May 2002:
Home | Problem Archive | Past Winners | Number Games | Contact Me
05/27/02)
A container is formed by rotating the first quadrant region of y = -x3+4x2+5x about the line x = -1. Find the volume of the container in liters, if the Cartesian coordinate unit is in centimeters.
Solution:
Let's start by the graph. The function is y = -x(x-5)(x+1) which intersects the x-axis in the first quadrant at x=5. Therefore the volume can be thought of as shells, with radius 1+x and height of y from x=0 to x=5. Hence, the integrand would be -2pi(1+x)x3+4x2+5x, from x=0 to x=5. Evaluating this would yield to 1125pi/2 or about 1167 cm3 or 1.167 liters. Not a lot of volume, is it?
______________________________________
05/20/02) Find the area of the yellow region in the following picture. Four semicircles are inscribed in a square of side 10.
I am still in wonder how to do this problem without calculus. I would be interested to know how to do this with geometry, but for the purpose of this site, we will do this with calculus. If we set up a coordinate system, a circle of radius 10 and center (-5,-5) intersects both the x- and y-axis and the area of the curve in the first quadrant would be a quarter of the wanted area. Taking the integral of y = sqrt(100-(x+5)2)-5 from x=0 to x=5(sqrt(3-1)). This would lead to an area of 25(pi/3+1-sqrt(3)) and the area asked in the problem would be 100(pi/3+1-sqrt(3)). This is close to 31.5, interesting right?
______________________________________
05/13/02) Two tangents are drawn at the function f(x)=(x-2)(x+3)(x+5) in points x=-4 and x=-1. The area enclosed by the two lines and the graph is divided into two parts by the function itself. What is the ratio of the larger portion to the smaller one?
A lot of calculations in this, right? A combination of ideas. First we find the equations of the lines. One is y = -x+2 at x =-4; the other is y = -10x-34 at x = -1. The intersection of the lines is fortunately at x = -4 itself. Now the integral part. The integral of f(x)-(-10x-34)dx from -4 to -1, which turns out to be 27/4. The larger portion is the integral of -x+2-f(x)dx from -4 to 2, which yields to 108. The ratio is therefore found to be 16. Oh, this problem is not hard, but painstaking, huh?
______________________________________
05/06/02) The largest cylinder that can be inscribed in a sphere is taken out of the sphere. What fraction of the original volume remains?
This problem seems hard, but a drawing would help. The radius of the sphere, R, is constant. In terms of R and h, the height of cylinder, the radius of the cylinder is the square root of R2-h2. The volume of the cylinder would then be pi*h*root(R2-h2). Applying derivatives, the maximum volume would occur when h equals 2R/root3. Now, we compare the volumes together. The volume of the sphere would be (4/3)pi*R3. The volume of the cylinder can be found in terms of R now that we know what h is in terms of R. As it turns out the ratio would be root3 to 1. Now this 1 is taken out of the sphere. So the fraction remaining is (root3-1)/root3. Neat, isn't it?
| Previous Month | Archive | Next Month |
|---|