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sin (pi/3 + x) = sin(pi/3)cos(x) + cos(pi/3) sin(x) If x is small, sin(x) approx = x, cos(x) approx = 1, so sin(pi/3 + x ) approx = sqrt(3)/2 + x/2. Its useful to remember that if x is small, sin(x) approx = tan (x) approx = x. (The tangent line to both these curves at x=0 is y=x.)

Which rule you use depends on what information you have been given. For example, if you have been given the length of three sides of a triangle and are trying to find the size of an angle, you would use the cosine rule - in this case the sine rule wont help you. In some cases, you can use...

ADFA = Australian defence forces academy. It's in Canberra, next to Duntroon. You graduate from there with a UNSW degree. ROSO = return of service obligation (basically the military put you through uni and pay your HECS, so you have to work for them for some time afterwards.)

Of course, if they specify a particular method then you must use it. But the OP's question didn't specify a particular technique, which is why I suggested using shells.

Have you learnt how to calculate volumes by shells? http://www.mathlearning.net/learningtools/Volumes/unroller.html Sometimes it is easier than using slices.

Upon further reflection, this problem with the trig substitution might be one reason why it is preferable to use a hyperbolic trig substitution. I just asked maple to solve this question, and it gave the answer, 1/2*x*(4+x^2)^(1/2)+2*arcsinh(1/2*x) so clearly maple prefers to use the...

That is an interesting question which I am still considering the answer to! The integral is certainly defined, because it is simply the upper branch of the hyperbola y^2 - x^2 =4. Since it is a continuous curve, I can't see any reason why the area underneath would not be well defined. It is...

e^ipi/2 + e^-ipi/2 = i-i = 0 You can't divide by zero, even in the complex plane.

You might want to reconsider what you just did there. Tan(z) most certainly isn't analytic at z=pi/2.

The acceleration is not a constant, its a function of x. (inverse square of the distance). So a = k/x^2, but when x=r, a = -g (since the gravitational acceleration is down). Hence k = -gr^2

It depends what you want. AP calculus seems to cover most of the 4 unit calculus material, but probably from a more theoretical viewpoint than you usually see at school. I've just been playing with the applet that illustrates various approaches to numerical integration...

I think I've posted before about MIT OCW for uni level maths. Apparently they have now got a website for high school students. You might find some of the stuff about AP calculus helpful. http://ocw.mit.edu/OcwWeb/hs/calculus/calculus/index.htm

I did an assignment on dyscalculia in special ed. I got the impression that it is probably a whole pile of different problems that get lumped into together because they all result in problems with learning maths. For example, some dyslexics have dyscalculia, but then there are also dyslexics...

is dyspraxia different from dyscalculia?

Re: integrate this slut Or you can try the substitution sqrt(x^2 +1) = sin(u)

Re: integrate this slut Let u = x; v' = x/sqrt(x^2 +1) EDIT: you will still need to make a trig substitution to evaluate the second integral.

The second circle will be tangent to both the first circle and the right hand side of the rectangle (somewhere). Let r be the radius of the second circle Drop perpendiculars from the centres of the two circles to the base of the rectangle. Let x be the distance between these two...

OK, we can add up the straight part of the guards path (when he is walking parrallel to the sides of the quadrilateral) and we get 18+18+40+45. Now we only have to figure out what happens at the corners R, P and Q. If the guard is to stay exactly 2 metres away from the corners, he must walk in...

You don't need to find any angles other than what's given.

Now, how about the other question?