Recent content by Vishak95

Ohh okay, thank you. Just to get my fundamentals correct, would be I be correct in saying this? \int \int \boldsymbol{F}\cdot \boldsymbol{n}dS = \int \int \int div(\boldsymbol{F})dV = \int \int (curl\boldsymbol{F})\cdot \boldsymbol{n }dA = \oint \boldsymbol{F}\cdot dr

Thanks for the response. That clears most of my doubts up. Just a further query though, the form of Stokes' Theorem in my textbook looks like this: \int \int (CurlF)\cdot n dA = \oint F \cdot dr Could you please explain why the n is not needed in dealing with this problem :P

Hi, Could someone please guide me with this question? I'm unsure as to what the curl has to do with finding flux.. PS. This isn't actually assessed work, it's from a past question paper that I am using to revise. Thanks heaps!

This question has got me quite confused, some guidance would be really appreciated. A room is to be kept at 23C using a heat pump from ambient. Heat gain to the room is 1.5kW per degree Celsius difference between the atmosphere and the room. 1) If the outside temperature is 32C, what is the...

Hi MHB. I'm having yet another doubt regarding differential equations. Can someone please help me out? Thanks. Consider the following differential equation: {y}''+{y}'= x^{2} I have found the homogeneous solution to be: y_{H}=c_{1} + c_{2}e^{-x} But when finding the particular solution...

Re: Find the global maxima of this function Awesome! Thank you Mark :D

Re: Find the global maxima of this function The limit appears to be going to \infty . So then I guess that there is no global maximum either?

Re: Find the global maxima of this function Great :D I think the only real boundaries would be x = 0 , and y = 1

Re: Find the global maxima of this function The critical points both appear to be saddle points (the derivative D = -1 for both)

Thank you so much for that explanation! I understand it now :)

Hi MHB. Can someone help me with this one please? I've worked out that the critical points are (0,0) and (2,1). But looking at the boundary x = 0, there seems to be no limit to the minimum value. Also, on the boundary y = 1, the value of f(x,1) = -1. So, would I be correct in saying that the...

I'm guessing that for k > 2, it will overflow and for 0 <or= k < 2 it will empty?

Ok, thanks. The thing that's still tripping me up is this part: D_{0} \in (0,4) And the fact that right now D0 isn't in the equation...

Okay, thanks :) So now for part (b) I've got the only equilibrium solution: D = k^{2} Which is stable. But I'm completely lost looking at part (c) :(

Ok, now I've got this: \frac{dV}{dD}= 100 \frac{dV}{dt} = \frac{dV}{dD}\cdot \frac{dD}{dt} Leading to: \frac{dD}{dt}= \frac{k-\sqrt{D}}{100}