September 2002:
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09/02/02) A cubic function has only one critical point at x = 2, at which its first and second derivative are zero. If the functional value at this point is 17 and the third derivative is simply 4, find the cubic function.
We can solve this problem with the usual approach of this website! We use y = ax3+bx2+cx+d. Now, we apply our information to this. Taking the third derivative, we find out that 6a = 4. Therefore, a is 2/3. Now, using this we take the first derivative and that would yield to 2x2+2bx+c = 0 when x = 2. The second derivative reveals that 8+2b = 0 (substituting 2 for x). Combining these, we obtain the values of 2/3,-4 and 8 for a,b and c respectively. Now, we implement the functional value to get d = 35/3. Relatively easy, wasn't it?
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09/09/02) Using Newton's Method up to three decimal digits of accuracy, determine the x for which a rectangle would have the largest area, when inscribed between the x-axis and y = cos x from x = -pi/2 to x = pi/2.
First, we have to determine what the base and height of the desired rectangle are in terms of x. The base would be 2x, and the height would be y or cos x. The area, thus, would be 2x * cos x. For this area to be maximized, we have to take its derivative and set it equal to zero. Doing so would yield to 2cos x - 2x * sin x = 0. Therefore, cos x = x * sin x. In other words, 1/x = tan x. Using Newton's Method requires us to have a function equal to zero. So, we change our expression to x * tan x -1 = 0. Now, we apply the method. N(x), for Newton, is equal to x - y(x)/y'(x). For an initial point, we can choose 1, since it seems reasonable. Applying the method 4 times assures us of the first 3 decimals, so x would be 0.860.
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09/16/02) Explain why the second derivative of the integral of -e1/x-2 dx from 0 to t would never equal 1/t2.
We know that the first derivative of the integral would just be -e1/t-2, by the first fundamental theorem of calculus. The second derivative would then be e1/t/t2, by applying chain rule. When equating this and 1/t2, we get that e1/t equals 1. This means that 1/t is equal to ln 1, which is 0. 1/t as we know can never equal zero, but approaches zero as t goes to infinity.
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09/23/02) A window consists of an elliptical shape inscribed in a rectangle. When located in Cartesian Coordinates in a way that its center lies on the origin, the elliptical shape forms the shape of y = k-x2 in the first and second quadrants, and y = x2-k in the third and fourth quadrants. The part outside the elliptical shape, but inside the rectangle is 32/3 m2 in area. Find the window's diagonal in meters.
By using the formulas, we can infer that the height of the window is 2k, and its width is 2*sqrt(k). Therefore, the rectangle's area is 4k*sqrt(k). The area of the oval shape would be 2 times the integral of k - x2 from -sqrt(k) to sqrt(k), which would be 8k*sqrt(k)/3. The part outside the oval shape, hence, would be 4k*sqrt(k)/3 and also equal to 32/3. This yields to a value of 4 meters for k. When the height of the rectangle is 8 and its width is 4, its diagonal would be 4*sqrt(5).
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09/30/02) Use two ways to calculate the limit of (xn-1)/(x-1) as x approaches 1.
One way, which is the more straightforward one is L'Hôpital's Rule. Using this rule, the limit would be equal to the value of limit of n*xn-1/1 as x gets close to 1. This is simply n*1 or n. The other way is to use factoring. The expression xn-1 is equal to (x-1)(xn-1+...+1) in which the longer expression in the parentheses has n terms. These n terms all have a value of 1, therefore, the limit would be n sets of 1 or n*1 = n.
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