A Applications of Matrices Q (struggling to do any of it) (1 Viewer)

[mreditor16]

Someone please lend a hand haha -

 

[VBN2470]

Haha, is this like the last question of the Chapter? I'm pretty sure I didn't do this q at all, when I looked at it, I just closed my book and when I asked my tutor how to do this q, he just ran away

 

[mreditor16]

Haha, is this like the last question of the Chapter? I'm pretty sure I didn't do this q at all, when I looked at it, I just closed my book and when I asked my tutor how to do this q, he just ran away

hahahah cries. Yes it is the last one. So no help from you? Should I bother with it lol?

 

[mreditor16]

hahahah cries. Yes it is the last one. So no help from you? Should I bother with it lol?

Also, in our final exam, typically how many marks worth of questions will be this difficulty, out of say a 100 mark paper?

 

[VBN2470]

hahahah cries. Yes it is the last one. So no help from you? Should I bother with it lol?

Don't bother with it, they will never ask this sort of q in any exam and they probably put it in there just for LOLs. I doubt most of the cohort wouldn't even know what a spherical triangle is so they probably didn't even touch this question at all.

 

[VBN2470]

Also, in our final exam, typically how many marks worth of questions will be this difficulty, out of say a 100 mark paper?

Literally 0, you will not find this difficulty level of q's in your final exam, the questions will be much easier to solve, only hard thing is time really, if the exam was 3 hours long (and not 2) a lot of people will be able to get much higher to nearly full marks for the final.

 

[VBN2470]

What you get in past final exams will pretty much be similar in difficulty level to your final (there is nothing too hard), so practice those papers and time yourself.

 

[mreditor16]

Don't bother with it, they will never ask this sort of q in any exam and they probably put it in there just for LOLs. I doubt most of the cohort wouldn't even know what a spherical triangle is so they probably didn't even touch this question at all.

Literally 0, you will not find this difficulty level of q's in your final exam, the questions will be much easier to solve, only hard thing is time really, if the exam was 3 hours long (and not 2) a lot of people will be able to get much higher to nearly full marks for the final.

What you get in past final exams will pretty much be similar in difficulty level to your final (there is nothing too hard), so practice those papers and time yourself.

Thanks so much, mate! You're a legend.

 

[InteGrand]

Someone please lend a hand haha -

Don't see why they're asking to use matrices, it's fairly easy without them (just need a decent diagram).

a) Construct the lunes as follows. Let one of the great circles (call it ) be the "equator", and let the other one start off the same as the equator, and then "tilt" it by an angle , so half of this second great circle is in the "northern hemisphere" tilted with the equator, and the other half is symmetrically in the "southern hemisphere".

Now, consider the lune in the northern hemisphere with angle to the equator (note that the other lune in the northern hemisphere then makes an angle , as the total angle measure of the hemisphere is 180 degrees). The area of this lune is times the area of the lune that goes one quarter of the way around the full sphere (since one quarter of the sphere is 90 degrees, i.e. radians). But the area of this "quarter-lune" is just one-quarter the surface area of the sphere, i.e. . Therefore, area of the lune of angle with equator is . Similarly, area of the other lune is .

Parts b and c not even needed to do part d)

d) Draw a decent diagram of the sphere containing three great circles. You'll see two spherical triangles formed, and these are congruent. Note that each spherical triangle is the intersecting area of three lunes, and there are three pairs of lunes, (so two of each is formed by the three great circles). Let the area of lune be . The angles between adjacent great circles are . Therefore, (from part a)). So adding the areas of all lunes present gives . This is an "excess" area compared to the surface area of the sphere, because the spherical triangles have been overcounted, as they are intersecting areas of lunes. If we denote the area of one of the spherical triangle by , then has counted each spherical triangle three times instead of once (so in total, 6T has been added instead of 2T we would want to add to get to the area of the sphere)

Hence in total, is in excess of the area of the sphere by . That is, . Solving for gives , as required.

 

[mreditor16]

WOW, thanks Integrand.

 

[InteGrand]

WOW, thanks Integrand.

Welcome