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centripetal acceleration v^2/r could be assumed unless specifically asked to derive; if asked to derive you should be ready with a standard textbook derivation, I personally prefer the one that assumes sin@--->@ as @--->0 as it is quick and simple.

Ambiguity is a main concern. Found the following notes: Ambiguous Unambiguous Unambiguous 1/2x............1/(2x).............(1/2)x 2^n-1.........2^(n-1)..........(2^n)-1 e^x^2........e^(x^2).........(e^x)^2 good form :sin^2(2x)

ND : you might have made a small computational error. Anyway well done, you're starting to bloom mathematically, on your way to a great medical career. For interest : f(a)=|...|^2 where a=cos@ f(a)=4(4a^3-c^2-4a+2) with stationary points given by, f'(a)=4(12a^2-2a-4)=0 giving a=2/3 or...

I espied my son's posts at the MQ thread trying to convince ND out of doing Med. As ND is a budding/promising mathematician maybe he should do med. after all.

How come you medics are so good at maths. Next time I have an intractable maths prob. I'll go to my local GP!

I believe, in the context of this problem -considering that since you are in no doubt that it is a minima, yes go ahead and take a punt - I might get shot down on this. But be careful, that one day examiners might come up with a question specifically to catch the -ve and +ve bluffers!

turtle : True it really comes out the same if you convert all trig. in terms of cos@. Thus you might have f(C(@))=|...|^2 where C(@)=cos(@) Thus if you differentiate with respect to @, you can get df(C(@))/d@=(df(C)/dC)*dcos@/d@=-(df(C)/dC)*sin@ (chain rule!) so we are really back to...

underthesun quote: -------------------------------------------------------------------------------- by looking at the graph of parabola, there would be no "maximum" for QT^2 since that the distance can get infinitely big for QT^2 at infinite values of t, hence t = sqrt(2) and t=-sqrt(2) are...

I was wondering why this question is finding it difficult to liftoff. Perhaps it is my notation |z| which refers to mod(z). Since |cos3@ - cos@ +2 + (sin3@ - sin@)i|=sqrt((cos3@ - cos@ +2)^2 + (sin3@ - sin@)^2) it might just be simpler to work with |....|^2. The real challenge to this...

ND : work with |cos3@ - cos@ +2 + (sin3@ - sin@)i|^2, differentiate with respect to @ as you suggested.

Must restate the problem. If z is a unit complex number, ie. |z|=1, find the z that will give |z^3-z+2| its greatest value. This question is from the genre of questions described in Tip 19 of Geha : Maximum and Minimum values of |z|. As Geha has remarked, and I quote, "these questions...

|z|=|x+iy|=sqrt(x^2+y^2) the modulus of z or its distance from the origin.

Following Q has some complex number, some algebra, some trigonometry, some calculus - a good practise question. Maximize |z^3-z+2| when |z|=1.

Quote: by looking at the graph of parabola, there would be no "maximum" for QT^2 since... If I was the marker yes. However, just in case you did try to prove minima by the usual way ie. second derivative, or increasing-decreasing table, I always tell my students (2,3 and 4U) that when you...

Underthesun: you've done well. This question is much a test of algebraic manipulation. Small error, corrected as follows : QT^2 = (2at - (-4a/t-2at))^2 + ((at^2 - (a/t^2)(4+4t^2 + t^4))^2 =(16a^2)(t+1/t)^2+a^2(4/t^2+4)^2 =(16a^2)((t+1/t)^2+(1/t^2+1)^2) =(16a^2)((t^2+1)^2)(1/t^2+1/t^4)...

A question worthy of q7 3U, or a definite Harder 3Unit. The chord AB is normal to the parabola x^2=4ay. Find the point A which minimizes the length of this chord.

ND :Isn't this just a 3u question? freaking_out: hew, i am glad that this type of crappy questions don't appear in exams. Pity the poor question! Yes it does have a retro feel to it, like Ye Olde 4 Unit, circa 1979. However even if it is just the lowly parabola- heavens whatsit doing...

I'll help with 1. As Underthesun derived eqn. of tangent is y = 2x/t + t^2 (1) Let u be another parameter point, its tangent eqn is y = 2x/u + u^2 (2) For the pt. of intersection, solve for x, x= (u+t)ut/2 and sub in (1), to get...

I do not have the Excel book, but could imagine the approach. It is a good short approach, if all that is asked is whether a line touches a curve, and you know the form of the tangent equation eg. xX_1/a^2+-yY_1/b^2=1. However, if the question asks whether the line misses or cuts the curve...

For part b) i), using the discriminant isn't really that tedious, considering it is a q8. write the line in y=mx+c ie. y=((X_0b^2)/(Y_0a^2))x-b^2/Y_0 We want to prove that c^2<(am)^2-b^2 ie. b^4/Y_0^2 < ((X_0b^2)/(Y_0a))^2-b^2 (****) How does one prove an inequality...