<p>well for the polar question....</p>
<p>x = r*cos(theta)</p>
<p>Plugging in x= -2 and the equation for r into the equation, you can solve for theta.</p>
<p>omfg i cant believe i got the interval of convergence wrong for taylor. That problems soo easy.. i guess the pressure got to me.</p>
<p>&<em>%)</em>#%@#%</p>
<p>yeah metra i feel you. its the same for me</p>
<p>can somebody explain why the limit test used to test for interval of convergance doesn't come out to zero, since every other term is 0 because all even derivatives of the series are zero, and shouldn't an+1/an be 0 ?? I'm confused .</p>
<p>i can't read wat the guy wrote :( for the explanation n stuff</p>
<p>For part 2b, the answer should be theta=2.786061, not 2.68636 as it says in our generous answer provider's file. I checked it using 2 different methods and spent quite a while on the test forcing my calculator to solve the crazy equation/system of equations. Any consensus?</p>
<p>
[quote]
For part 2b, the answer should be theta=2.786061, not 2.68636 as it says in our generous answer provider's file. I checked it using 2 different methods and spent quite a while on the test forcing my calculator to solve the crazy equation/system of equations. Any consensus?
[/quote]
Just punched it in myself and your number seems correct - probably made a mistake copying something when I did it in a hurry. Will update the zip accordingly.</p>
<p>Also, in 4b, Okrogius sets dy/dx = 0, when from the slope field it is apparent that at (0,1) dy/dx=-1. My answer is 2ln(3/2)+1. Consensus? Thanks for the quick reply on the other one, Okrogius. I appreciate your effort scanning and blacking out and serving all this stuff! We will go over the answers in class eventually anyway, but I didn't feel like waiting.</p>
<p>hey sleet... the problem says that the curve passes through the point (0,1), but they're looking for value of y when x = ln(3/2). since the problem says that there is a local min at x=ln(3/2) this means that dy/dx = 0 at that point. then the rest is just plug and chug as his work shows. i think the info about (0,1) is just to try to help u visualize where the minimum would be on the graph, not to actually be used to solve the problem. get it?</p>
<p>Hey i just traced to get 2.786, do u think ill lose points?</p>
<p>rcb, as long as you're accurate to the nearest 3rd decimal place, you will get full credit for your answer. the thing is, just putting 4 decimal places and truncating the rest guarantees that you got the answer right (as long as you didnt copy it down wrong). if you rounded wrong to the nearest 3rd decimal place, you will get that point taken off. BUT you got it rounded right so ur good :-).</p>
<p>Okay, got it - opspeed. I just thought you could plug the (0,1) into the slope field and kinda ignored the local min part of the problem - foolish thing to do.</p>
<p>For the overestimating/underestimating I think I put overestimates...I dunno the second deriv. looked positive, but it was a weird concept question....crap, I think for the Taylor series one I accidentally put -5 to 1 instead of -1 to 5......I can't believe I did that...do you think that i can still get some points for that?</p>
<p>with me i HAD THE RIGHT INTERVAL OF CONVERGENCE! then i screwed up my abs. values. i am so mad.....seriously</p>
<p>Can somebody explain why lim n-->infinity |an+1/an| not equal to zerom since every other term is a 0 ?</p>
<p>okay heres the thing. youre right that every other term is zero but then when you write the series, you are just going by every second one. thats why they asked you for the coefficient of an even term. for the interval of convergence you can still use the ratio test as long as you realize that you are doing the ratio between every two terms. then you find where the x part is less than 1 (i freaking messed up simplifying my absolute value here. DAMN!). every other term are the only ones that count and so you find the interval of convergence for those terms.</p>
<p>i assumed that the function was y=e^x... for the one problem for the min... i was running out of time...</p>
<p>will i get credit for saying that that means the y value at the min is 1.5? cuz that does fit the conditional statement f(0)=1, and i explicitly stated my assumption...</p>
<p>hmmm, interestingly my calculus teacher told me yesterday that the interval of convergence DOES include x=-1. i didn't really argue her nor did she show me her work, so maybe she made a mistake somewhere. what i did was use the coefficient i got in the previous answer and then just added the (x-2)^2n term to it then did the ratio test. i may have to correct her tomorrow... or be corrected lol.</p>