HW2-solutions - merino(aem2588 – HW2 – chen –(55405 1...

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Unformatted text preview: merino (aem2588) – HW2 – chen – (55405) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A machinist has to manufacture a circular metal disk with area 930 π sq. cms. How close to the exact radius must the machinist control the radius if he is allowed an error tolerance of ± 8 π sq. cms. in the area of disk? 1. within approximately 0 . 1328 cm 2. within approximately 0 . 1288 cm 3. within approximately 0 . 1318 cm 4. within approximately 0 . 1298 cm 5. within approximately 0 . 1308 cm cor- rect Explanation: The area of a circular disk of radius r is πr 2 . When the area is 930 π , therefore, the exact radius, r ext , of the disk is r ext = √ 930 . If, however, the machinist is allowed to make the disk so that its area A is within ± 8 π of 930 π then A must satisfy the inequalities − 8 π ≤ A − 930 π ≤ 8 π . Thus the radius r the machinist makes the disk must satisfy the inequalities − 8 ≤ r 2 − 930 < 8 , in other words, √ 922 < r < √ 938 . Hence the area of the disk that the machinist makes will be within ± 8 π sq. cms. of 930 π sq.cms. when r ext − r < √ 930 − √ 922 ≈ . 1314 , and r − r ext < √ 938 − √ 930 ≈ . 1308 . Consequently, the radius r must be within approximately 0 . 1308 cms of the exact value √ 930. 002 10.0 points Find the value of b , b ≥ 0, for which lim x → braceleftBig √ 5 x + b − 5 x bracerightBig exists. 1. b = 27 2. b = 28 3. b = 26 4. b = 25 correct 5. b = 24 Explanation: We are told that lim x → braceleftBig √ 5 x + b − 5 x bracerightBig = A for some value A , but we aren’t told what the particular value of A is. The question requires us to see exactly what value b must take for A to exist. To begin, note that √ x + y − √ z = x + y − z √ x + y + √ z . merino (aem2588) – HW2 – chen – (55405) 2 Thus √ 5 x + b − 5 x = 5 x + b − 25 x ( √ 5 x + b + 5) = 5 x x ( √ 5 x + b + 5) + b − 25 x ( √ 5 x + b + 5) . Now lim x → √ 5 x + b + 5 = √ b + 5 , so by properties of limits, lim x → 5 x x ( √ 5 x + b + 5) = lim x → 5 √ 5 x + b + 5 = 5 √ b + 5 . Consequently, once again by the properties of limits, lim x → braceleftBig b − 25 x ( √ 5 x + b + 5) bracerightBig = lim x → parenleftBig √ 5 x + b − 5 x − 5 x x ( √ 5 x + b + 5) parenrightBig = A − 5 √ b + 5 . But lim x → braceleftBig b − 25 x ( √ 5 x + b + 5) bracerightBig = ( b − 25) lim x → braceleftBig 1 x ( √ 5 x + b + 5) bracerightBig . Since lim x → 1 x ( √ 5 x + b + 5) doesn’t exist, however, the only way lim x → braceleftBig b − 25 x ( √ 5 x + b + 5) bracerightBig = A − 5 √ b + 5 can hold is if b − 25 = 0 ....
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This note was uploaded on 02/23/2011 for the course MATH 408C taught by Professor Knopf during the Spring '10 term at University of Texas.

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HW2-solutions - merino(aem2588 – HW2 – chen –(55405 1...

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