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	So that's 600 ft that we would have after it would have climbed 600 ft	8,595	261	[]
volunteer	after 1 2nd	8,595	262	[]
student	Mhm.	8,595	263	[]
student	Yeah, I see.	8,595	264	[]
volunteer	climbed 600 ft above 1000.	8,595	265	[]
volunteer	OK.	8,595	266	[]
student	Yeah, perfect.	8,595	267	[]
volunteer	So as we can, we can see that the expression works.	8,595	268	[]
volunteer	for any time T after t equals 0.	8,595	269	[]
volunteer	if we had at 2 seconds	8,595	270	[]
student	Mhm.	8,595	271	[]
volunteer	the rocket should be 1000 +	8,595	272	[]
volunteer	600 times 2 seconds.	8,595	273	[]
volunteer	which would 2200 ft.	8,595	274	[]
volunteer	so we would 2200 ft where the rocket is at 2 seconds.	8,595	275	[]
volunteer	over.	8,595	276	[]
volunteer	502,200 divided by would be the tangent of the angle.	8,595	277	[]
student	Yes	8,595	278	[]
volunteer	So I, I detected	8,595	279	[]
volunteer	I	8,595	280	[]
volunteer	please ask again if I'm, if it's not clear, um,	8,595	281	[]
student	No, it's really clear	8,595	282	[]
volunteer	so I think we've got the hardest part done here because	8,595	283	[]
volunteer	now	8,595	284	[]
volunteer	we have an expression for the tangent of the angle, and it's in terms of time	8,595	285	[]
volunteer	So what we can do is we can	8,595	286	[]
volunteer	use	8,595	287	[]
volunteer	to differentiate both sides of this equation by	8,595	288	[]
volunteer	time.	8,595	289	[]
volunteer	and we should be able to get how that angle	8,595	290	[]
volunteer	changes with respect to time.	8,595	291	[]
student	OK	8,595	292	[]
volunteer	Hey	8,595	293	[]
volunteer	we ask more questions if, if I'm	8,595	294	[]
volunteer	if it's not clear	8,595	295	[]
student	Yeah	8,595	296	[]
volunteer	OK	8,595	297	[]
volunteer	So now	8,595	298	[]
volunteer	um what would be the	8,595	299	[]
volunteer	so if I'm differentiating the tangent of theta, what would I get if I'm differentiating that with respect to time.	8,595	300	[]
student	If you're what	8,595	301	[]
volunteer	So if I take the derivative of the tangent function, what would I get?	8,595	302	[]
volunteer	What, what kind of	8,595	303	[]
student	Um,	8,595	304	[]
student	uh, 2 square.	8,595	305	[]
volunteer	beautiful	8,595	306	[]
volunteer	So, if I can write without	8,595	307	[]
volunteer	messing this up here	8,595	308	[]
volunteer	you're exactly right	8,595	309	[]
volunteer	We would have	8,595	310	[]
volunteer	sink and square theta.	8,595	311	[]
volunteer	and by the chain rule, since we're	8,595	312	[]
volunteer	multiplying, we're differentiating with respect to time. We must say the theta DT.	8,595	313	[]
student	Mhm	8,595	314	[]
volunteer	So we've, we've taken that	8,595	315	[]
volunteer	derivative with respect to time on that side. And now we need to do the same thing on the right side.	8,595	316	[]
volunteer	And if we do that	8,595	317	[]
volunteer	we actually see	8,595	318	[]
volunteer	we could use the um	8,595	319	[]
volunteer	see the best way to do it	8,595	320	[]
volunteer	Several ways to do that	8,595	321	[]
volunteer	We could just, um, simplify this as	8,595	322	[]
volunteer	if we wanted to, probably the easiest would be to have it as 15th	8,595	323	[]
volunteer	We want to take the derivative with respect to time. I'm gonna have to raise.	8,595	324	[]
volunteer	All right, we've got an expression which has a denominator 5000.	8,595	325	[]
volunteer	So,	8,595	326	[]
volunteer	I want to take the derivative with respect to time.	8,595	327	[]
volunteer	If I do, I can break this into two	8,595	328	[]
volunteer	variables	8,595	329	[]
volunteer	The first one will be just 1/5.	8,595	330	[]
volunteer	and the second one would be	8,595	331	[]
volunteer	um	8,595	332	[]
student	I	8,595	333	[]
volunteer	6600 over 5000, which would be 6/50.	8,595	334	[]
volunteer	times T	8,595	335	[]
volunteer	So all I did was just divide	8,595	336	[]
volunteer	I took the numerator and broke it into two parts.	8,595	337	[]
student	say you have to brush your teeth.	8,595	338	[]
volunteer	So far, so good	8,595	339	[]
student	Yeah	8,595	340	[]
volunteer	OK	8,595	341	[]
student	was my teeth	8,595	342	[]
volunteer	All right.	8,595	343	[]
volunteer	Now if we take	8,595	344	[]
student	Sorry about the	8,595	345	[]
volunteer	oh, you're fine	8,595	346	[]
volunteer	All right, let's just rewrite what we have on the left. Sekin square.	8,595	347	[]
volunteer	beta	8,595	348	[]
volunteer	the theta DT.	8,595	349	[]
volunteer	and the derivative of a constant is zero. What would I get for the derivative of the second term?	8,595	350	[]
student	Um, you would get, wait, let me see. God bless your teeth.	8,595	351	[]
student	What do you want?	8,595	352	[]
volunteer	That's it	8,595	353	[]
student	OK, so you would get um 6/50 and then	8,595	354	[]
student	yeah, the DT well the time?	8,595	355	[]
volunteer	Say it again, please	8,595	356	[]
volunteer	Yeah.	8,595	357	[]
student	Wait, is the tea the time right?	8,595	358	[]
student	OK.	8,595	359	[]
volunteer	So	8,595	360	[]

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