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(1) Al-Kuhi studied the centers of gravity of the following series of shapes. Fix a semicircle ABG with diameter AG, center (of the corresponding circle) D along AG, and BD perpendicular to AG (see figure below). Then form the triangle ABG and the (segment of the) parabola ABG, as pictured. Al-Kuhi considered both these plane figures and the solids obtained by rotating them about the line BD (sweeping out a hemisphere, a cone, and a paraboloid). He announced in a letter to al-Sabi that the centers of gravity of the various planar and solid figures lie along BD at points P dividing BD in the following ratios: (a) Triangle: DP/DB = 1/3. (b) Cone: DP/DB = 1/4. (c) Parabola: DP/DB = 2/5. (d) Paraboloid: DP/DB = 2/6. (e) Semicircle: DP/DB = 3/7. (f) Hemisphere: DP/DB = 3/8. Such calculations are great feats prior to the introduction of calculus, but they are not all correct. Use calculus to derive the correct results for the planar figures (triangle, parabola, semicircle). Here is an outline of one way to proceed: we may assume the circle has radius 1, and that the points A and D are (0, 0) and (1, 0). For any of the figures considered, let f(x) denote the height at x, and let A denote the total area, so that A = ∫ f(x)dx. Calculus tells us that the point P is then (1, p) where p = [∫ (f(x)^2)/2 dx] / A Use this formula to compute p for each of the three shapes. For the semicircle, you will find that your answer does not agree exactly with that of al-Kuhi: what value of ̀π would be implied by the semicircle answer? (Our reading from the Sourcebook, 572-573, is in fact al-Kuhi’s response to a contemporary who questioned the result of his calculation, since it seemed at variance with Archimedes’ approximations to π. See 568-572 of the Sourcebook for more of this exchange.) (2) This week we will study Ibn al-Haytham’s work on the volume of a paraboloid. Consider a parabola and a line perpendicular to its axis that intersects the parabola in points P and Q and thereby determines a parabolic segment. Consider the solid obtained by rotating the parabolic segment about the line. Also consider the cylinder obtained by rotating the rectangle whose top is PQ and whose bottom is obtained by translating PQ so that it is tangent to the parabola at the vertex. (See picture.) Show that the volume of the paraboloid is 8/15 of the volume of the cylinder in the following steps: (a) Write down an equation of the parabola and an equation of the line through P and Q. (b) Compute the volume of the cylinder (in terms of the parameters from the previous part). (c) Set up and compute an integral that computes the volume of the paraboloid, and compare it to your cylinder calculation.
Sri K.
In this question, you will work step-by-step through an optimization problem. Many airlines restrict the size of carry-on items in terms of their combined length, width, and height (L + W + H). I am due to fly on an airline that states 'The sum of the length, width, and height of your item must be at most 70 inches. Suppose I want to take a box (cuboid) on my trip that will allow me to carry on as much as possible, and I also want this box to have a square base. The questions we will answer using optimization are: What dimensions should the box have, and what would be the resulting maximum volume? If the base of the box is square, whose sides have a length x, and the height of the box is h (both measured in inches), enter expressions for the volume of the box, V, and the sum of the length, width, and height of the box in terms of x and h. Since the sum of the length, width, and height must be less than or equal to 70 inches, and it is clearly best to choose this sum equal to 70 inches, write down the constraint equation. Rearrange the constraint equation to give the height of the box, h, in terms of the length of the base, x. What is the objective function for this problem? Using the constraint equation, rewrite the objective function in terms of x alone. Differentiate V with respect to x, to find the derivative dV/dx. Find the values of x for which we have a potential relative extreme point of V. One of these is at x = 0, and the other occurs when x is equal to what value? (Give your answer, and those that follow, correct to two decimal places.) For this question, you need not show that this is actually a relative maximum point. (But you should know how you would do this.) Now that you know the length of the base of the box, what is its height? height : inches What is the maximum volume of the box? volume : cubic inches.
Adi S.
A rancher wants to fence an area of 1,500 square yards in a rectangular field that borders a straight river, and then divide in half with a fence perpendicular to the river. He needs no fence along the river. The dividing fence costs half as much as the surrounding fence. How can he do this so that the cost of the fence is minimized? Follow the steps: (a) Let the length (the side being divided) to be x and the width to be y, and assume that the fence costs $1 per yard. Then the quantity to be minimized is (expressed as a function of both x and y) C= 3/2(x + 2y). (Use fraction for coefficients.) (b) The condition that x and y must satisfy is y= 1500/x. (c) Using the condition to replace y by x in C, C can then be expressed as a function of x: C(x)= (3x^2 + 9000)/2x. (d) The domain of C is ( 0 , infty ). (Use "infty" for ∞.) (e) The only critical number of C in the domain is x= 54.8. (Keep 1 decimal place (rounded)). We use the Second-Derivative Test to classify the critical number as a relative maximum or minimum, or neither: At the critical number x= 54.8, the second derivative C''( 54.8 ) is positive. Therefore at x= 54.8, the function has a relative minimum. (f) Finally, plug x= 54.8 into the condition of x and y we obtain y= 27.4. (Keep 1 decimal place (rounded)). Therefore the length and width of of the rectangular field that will minimize the total cost for materials are x= 54.8 yards and y= 27.4 yards, with the divider equals 27.4 yards.
Shyam P.
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