April 2002:
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04/29/02) Write a generalized condition or equation for a cubic function whose inflection point is also its extrema. Use your own notations.
Let us assume the cubic is ax3+bx2+cx+d. The derivative would thus be 3ax2+2bx+c and the second derivative would be 6ax+2b. Then, we equate each to zero and solve for x by using quadratic formula for the first one and simple algebra for the second one. Now we know the x has to be the same for both; so beautifully we just need to make sure the term under the square root is zero in our quadratic formula. Therefore, 4b2-12ac = 0, or more simply b2-3ac is equal to zero. This is the condition we need to have in order for the cubic to be as questioned. Yet, it is interesting that the extrema would not be a relative maximum or minimum, but just an inflection point, in which the tangent line has a slope of zero!
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04/22/02)
Find the derivative of y = (x-1)2:
i) with respect to 2x.
ii) with respect to ln(x).
iii) with respect to 2x3.
Seems hard, doesn't it? Now, let's learn a trick. We know that (dy/dx)/(dt/dx) is the same as dy/dt since dx would never equal zero. Therefore, for finding the derivative of any function with respect to any t, we just use this formula. Let us apply it. In part i, we have (d(x-1)2/dx)/d(2x)/dx, which would yield to (2x-2)/2 or just x-1. In part ii, it would be (d((x-1)2/dx)/(d(ln x)/dx) that is (2x-2)/(1/x) or 2(x2-1). Finally using the same process part iii would yield to (2x-2)/6x2 or (1/3x)-(1/3x2)
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04/15/02) AP Calculus Sample Problem From point (3,5) two tangents are drawn to y = x2. Find the slopes of these two lines.
This is not a very hard problem. You have to carefully realize that each point at the parabola has coordinates (h,h2). Now using derivatives, you know that the slope of any tangent at y would be 2x or in this notation 2h. Now we imply the slope formula. (5-h2)/(3-h)=2h. Solving for h, we get two values of 1 and 5. The slopes are 2h, so they are 2 and 10 respectively.
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04/08/02) Part I: Is it legal to say that the integral of (1/x)dx is ln(kx)+C where k and C are constants? Prove your point from both a derivative and integral point of view. Emphasize on the k part!
Part I: Yes, indeed it is. From Mr. derivative's point of view, the derivative of ln(kx)+C is (1/kx)*k which is 1/x. From Mrs. Integral's point of view, ln(kx1)-ln(kx2) is just ln(x1)-ln(x2) according to the rules of logarithm. Therefore, the Fundamental Theorem of Calculus would prove this true.
Part II: Oh, a little trig, ready? The derivative would be 1/(sin x+ cos x) times cos x-sin x, because of chain rule. Now playing around with this we can multiply the numinator and denominator by sin x-cos x. That would yield to (1-2sin x*cos x)/sin2x - cos2x). Using double angle formulas, this would be (1-sin 2x)/(cos 2x).
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04/01/02) Find the integral of csc6(x)-cot6(x)dx. Show all your work.
Let's do some trigonometry, without which it is not possible to solve this problem. The integrand can be rewritten as
(csc2(x))3-(cot2(x))3. Now we use the algebraic expansion for difference of cubes. According to that, this can be rewritten as
(csc2(x)-cot2(x))(csc4(x)+csc2(x)*cot2(x)+cot4(x)). The first term is 1, according to pythagorean theorem. The second term can be simplified even more to
(csc2(x)-cot2(x))2+3csc2(x)*cot2(x). Hence, the whole expression is simplified to 1+3csc2(x)*cot2(x). It is way easier to take the integral of this rather than the first expression. By substituting u for cot(x) we obtain the solution to be x-cot3(x)+C. Wasn't that sweet?
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