I don't understand question 6 for the june 2014 paper as I don't know how to simplify the gradient to solve the proof or find the center of the chord for part c.
Also, for question 9, in the reduction formula I understand how to do it, but I just wanted some clarification on what x^2=x^2-1=1 meant? - In the mark scheme it shows this is used to separate the bracket into 2 and I was just hoping you could explain why and how to use it?
Hopefully, this makes more sense than what the mark scheme did.
this may be a stupid question but when I'm writing an equation of a line in this form what do I do if one of the direction parts is 0...do I write a zero on the denominator?
this may be a stupid question but when I'm writing an equation of a line in this form what do I do if one of the direction parts is 0...do I write a zero on the denominator?
Does anyone recall any tips/tricks for getting the reduction formula that you encountered in questions from the textbook/other sources? It often turns into a case of either spotting the trick or not making any progress so I was just wondering if people have some techniques that usually work well for reduction questions of different types
Does anyone recall any tips/tricks for getting the reduction formula that you encountered in questions from the textbook/other sources? It often turns into a case of either spotting the trick or not making any progress so I was just wondering if people have some techniques that usually work well for reduction questions of different types
It often comes down to creatively factorising the integral in your IBP expression. The two last year (R paper and standard) involved adding and subtracting an expression and I've seen some past questions do this as well I think. If you have to simplify something like x^2(x^2 + 7)^(n - 1) x^2(x^2 + 7)^(n - 1) = x^2(x^2 + 7)^(n - 1) + 7(x^2 + 7)^(n - 1) - 7(x^2 + 7)^(n - 1) Which is identical to (x^2 + 7)(x^2 + 7)^(n - 1) - 7 (x^2 + 7)^(n - 1) So (x^2 + 7)^n - 7(x^2 + 7)^(n - 1)
For trig, identities are your friends; in particular sin^2 and cos^2. This is equally true for hyperbolics since you have all the similar identities. Natural logs are usually straightforward matters; if you have something of the form (x^k) * (lnx)^n, then just set dv/dx = x^k and it follows naturally
edit: "creatively" is used in the loose sense, I know it's nothing particularly special but that kind of manipulation isn't usually demanded
Could someone explain what their method would be for part (d) of this question please? Here was my method but it was fairly algebra-heavy and I'm rusty on vectors so I'm wondering if there's an easier way:
Find the plane that l2 lies in by using r.n=b.n (plane has same normal vector but passes through B) Sub components of D into Cartesian equ of this plane to get a linear equation with a in terms of b Use Pythagoras with the given distance and sub in previous equation to get a quadratic in terms of b Solve for b then sub back into equation for a.
For question 4b on the June 2014 R paper, I got 1152/35 as my answer, when the real answer should be (1152/35)sqrt3. Which means I figured out I0 wrongly, as I got it to equal 1. Can someone please explain how to get sqrt3 as my I0?
For question 4b on the June 2014 R paper, I got 1152/35 as my answer, when the real answer should be (1152/35)sqrt3. Which means I figured out I0 wrongly, as I got it to equal 1. Can someone please explain how to get sqrt3 as my I0?
I-zero = the integral of one (limits of 0 --> sqrt3)
Could someone explain what their method would be for part (d) of this question please? Here was my method but it was fairly algebra-heavy and I'm rusty on vectors so I'm wondering if there's an easier way:
Find the plane that l2 lies in by using r.n=b.n (plane has same normal vector but passes through B) Sub components of D into Cartesian equ of this plane to get a linear equation with a in terms of b Use Pythagoras with the given distance and sub in previous equation to get a quadratic in terms of b Solve for b then sub back into equation for a.
I am too not completely up to speed with vectors. I did very similar, I got a = 3 and b = -6. I worked out the vector DB to get a direction vector of l2 and then took the dot product with the normal to the plane, equating this to zero as the normal to the plane is also normal the l2. This gives an equation in terms of a and b. I then also did pythagoras and eliminated a to calculate b.