 - 1) - 3| < \epsilon[/math] we then have: [math]|(x^2 - 1) - 3| < \epsilon \rightarrow|x^2 - 4| < \epsilon \rightarrow|x^2 - 2^2| < \epsilon \rightarrow|x + 2||x - 2| < \epsilon \rightarrow|x + 2| < \frac{\epsilon}{|x - 2|}[/math] this suggests: [math]\delta = \frac{\epsilon}{|x - 2|} \Leftrightarrow \epsilon = |x - 2| \delta[/math] testing new [math]\delta[/math]: [math]|(x^2 - 1) - 3| < |x - 2| \delta \rightarrow|x + 2||x - 2| < |x - 2| \delta \rightarrow|x + 2| < \delta[/math] and proved. Sorry for the error, very careless of me! EDIT: I can't edit my original post, but the correction is above. Any feedback as to the proof?

Yeah I got it And the other proof, that alright?

Hi there, I've been doing some questions out of a book, and came across this question which I found interesting. I was wondering if there are other mathematical techiques that can be used to solve a problem like this? Question: The rotational period of Earth is 23.933 hours. A space shuttle revolves around Earth's equator every 2.231 hours. Both are rotating in the same direction. At the present time, the space shuttle is directly above the Galapagos Islands. How long will it take for the space shuttle to circle Earth and return to a position directly above the Galapagos Islands? Solution: The time taken to travel is [math]\frac{\theta}{\omega}[/math]. We know that the shuttle will complete one full orbit, and by the time it completes this orbit in 2.231 hours the Earth will have rotated a certain angle much less than that of the shuttle. We know that the shuttle will come back to the same point it was at in the beginning after completing a full orbit and travelling a further [math]\theta[/math] to catch up with the original point. Taking [math]\omega _{e}, \omega _{s}[/math] to be the angular velocity of the Earth and shuttle respectively gives: [math]\frac{\theta}{\omega _{e}} = \frac{2 \pi + \theta}{\omega _{s}}\rightarrow \frac{2 \pi + \theta}{\omega _{s}} - \frac{\theta}{\omega _{e}} = 0\rightarrow \theta = -\frac{2 \pi \omega _{e}}{\omega _{e} - \omega _{s}}[/math] And then solving for time [math]t[/math], by dividing by [math]\omega _{e}[/math]: [math]t = \frac{-\frac{2 \pi \omega _{e}}{\omega _{e} - \omega _{s}}}{\omega _{e}}\rightarrow t = -\frac{2 \pi}{\omega _{e} - \omega _{s}}[/math] For this example my solution gave [math]t \approx 2.46[/math] hours. Thanks

Hi guys, Thanks for the replies, very helpful. I'm sure it's right now! mathematic, the definition of continuity is that the function is continuous everywhere on it's domain, and so the limit everywhere on the domain exists, so essentially the question was that I was told the limit was [math]f(a)[/math] as the limit approached [math]0[/math], and so all I had to do was take [math]h[/math] to [math]0[/math] in the limit to prove that it was continuous. I stuggle with the logic, here is a harder proof I attempted before and I was hoping to get some verification on my attempt: Prove the statement using the [math]\epsilon , \delta[/math] definition of a limit: [math]\lim_{x \to -2}(x^2 - 1) = 3[/math] if [math]0 < |x -(-2)| < \delta[/math] then [math]|(x^2 - 1) - 3| < \epsilon[/math] we then have: [math]|(x^2 - 1) - 3| < \epsilon\rightarrow |x^2 -2| < \epsilon\rightarrow |x + 1||x - 1| < \epsilon\rightarrow |x - 1| < - \frac{\epsilon}{|x + 1|}[/math] this suggests: [math]\delta = \frac{\epsilon}{|x + 1|} \Leftrightarrow \epsilon = |x + 1| \delta[/math] testing our new [math]\delta[/math] : [math]|(x^2 - 1) -3| < \delta |x + 1|\rightarrow |x + 1||x - 1| < \delta |x + 1|\rightarrow |x - 1| < \delta[/math] and we have proved the statement. I have a feeling that the above is fishy because of the fact that I derived the new [math]\delta[/math] from the given statement? Hope someone can help out. Thanks!

Hi there, I have been doing some calculus exercises, and was wondering if I had this right. Prove that [math]f[/math] is continuous at [math]a[/math] if and only if [math]\lim_{h \to 0}f(a+h) = f(a)[/math] I'm not sure if I got this correct, but I assumed that as we take [math]h[/math] to [math]0[/math] then [math]f(a+h) = f(a+0) = f(a)[/math] and so we are done? Thanks