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	the tangent is the high.	8,595	661	[]
volunteer	and	8,595	662	[]
volunteer	they did the same thing we did.	8,595	663	[]
volunteer	um	8,595	664	[]
volunteer	so I	8,595	665	[]
volunteer	see what they did	8,595	666	[]
volunteer	OK, notice that when they	8,595	667	[]
volunteer	they said when when H equals 1000.	8,595	668	[]
volunteer	when the height 1000, which is at times 0, OK?	8,595	669	[]
student	Mhm	8,595	670	[]
volunteer	So they're actually, they're actually taking	8,595	671	[]
volunteer	um	8,595	672	[]
volunteer	they're actually kind of setting the time equal to 0, if I'm, when I'm looking at this,	8,595	673	[]
student	So we missed that part	8,595	674	[]
volunteer	I think they missed it, but let, let me just say why I think that	8,595	675	[]
volunteer	So the problem	8,595	676	[]
volunteer	as you put it up there at the beginning,	8,595	677	[]
volunteer	it said	8,595	678	[]
volunteer	let me go all the way, I have to go all the way back to the original statement, but	8,595	679	[]
volunteer	um,	8,595	680	[]
student	OK	8,595	681	[]
volunteer	this is	8,595	682	[]
volunteer	in my opinion,	8,595	683	[]
volunteer	this is actually just showing it at that initial 2 at time equals 0.	8,595	684	[]
volunteer	Um.	8,595	685	[]
volunteer	but it doesn't answer the question as, as	8,595	686	[]
volunteer	when time equals other values, OK?	8,595	687	[]
student	Mm	8,595	688	[]
volunteer	Do you see that	8,595	689	[]
volunteer	See what I'm saying	8,595	690	[]
student	Yeah	8,595	691	[]
volunteer	The question didn't say what was the rate of change at time equals 0. It just said come up with an expression	8,595	692	[]
volunteer	for it. So if we actually plug time equals 0 in	8,595	693	[]
volunteer	to our	8,595	694	[]
volunteer	um	8,595	695	[]
volunteer	we should get the same answer	8,595	696	[]
volunteer	Do you see what I'm, I'm questioning there.	8,595	697	[]
student	Yeah, I see what you're trying to say.	8,595	698	[]
volunteer	Because	8,595	699	[]
student	I'm	8,595	700	[]
volunteer	it said come up with it, with it as an expression of um	8,595	701	[]
volunteer	with respect to time	8,595	702	[]
volunteer	Well	8,595	703	[]
volunteer	it's fine if we plugged in t equals 0.	8,595	704	[]
volunteer	then it just becomes	8,595	705	[]
volunteer	5000 squared / 5000 square, doesn't it?	8,595	706	[]
volunteer	Um	8,595	707	[]
volunteer	no, becomes, becomes 5000 squared over	8,595	708	[]
volunteer	5000 square + 1000 square.	8,595	709	[]
volunteer	So all I'm saying	8,595	710	[]
volunteer	is	8,595	711	[]
volunteer	their, their answer is right.	8,595	712	[]
volunteer	for	8,595	713	[]
volunteer	if you're just looking at the point where the rocket's just starting the launch, that the rate of change	8,595	714	[]
volunteer	is	8,595	715	[]
volunteer	going to be at that point	8,595	716	[]
volunteer	But if you wanted to find out what the rate of change is of the angle at other times, it would be a different answer.	8,595	717	[]
volunteer	Go ahead and plug in T equals 0 into what we did at the very bottom and see if I get the same answer as they did, OK?	8,595	718	[]
student	OK	8,595	719	[]
volunteer	Do you, do you see what I'm saying?	8,595	720	[]
student	Yeah, I see what you're saying	8,595	721	[]
volunteer	OK, so I'm going to	8,595	722	[]
volunteer	get my calculator out here.	8,595	723	[]
volunteer	I'm gonna do a fraction	8,595	724	[]
volunteer	the 2nd 1, 2nd fraction first there.	8,595	725	[]
volunteer	I'm gonna use 1000 square.	8,595	726	[]
volunteer	divided by	8,595	727	[]
volunteer	5000 square	8,595	728	[]
volunteer	plus	8,595	729	[]
volunteer	and since time equals 0, it would be 1000 square.	8,595	730	[]
volunteer	So 1000	8,595	731	[]
volunteer	square	8,595	732	[]
volunteer	and I get 25, 26, which is exactly	8,595	733	[]
volunteer	well, let's	8,595	734	[]
volunteer	so 25, 26 for that answer times 650.	8,595	735	[]
volunteer	So I'm gonna take times	8,595	736	[]
volunteer	6	8,595	737	[]
volunteer	50	8,595	738	[]
volunteer	and	8,595	739	[]
volunteer	so	8,595	740	[]
volunteer	our answer is exactly the same as theirs	8,595	741	[]
volunteer	when you plug in T equals 0.	8,595	742	[]
volunteer	I, I worked through the math, OK?	8,595	743	[]
volunteer	You get, you get the exact same radiance per second.	8,595	744	[]
volunteer	Are you still there	8,595	745	[]
student	Yeah, I'm here	8,595	746	[]
volunteer	OK. So if you plug in T equals 0 into what we did.	8,595	747	[]
volunteer	It gives it, it gives the same answer they get.	8,595	748	[]
student	OK	8,595	749	[]
volunteer	But this, this answer that you and I worked on is	8,595	750	[]
volunteer	far better	8,595	751	[]
volunteer	because it allows you to	8,595	752	[]
volunteer	find out the rate of change of the angle	8,595	753	[]
volunteer	at any time after	8,595	754	[]
volunteer	the launch	8,595	755	[]
student	Sure.	8,595	756	[]
volunteer	Their answer is only for the	8,595	757	[]
volunteer	for	8,595	758	[]
volunteer	at the instant of the launch, how fast does the angle changing?	8,595	759	[]
volunteer	But if you wanted to find how fast the angle was changing after 1 2nd, you would have to plug in one into our	8,595	760	[]

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