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Yeah it can.. Find the equation of the tangent to y = x at (1,1) dy/dx = 1 At (1,1) dy/dx = 1 y - 1 = 1(x - 1) y - 1 = x - 1 y = x oh wait a second.. you said curve... so much for me being a smart ass ( i was just kidding anyway)

Just a question: Find the equation of the tangent to the curve y = ex + 1 at the point (1, e + 1) This is my working... y = ex + 1 dy/dx = ex At (1, e + 1) dy/dx = e1 = e y - (e + 1) = e(x - 1) y - e - 1 = ex - e y - 1 = ex y = ex + 1 well the answer says.. y = ex + 1...

Yeah i was thinking what he did was a little weird.. cuz if you were to do x = a(p+q) y = apq x2 = a2(p2 + 2pq + q2) x2 =a2[(p + q)2) x2 = a2(x2/a2) x2 = x2 which obviously gets you no where.... can someone try doing the question please?

um i think pq = -4.. cuz its not the tangents... its OP and OQ which are mOP = ap2[/sup} / 2ap mOQ = aq2[/sup} / 2aq mOP * mOQ = -1 pq/ 4 = -1 pq = -4 ap²+aq² = a(p²+q²) = a[(p+q)²-2pq] = a[(x/a)²+8]. but if you do it like that....... and you put it into x2 = a(a[(x/a)2+8] +...

Two two points P(2ap, ap2) and Q(2aq, aq2) are on the parabola x2 = 4ay. i) The equation of the tangent to x2 = 4ay at an arbitary points (2at, at2) on the parabola is y = tx - at2. Show that the tangents at the points P and Q meet at R, where R is the point (a(p+q),apq). ** I can do...

oh yeah that 50% thing...i was talking bout ext 2. maths.. dunno why i left that out though

The marks i was talking about were HSC marks.. I get confused with all this moderated, aligned and ect. marks. The point was really to see if a 90+ UAI would still be available with good marks of all subjects, except english. I dont know if ill be able to get marks as such as these, probably...

Hi, i do the following subjects: English Std, Maths Ext. 1, Maths Ext. 2, Physics, SDD and ITVET. I was wondering if it would be possible to get a UAI of 90+ (90 is fine) with marks (scaled) such as: English Std: 70 Maths Ext 1.: 90 Maths Ext 2.: 80 Physics: 90 SDD: 90 ITVET: 90...

Thanks anyway though.

A question asks: Find d/dx(x sin2x) and hence find 0∫π/4 x cos2x dx. Heres my working out Let f(x) = x sin 2x f'(x0 = sin2x + 2xcos2x 0∫π/4 (x cos2x) dx = 1/2 0∫π/4 (2x cos 2x) dx = 1/2 0∫π/4 (f'(x) - sin2x) dx = 1/2 [sin 2x + cos2x / 2]sub]0[/sub]π/4 = 1/2 [1 + 0 - 0 - (1/2) =...

Oh ok then thanks alot!

Hmm if you use the y-limits, why would it need to be halved? Like if you see my attached picture when you rotate i see how you get both sides of the y-axis with half a rotation. But the part that is halved is not based on the y-limits it is the x-limits that are halved. The actual x-limits are x...

Hehe wow, thanks for all your help. Glad i didnt get you fed up or anything. So just to confirm, when you rotate about either axis, you always use the x limits?

hey hehe thanks for your attempt. The answer in the back of the book is just pi units3, and also it looks like your method is incorrect. The method i followed didnt work for this question, so i cant advise you on what to use...CUZ I DONT KNOW MYSELF!!

Um the graphs that you drew were wrong. You drew x^2 = y - a sideways parabola. x^2 = y is exactly the same as y = x^2 so it is a standard parabola. If you use the x-values x = 1 and x = -1 you end up getting either 2pi or 4 pi..i cant remember. Im really confused now REALLY confused...

Hehe yeah i kow the points of intersection are x = -1 and x = 1 BUT the y values for the points of intersection are y = 1, y = 1. And for this question its a rotation about the y-axis, so arent you supposed to use the y values?

Hey um i think you misread wrong again.. V=€ˆ20(2 - y - y)dy =€ˆ20(2 - 2y)dy =€[2y - y2]20 =€[4-4] = 0? Thats what i did integrating 2 - 2y you get 2y - 2y2/2 which gives 2y - y2 which is what i did. Anyway the answer isnt 2pi its just pi.

yeah i did that... If you see 2 - y - y x2 = y and x2 = 2-y But what im confused about is i did the volume of one curve rotated about the axis, minus the volume of the other curve rotated about the axis. My book explains that the volume is found like that, but finding the volumes of...

Um well........i didnt say the radius was 2 - y - y... My book explained that the volume of an area between two curves rotated about an axis, is the volume of one curve rotated about one axis, minus the volume of the other curve rotated about the same axis. Thats what i followed. It didnt say...

I dont get what you did.. would you mind explaining?