I didn't really understand why , in this vector question, there are TWO values for a and b. If someone could help me I would be really grateful! thanks!
I didn't really understand why , in this vector question, there are TWO values for a and b. If someone could help me I would be really grateful! thanks!
Definition:
dxdy
means "the derivative of y with respect to x".
So when you differentiate a y term with respect to x, you need to put that.
About the vectors q, it just means that the lines are parallel for 2 values of a and b.
I didn't really understand why , in this vector question, there are TWO values for a and b. If someone could help me I would be really grateful! thanks!
I haven't got time to watch the video right now for 2), but for 1) when you have to do an implicit differentiation you're basically making use of the chain rule - remember when you first learnt the chain rule you wrote y as a function of u and u as a function of x, and found that:
dxdy=dudydxdu
With implicit differentiation you normally have something like f(y) = g(x)
so when you differentiate the LHS you can think of f being a function of y and y being a function of x, so the chain rule would tell you that:
hello , i don't see how that could potentially work because if you diff 2x+3 you get just 2 , isn't it so? So it definitely cannot be forming that pattern of diff/ original function , isn't it so? Please correct me...
hello , i don't see how that could potentially work because if you diff 2x+3 you get just 2 , isn't it so? So it definitely cannot be forming that pattern of diff/ original function , isn't it so? Please correct me...
Oh I thought it was an interview to get into cambridge and oxford?
Hello , if u don't mind , I wanted some maths help from u :
U know that in C4 differentiation chapter, we have to know how to do 'connected rates of change" problems. What tips would u give me with regards to them and also are the problems involving cones the hardest we could get on them?
Also I was wondering about these sorts of questions where they give a different thing as our substitution, for example x=2sinu and stuff like that ...any tips with such problems and what is the aim of giving such difficult things to sub in ...
Oh I thought it was an interview to get into cambridge and oxford?
Hello , if u don't mind , I wanted some maths help from u :
U know that in C4 differentiation chapter, we have to know how to do 'connected rates of change" problems. What tips would u give me with regards to them and also are the problems involving cones the hardest we could get on them?
Well, remember that 'proportional' means that there's a constant k involved, and to extract the maths by reading the question carefully.
Have a look at Q8 in the first paper (2010) and Q5 in the second (2009).
Also I was wondering about these sorts of questions where they give a different thing as our substitution, for example x=2sinu and stuff like that ...any tips with such problems and what is the aim of giving such difficult things to sub in ...
Firstly, don't worry too much as they always give you the substitution, and you only need to figure it out for yourself in FM.
Secondly, they just want to test you to the maximum extent.
So, suppose they've asked you to integrate
∫a2−x21dx
using
x=asinu
First, you would differentiate to get dx in terms of d theta, and then substitute in.
I did the following and just wanted to check with u if I have done it right:
x=2cost y=sin2t
dx/dt= -2sint
so integral sign (sin2t) (-2sint)
then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!? thanks laura
I did the following and just wanted to check with u if I have done it right:
x=2cost y=sin2t
dx/dt= -2sint
so integral sign (sin2t) (-2sint)
then [sin(3t) / (3) - sint ] and for the limits I got t= 90 degrees and 0 . I subbed them in but got -4/3....what do i do next? am i going the right way is this one of those questions where the limits will swap around because we surely cannot have a negative area!? thanks laura