December 2002:
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12/02/02) Find the equation of the tangent line of x3-3x2+1 at a positive coordinate where the function's normal is parallel with the line that connects the origin to the graph's local minimum.
Is it too much information? Let's digest it gradually. First we should find out the equation of the line that's mentioned in the problem. The function's local minimum has a derivative of 0 and a positive second derivative. This is at (2,-3), and the line connecting it to the origin has an equation of y = (-3/2)x. Now, if the graph's normal is parallel to this at a point, its tangent should be perpendicular to this. So the slope of the tangent line at that point should be 2/3. The function's derivative, which is 3x2 - 6x, is thus equal to 2/3. This yield to two answers, but we only want the positive one, which is 1 + sqrt(11)/3. Now we have to find the equation of the tangent line. The x-coordinate is 1 + sqrt(11)/3 and slope is 2/3. All we need is its y-coordinate which is -16*sqrt(11)/27 -1. The equation would be :
y + 16*sqrt(11)/27 +1 = (6x - 2*sqrt(11) -6)/9
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12/09/02) Design a general formula for the volume of a parallelogram with a vertex at the origin being revolved about the y-axis.
The picture to the right illustrates one possiblity of the problem. To find the volume of this parallelogram revolved about the y-axis, we need to assign the two sloped lines equations in terms of x. One would be x = ay/h and the other x = ay/h + b. The volume would then be pi times the integral of (ay/h + b)2 - (ay/h)2 using the washer formula. Doing the calculus, we get pi*bh(b+a)
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12/16/02) A man of height 6 ft is 5 ft away from a street light of height 10 ft, when he starts to walk with an initial velocity of 3 ft/s and acceleration of 2 ft/s2. Find the rate of change of his shadow 10 seconds after he starts walking.
Bonus: Put these in order from largest to lowest as he continues to walk: his distance from the light, the rate of change of this distance, the length of his shadow, and the rate of change of the shadow.
Let the distance of the man from the streetlight be d and the length of his shadow be L. Using similar triangles we find that 6/10 = L/(L+d), or more simplified, 3d = 2L. This also yields to 3dd/dt = 2dL/dt. The information about the man's movement shows that d = 5 + 3t + (1/2)*2*t2 = 5 + 3t +t2. Hence, dd/dt would be 3 + 2t. Now we substitute dd/dt in the relationship found earlier and obtain dL/dt = 3t + 9/2. Thus, at t = 10, the rate of change of the shadow's length equals 3(10) + 9/2 or 34.5. The bonus can simply be shown by graphing d, L, dd/dt and dL/dt, as illustrated in the left.
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12/23/02) Lissajous figures are well-known waves in physics, forming as results of parametric equations in terms of time. Assuming that y = 5sin t and x = 5cos t, show that the Cartesian graph's second derivative never equals zero.
First, we find dy/dx or the first derivative. This is equal to dy/dt/dx/dt or 5cos t/ -5sin t which is equal to -cot t. The second derivative is equal to d(dy/dx)/dt / dx/dt, which equals csc2 t / -5sin t or -1/5sin3 t. For no value of t does this equal zero.
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12/30/02) Find the total area between the graph of |2003 x e-x2| and the x-axis.
This is equal to the integral of |2003 x e-x2| from negative infinity to infinity. Since the function is even, it is symmetric, and we can find the area from 0 to infinity and double it. Then, we don't have to include the absolute value as well. Using the techniques of improper integrals, this would be equal to 2(2003/2) or 2003.
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