November 2002:
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11/04/02) No Problem Posted
11/11/02) Find my equation if I have these conditions:
The fact that it is symmetric about the y-axis makes the equation an even one, and thus it does not have a factor of x3 or x. The general form would then be ax4 + bx2 + c. The derivative would be 4ax3 + 2bx. Using the fact that this is equal to zero at the extrema and incorporating the functional value of (1,15/4) the equation is found to be x4/4 - 3x2/2 + 5. Ain't it fun?
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11/18/02) Let's put it all together. Find the limit, as x goes to zero, of the derivative of the integral of sin(k)dk from k=0 to k=x2. Show all the work!
Let's go step by step before panic! First the integral. The integral would simply be -cos(k) from 0 to x2 which equals 1-cos(x2). The derivative of this result would be 2x*sin(x2). Taking this expression's limit as x approaches 0, we simply get 0. It just sounds hard, right?
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11/25/02) A cone of height 10 m and base radius 5 m, is cut so that an upper cone of height 6 m is removed from it. The remainder becomes a container to which water pours in at a speed of 8 m/s. Find the rate of change of its radius at an instant where the level of water is 1 m.
Let us start by finding the container's volume formula. Since a cone has been removed from a big cone, the formula would be
(pi/3)*(10*52 - (1/3)*(10-h)*r2). Another relationship can be deduced from similar triangles that (10-h)/r = 10/5. So 2r is equal to 10-h. This substitution can be put into the equation for volume. Now we take the derivative of the volume formula, and get dV/dt = -2*pi*r2 dr/dt. When h is 1 m, we know r has to be 9/2 m due to the relationship explained above. So, the rate of change of radius, dr/dt, would be -16/(81*pi) or about -0.06288 m/s.
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