First Year Mathematics A (Differentiation & Linear Algebra) (1 Viewer)

[iforgotmyname]

Re: MATH1131 help thread

Lol idk we don't do the same course. But I thought you guys also had 4 questions. (2 alg 2 calc)

I think she missed the last page..._

 

[Flop21]

Re: MATH1131 help thread

I think she missed the last page..._

oh shit yeah, I missed the "4", thought it was just one big Q3

 

[leehuan]

Re: MATH1131 help thread

The paper is out of a grand total of 80. Each question is of equal value, thus each question has a mark allocation of 20.

Unfortunately, and disappointingly, the individual mark allocations for each part are not given at uni.

 

[Flop21]

Re: MATH1131 help thread

I have no idea why the question got deleted but to expand on this.







Here, we have a=[1;1], t=[1;2], c=[2;5]

So how would you find something that is perpendicular to the line that passes through another specific point?

(I know its perpendicular if dot product = 0).

 

[iforgotmyname]

Re: MATH1131 help thread

So how would you find something that is perpendicular to the line that passes through another specific point?

(I know its perpendicular if dot product = 0).

Specific point + lamda<vector with dot prod=0>

 

[turntaker]

Re: MATH1131 help thread

why is maths so shit

 

[leehuan]

Re: MATH1131 help thread

So how would you find something that is perpendicular to the line that passes through another specific point?

(I know its perpendicular if dot product = 0).

One way (albeit not the best way) to find a vector perpendicular to a line is to just let your arbitrary vector x=(a,b,c)

Then x.u=0

But you can easily equate the components out and find any set of values for a, b and c that works.

Note: There is an arbitrarily large amount of vectors perpendicular to a line in R³. If you wanted to find a vector perpendicular to TWO lines this would be much harder, and the cross product would be employed.

 

[leehuan]

Re: MATH1131 help thread

why is maths so shit

Did you even understand what IFMN or myself said?

 

[turntaker]

Re: MATH1131 help thread

Did you even understand what IFMN or myself said?

Yea but I literally forgot vector representation of lines, i have to go back and properly read it. but i ceebs so much.

 

[leehuan]

Re: MATH1131 help thread

Yea but I literally forgot vector representation of lines, i have to go back and properly read it. but i ceebs so much.

It can be inferred if you know what the significance of the parameter lambda means.

Because lambda is just any real number, it means it can be 1, 2, -100, 696969696969 or whatever. That's why the vector associated with lambda is the direction vector - because all the values of lambda are plotted, it maps out a line.

The direction vector being the same for parallel lines is just a consequence of what direction means.

(However, obviously it can also be a scalar multiple, as that only alters the length of the vector. Not which way it's pointing)

 

[Flop21]

Re: MATH1131 help thread

Turntaker when's your math test?

 

[iforgotmyname]

Re: MATH1131 help thread

One way (albeit not the best way) to find a vector perpendicular to a line is to just let your arbitrary vector x=(a,b,c)

Then x.u=0

But you can easily equate the components out and find any set of values for a, b and c that works.

Note: There is an arbitrarily large amount of vectors perpendicular to a line in R³. If you wanted to find a vector perpendicular to TWO lines this would be much harder, and the cross product would be employed.

Yeah, you need to get the cross product between 2 lines then do <cross prod>*<x-(point)> to convert into point normal form

Then you need to do convert into cartesian again. Not worth the effort to mark ratio, so if this shit does pop out then move on and cone back when you have time

 

[turntaker]

Re: MATH1131 help thread

Turntaker when's your math test?

Tuesday

 

[leehuan]

Re: MATH1131 help thread

Yeah, you need to get the cross product between 2 lines then do <cross prod>*<x-(point)> to convert into point normal form

Then you need to do convert into cartesian again. Not worth the effort to mark ratio, so if this shit does pop out then move on and cone back when you have time

This is for if you want to find a line, passing through a, normal to Π.

However, luckily, conversion from the point normal form n.(x-a)=0 to Cartesian form ax+by+cz=d is REALLY easy, as a, b and c are the components of n and d is just n.a

 

[Flop21]

Re: MATH1131 help thread

y = x^(sinx)

find dy/dx

 

[Flop21]

Re: MATH1131 help thread

Tuesday

wow plenty of time, mine's tomorrow ha fml

 

[leehuan]

Re: MATH1131 help thread

Note: The CASIO fx-100 AU PLUS calculator, famous for allowing Ext 2 students to do complex numbers in the exam room, ALSO calculates vector dot product and vector cross product for you. This is a good way to check your answer in the exam room, provided you know how to use it.

 

[iforgotmyname]

Re: MATH1131 help thread

Note: The CASIO fx-100 AU PLUS calculator, famous for allowing Ext 2 students to do complex numbers in the exam room, ALSO calculates vector dot product and vector cross product for you. This is a good way to check your answer in the exam room, provided you know how to use it.

Didnt know it calculated dot and cross as well. Just brought it to pass elec... Dem features is good

 

[iforgotmyname]

Re: MATH1131 help thread

y = x^(sinx)

find dy/dx

Hint,ln both sides

 

[leehuan]

Re: MATH1131 help thread

y = x^(sinx)

find dy/dx

wow plenty of time, mine's tomorrow ha fml

Mine is next Friday...