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fp1 complex numbers

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Reply 180
Avatar for bigmeat
bigmeat
Original post by NeverLucky
I think what you're saying is correct. However, for part b, l2l_2 is perpendicular to l1l_1. So if l1l_1 has a positive gradient, then l2l_2 has a negative, reciprocal gradient.

i.e. m1×m2=−1m_1 \times m_2 = -1


i understand but in this video the guy doesn't put the 2 points on the same part of the graph(close together) he puts one on the first quadrant and the other which has (x,-y) which is on the "bottom" part of the parabola.
see
http://www.examsolutions.net/a-level-maths-papers/worked-solution/worked-solution.php?paper_id=86&solution=8.2

i guess though i could still try and carry on.
Original post by bigmeat
i understand but in this video the guy doesn't put the 2 points on the same part of the graph(close together) he puts one on the first quadrant and the other which has (x,-y) which is on the "bottom" part of the parabola.
see
http://www.examsolutions.net/a-level-maths-papers/worked-solution/worked-solution.php?paper_id=86&solution=8.2

i guess though i could still try and carry on.


It honestly doesn't really matter where he puts the points - the points on the graph are arbitrary. It really is just for the viewer to get an idea of the method on how to do the question. He could have just as easily put the points next to each other but of course that's not going to be neat.
(edited 7 years ago)
Reply 182
Avatar for bigmeat
bigmeat
Original post by NeverLucky
It honestly doesn't really matter where he puts the points - the points on the graph are arbitrary. It really is just for the viewer to get an idea of the method on how to do the question. He could have just as easily put the points next to each other but of course that's not going to be a neat.


oh right, i though i did it wrong from the first step thanks ^-^
Original post by Chittesh14
It's all part (b)s today lol. Another one....
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Working out in case you want to see it

Spoiler



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(edited 7 years ago)
Original post by RDKGames
Working out in case you want to see it

Spoiler


Posted from TSR Mobile


I'm just playing games for like 4 hours lol. I'm going to do Maths later, but thank you for posting this. I'll take a look if I need it - most probably will.
Reply 185
Avatar for flyingd
flyingd
Original post by RDKGames
Working out in case you want to see it

Spoiler


Posted from TSR Mobile


what's the difference between cartesian and parametric?
Original post by flyingd
what's the difference between cartesian and parametric?


Cartesian is only in terms of x and y.
Parametric form is mostly in terms of coordinates that you have to rearrange and substitute in order to find the Cartesian form.
Reply 187
Avatar for flyingd
flyingd
Original post by Chittesh14
Cartesian is only in terms of x and y.
Parametric form is mostly in terms of coordinates that you have to rearrange and substitute in order to find the Cartesian form.


ok thanks
Original post by flyingd
what's the difference between cartesian and parametric?


Cartesian are your normal everyday equations involving x and y; where both are dependent on each other. Parametric equations introduce a third variable, usually t, where both x and y are dependent on it; as you manipulate t, both x and y change accordingly. This third variable usually has a range that you can set, but it's not necessary. You can change between the two forms by means of substitution.

Example:
y=x2+2x+1y=x^2+2x+1

can be represented in the parametric form of:

[br](x,y)[br]x=t[br]y=t2+2t+1[br](x,y)[br]x=t[br]y=t^2+2t+1

where t is real. Try some out on Desmos and you can see how parametrics are much more useful than cartesian coordinates in some cases. :smile:

Parametrics open up doors to some cool looking graphs such this one that I've created when I first learnt about parametrics: https://www.desmos.com/calculator/jekx6csxzm
(edited 7 years ago)
Original post by flyingd
http://www.examsolutions.net/maths-revision/further-maths/coordinate-geometry/rectangular-hyperbola/tangent-normal/cartesian/example-1.php

^^ is there another way to do this? i don't know how to do implicit differentiation but i do know how to use the chain rule


Just normally do it. Like xy = c^2...
y = c^2/x
y = c^2x^-1
dy/dx = -c^2/x^2

and then sub in the x value (a) find gradient use y-y1 = m(x-x1) at that point etc... done easy!
Original post by flyingd
http://www.examsolutions.net/maths-revision/further-maths/coordinate-geometry/rectangular-hyperbola/tangent-normal/cartesian/example-1.php

^^ is there another way to do this? i don't know how to do implicit differentiation but i do know how to use the chain rule


What Chittesh said is the most straight forward way. You can also use parametrics which is strightly more long winded but can be useful if that y was y2.

ImageUploadedByStudent Room1469613646.014366.jpg


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Reply 191
Avatar for flyingd
flyingd
Original post by Chittesh14
Just normally do it. Like xy = c^2...
y = c^2/x
y = c^2x^-1
dy/dx = -c^2/x^2

and then sub in the x value (a) find gradient use y-y1 = m(x-x1) at that point etc... done easy!


Original post by RDKGames
What Chittesh said is the most straight forward way. You can also use parametrics which is strightly more long winded but can be useful if that y was y2.

ImageUploadedByStudent Room1469613646.014366.jpg


Posted from TSR Mobile

thanks both
Original post by flyingd
thanks both


Anytime.


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What Chittesh said is the most straight forward way.


No it isn't, implicit differentiation is more straight forward, and always the method I'd recommend for these type of questions, re-arranging and then differentiating is just bad form.
Original post by Zacken
No it isn't, implicit differentiation is more straight forward, and always the method I'd recommend for these type of questions, re-arranging and then differentiating is just bad form.


Some people don't understand implicit differentiation that well? I think it depends more on the person...


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Reply 195
Avatar for flyingd
flyingd
Original post by Zacken
No it isn't, implicit differentiation is more straight forward, and always the method I'd recommend for these type of questions, re-arranging and then differentiating is just bad form.


guess i should learn it then? would i find this in c3 or c4?
Original post by flyingd
guess i should learn it then? would i find this in c3 or c4?


You'll find it in C4; it's quite straightforward once you learn it really.
Original post by Zacken
No it isn't, implicit differentiation is more straight forward, and always the method I'd recommend for these type of questions, re-arranging and then differentiating is just bad form.


I meant that in the context where implicit differentiation was out of the question. Obviously that would be the go-to option for everyone but he clearly stated how to do the question without it.
Reply 198
Avatar for flyingd
flyingd
Original post by NeverLucky
You'll find it in C4; it's quite straightforward once you learn it really.


ok thanks i'll have a look
Original post by flyingd
ok thanks i'll have a look


If you're going into Year 12, I'd say stick with what I done, I prefer it much more. I haven't learnt implicit differentiation myself but that's because I just don't want to lol.


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