March 2002:
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03/25/02) A bowl's shape looks like the graph of y=(x8-x6)1/4 rotated about the x-axis from x=1 to x=4. What would the volume of this bowl be? Show how you solve it.
Integral problem right? The volume would be the integral of pi times y(x)2dx from x=1 to x=4. Or simply, integral of the square root of x8-x6 from 1 to 4. Now, the question is how you take the integral of that? By taking out an x6 out of the square root, we obtain x3 times the square root of x2-1. x3 is the same thing as x times x2. Therefore, if we substitute u for x2-1, we have the du in the integral, which is a factor of x. We also have x2 which can be written in terms of u. Now the integral is simply (1/2)*(u+1)*u1/2 from u=0 to u=15. Taking the integral would yield to a volume of 50*root(15) or about 608.37 units cubed. Another good way to solve the integral is trigonometric substitution. Isn't it a big bowl?! Topic:
Applications of Integrals
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03/18/02) Find the result of dividing the 99th derivative of f(x)=x100+1/x by 99!.
Let us look at this problem piece by piece. The 99th derivative of x100 is simply 100!x, since each time the power is being multiplied by the function and decreased by 1. The other piece, 1/x, has also an interesting pattern. Its derivatives alternate between positive and negative and the power of x in the denominator is increased by 1. Therefore, its 99th derivative would be -99!/x100. Now when we divide our whole solution by 99!, we obtain 100x-1/x100 which is indeed the answer to this problem. Take a look at the problem 12/31/01 in the solutions for more explanation.
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03/04/02 Updated For Two Weeks) How does the graph of this equation look like: (y2+(x-3)2-4)2 + (y2+x2-1)2 = 0 ? At what value(s) for x does the derivative equal to zero? Does it have any inflection point? Explain.
A very sweet idea underlines this problem. When the sum of the squares to two statements equals zero, they BOTH have to be zero, or the equation wouldn't work. Hence, y2+x2-4 and y2+x2-1 have to be zero. The two equations lead to two circles tangent at each other at x = 1. However, the equation does NOT work if the two quantities for x and y are not equal in the two graphs. Therefore, the only possible place for the function is at x = 1, or the point(1,0). This graph is lovely; IT IS JUST A POINT! There is no change in the y-value, so there is no derivative and no inflection point. Love that, right?
Topic:
Implicit Equations and Derivatives
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