1322FinalSolutionF17.pdf - MAT1322D Solution to Final...

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Mathematical Applications for the Management, Life, and Social Sciences
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Chapter 13 / Exercise 15
Mathematical Applications for the Management, Life, and Social Sciences
Harshbarger
Expert Verified
MAT1322D Solution to Final Examination December 2017 1 Solution to the Final Examination MAT1322D, Fall 2017 Part I. Multiple-choice Questions (2 12 = 24 marks) FADAFEECBBDD 1. The area of the region above the graph of y= x2and under the graph of y= 8 x2in the first quadrant is Solution. (F) Let x2= 8 x2. 2x2= 8, x= 2. The area is 232222200032(8)2(4)2 433xxxxdxxdxx. 2. Let Rbe the region above the parabola y= x2and under the line y= 2x. Solid Bis obtained by revolving Rabout the line y= 4. Then the volume of Bis calculated by the integral Answer. (A) rin= 4 2x rout = 4 − x2 y= 4 y= x2 y= 2x X Y O
We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Mathematical Applications for the Management, Life, and Social Sciences
The document you are viewing contains questions related to this textbook.
Chapter 13 / Exercise 15
Mathematical Applications for the Management, Life, and Social Sciences
Harshbarger
Expert Verified
MAT1322D Solution to Final Examination December 2017 2 3. Suppose a pool has the shape of an inverted truncated pyramid. The top of the pool at the ground level is a square with side length 20 meters, and the bottom of the pool is a square with side length 10 meters. The depth of the pool is 5 meters. The pool is filled with water with density kg / m3. Let xbe the distance between a horizontal layer of water and the top of the pool. Denote the acceleration of gravity be gm / sec2. Then the work, in Joules, needed to pump the water in the pool to a point 2 meters above the ground is calculated by the integral Solution. (D) A horizontal layer of water in the pool is a square with side-length L(x) = 10 + 2 (5 x) = 20 2x. The volume of this layer with thickness dxis V(x) = (L(x))2= (20 2x)2dx. The weight of this layer of water is w(x) = gV(x) = g(20 2x)2dx. The work needed to pump this layer of water to a point 2 meters above the ground is W(x) = w(x)(x+ 2) = g(20 2x)2(x+ 2)dx. The total work is W= 520(202 ) (2)gxxdx.

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