Transcript 2013-07-01 2184321

Transcript 2013-07-01 2184321 -

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<meta name="viewport" content="width=device-width; initial-scale=1.0; "\>Event ID: 2184321 Event Started: 7/1/2013 3:30:00 PM ---------- "Please stand by for realtime captions." >> I have the exams for a couple of you. So [ INDISCERNIBLE, VOLUME TOO LOW. ] . >> Some of you did well, and obviously not everybody did well, but, you know there was some really good scores, and some [ INDISCERNIBLE ] if you did not do well, if you were below a 70, especially, you know, if you were significantly below 70, then obviously you'll want to, I want to encourage you to come see me and get some extra help, because, you know, there's just not that much time. We're already in week four. Can you believe that? Halfway done after Wednesday. You get fourth of July off this week, so only this class meets three days, first few days of the week. But anyway, if you are struggling, in particular, like, you're struggling with the chain rule for example, or like, or some of those derivatives, you didn't do very well. Even if you got above a 70, you missed quite a few points on problem number six, then, you know, I want to make sure that I help did you, because it's really important that you understand what the chain rule is and how it works. >> For example, I noticed a little misconception there on some of the problems. We can certainly go over any of these. If there are particular ones that you like to look at, I can post the key to Blackboard/, I guess [ INDISCERNIBLE, [ INDISCERNIBLE, VOLUME TOO LOW. ] I can provide you with a copy of the key or [ INDISCERNIBLE ] version. I haven't gotten it scanned in yet, but with any of the ones from number six that you'd like to look at, I'd be happy to go over those. In particular that problem, but really any of the problems. Yes? No? >> [ INDISCERNIBLE, VOLUME TOO LOW. ] . >> Okay. I wrote some funny comments. Sometimes I do that. I will pick one to do because I think probably, I'm trying to think which one was the hardest. Well C was an interesting one. Why don't we look at C and D together. We'll look at the [ INDISCERNIBLE ] so C says, does h of t equal negative two times two minus contingent of t over five see can't of t. Now, some folks cant of t. Now, some folks I think most people chose to do this as [ INDISCERNIBLE ] and that's fine. But there's other ways you can do it. I think there was one person who did it usually the [ INDISCERNIBLE ] rule. By saying I'm going to write this as negative 2/5 co-sign of t, secant t in the denominator is the name as co-sign t not in the denominator, and two minus tangent of t. And again, but you actually don't even need that. So if you, but I think most people did it that way, and that was fine. But you could have also just multiplied this out and got negative 4/5 co-sign of t plus 2/5 tangent times co-sign, what's that? >> [ INDISCERNIBLE, SPEAKER'S VOLUME TOO LOW ]
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This note was uploaded on 09/28/2013 for the course EGR 125 taught by Professor Edwarderisman during the Summer '13 term at College Of Lake County.

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