Split (1)

train · 116k rows

role stringclasses 2 values	content stringlengths 0 2.1k	session_id int64 10 21.7k	sequence_id int64 0 2.38k	annotations listlengths 0 8
volunteer	I'm sorry, say again	8,595	1,561	[]
student	The uh 12 cubic centimeters.	8,595	1,562	[]
volunteer	So it is, let me go back up to the top and read it with you.	8,595	1,563	[]
volunteer	So the volume remains constant at 12 pi cubic meters.	8,595	1,564	[]
volunteer	So what does that say about DVDT?	8,595	1,565	[]
student	Uh, that is constant. It's 0.	8,595	1,566	[]
volunteer	Yes, that's the key right there. So that's a little bit of a trick.	8,595	1,567	[]
volunteer	DVDT is 0 because they're telling you that it is a constant.	8,595	1,568	[]
student	No	8,595	1,569	[]
volunteer	And we know that the derivative of a constant is zero, so that first term	8,595	1,570	[]
volunteer	gets blown away, right	8,595	1,571	[]
student	Right	8,595	1,572	[]
volunteer	OK, let's see how we're doing. So you already know you got age. Let's see, what do we have V and DHBT?	8,595	1,573	[]
student	The position is negative, um, 1 m. So just -1.	8,595	1,574	[]
volunteer	Um	8,595	1,575	[]
volunteer	right, shrinking. And what is	8,595	1,576	[]
volunteer	uh B	8,595	1,577	[]
student	And then the V is the 12 pi.	8,595	1,578	[]
volunteer	You got it. So let's plug in and see what you get for t at.	8,595	1,579	[]
volunteer	as the A equals.	8,595	1,580	[]
student	OK	8,595	1,581	[]
student	Mm	8,595	1,582	[]
student	I'll just write it on here	8,595	1,583	[]
student	OK	8,595	1,584	[]
student	over	8,595	1,585	[]
student	So is it basically to for high overnight.	8,595	1,586	[]
volunteer	Mhm.	8,595	1,587	[]
student	You have to fur further, yeah, we can further do it but I really.	8,595	1,588	[]
volunteer	Mhm.	8,595	1,589	[]
student	free and cover.	8,595	1,590	[]
student	and then	8,595	1,591	[]
student	not forgetting the	8,595	1,592	[]
student	so.	8,595	1,593	[]
student	need	8,595	1,594	[]
student	ter s and meters	8,595	1,595	[]
student	That.	8,595	1,596	[]
volunteer	So let's see what you you're talking the BOD things so	8,595	1,597	[]
volunteer	everything's in uh	8,595	1,598	[]
volunteer	square meters, right?	8,595	1,599	[]
student	per hour	8,595	1,600	[]
volunteer	Um	8,595	1,601	[]
student	square meters	8,595	1,602	[]
volunteer	let's see	8,595	1,603	[]
volunteer	We, let's see what they say.	8,595	1,604	[]
volunteer	Yeah, meters and square meters per	8,595	1,605	[]
volunteer	uh	8,595	1,606	[]
student	per hour I mean	8,595	1,607	[]
volunteer	where did they say the time?	8,595	1,608	[]
volunteer	Oh, I'm gonna	8,595	1,609	[]
student	like just below	8,595	1,610	[]
student	where it says um	8,595	1,611	[]
volunteer	call per hour, I see, right.	8,595	1,612	[]
student	yeah, for uh	8,595	1,613	[]
student	OK.	8,595	1,614	[]
volunteer	So the square meters per hour, right?	8,595	1,615	[]
student	And it is square meters because we're finding that DADT, which is the derivative of the uh area, right?	8,595	1,616	[]
volunteer	Correct	8,595	1,617	[]
student	OK.	8,595	1,618	[]
volunteer	So that's kind of what we're just basing it on, yeah.	8,595	1,619	[]
volunteer	So	8,595	1,620	[]
volunteer	let me just step back for a minute and think about what you got here.	8,595	1,621	[]
volunteer	So we know that if the if the height is shrinking	8,595	1,622	[]
student	Mhm	8,595	1,623	[]
volunteer	and the thing is staying constant. I'm just thinking out loud here. The area must be going up and you have a positive area, right?	8,595	1,624	[]
volunteer	Cause	8,595	1,625	[]
volunteer	so what I'm, I'm just thinking intuitively about your answer, which I think it makes sense.	8,595	1,626	[]
volunteer	Um.	8,595	1,627	[]
student	OK	8,595	1,628	[]
volunteer	they're telling you that the volume is constant	8,595	1,629	[]
volunteer	but that the height is shrinking.	8,595	1,630	[]
volunteer	Therefore, it seems logical that the area is increasing or the radius, you know, radius and area are increasing, and that's what you came up with. You came up	8,595	1,631	[]
volunteer	with the fact that	8,595	1,632	[]
volunteer	in order for the volume	8,595	1,633	[]
student	Well sorry you like the little Betty didn't hear what you say.	8,595	1,634	[]
volunteer	you want me to start over	8,595	1,635	[]
volunteer	Oh, OK	8,595	1,636	[]
volunteer	Um	8,595	1,637	[]
volunteer	so I'm just looking at your answer, which I believe is correct.	8,595	1,638	[]
volunteer	Reme I'm just thinking about what the whole problem says, which could be uh	8,595	1,639	[]
volunteer	could be kind of the answer to part E here, but	8,595	1,640	[]
student	OK.	8,595	1,641	[]
volunteer	um	8,595	1,642	[]
volunteer	so they told you the volume was constant.	8,595	1,643	[]
volunteer	in the problem	8,595	1,644	[]
volunteer	But yet they told you the height was shrinking.	8,595	1,645	[]
volunteer	right?	8,595	1,646	[]
volunteer	But	8,595	1,647	[]
student	Mhm. Yeah.	8,595	1,648	[]
volunteer	so that means in order for the volume to be to be constant and the high shrinking that the area, the cross-sectional areas must be getting bigger, right?	8,595	1,649	[]
student	Yeah.	8,595	1,650	[]
volunteer	And that's what your answer your answer says, the areas	8,595	1,651	[]
volunteer	increasing	8,595	1,652	[]
volunteer	even though the, the height is decreasing at 1 m per second	8,595	1,653	[]
volunteer	or per hour	8,595	1,654	[]
volunteer	The area is increasing at 45/3 square meters per hour.	8,595	1,655	[]
volunteer	So.	8,595	1,656	[]
volunteer	So first of all, can you, do you know what the answer to this question is or this one?	8,595	1,657	[]
student	Yeah, there's a, it's I can check it.	8,595	1,658	[]
volunteer	Yeah, let's make sure we're	8,595	1,659	[]
student	Yeah	8,595	1,660	[]

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