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Unformatted text preview: Math 20 Midterm 1 guideline Winter 2010 0.1 Sample midterm solutions Please email me if you spot an error in these solutions. Keep checking back at the website for new postings with error corrections. 1. Solve the differential equation, i.e. find the unknown function. Report the most general solution, if appropriate. (a) f ( x ) = 8 x 3 + 12 x + 3, f (1) = 6. Solution: We find an antiderivative using the ‘power rule’. Here: f ( x ) = 8 x 4 4 + 12 x 2 2 + 3 x + C, or f ( x ) = 2 x 4 + 6 x 2 + 3 x + C is the most general antiderivative. In this problem we must resolve C , as f (1) = 6 so 6 = f (1) = 2 + 6 + 3 + C, and so C = 5. We conclude f ( x ) = 2 x 4 + 6 x 2 + 3 x 5 . (b) f 00 ( x ) = 6 x + sin x . Solution: We take two antiderivatives f ( x ) = 6 x 2 2 cos x + c 1 , and then f ( x ) = 6 x 3 2 · 3 sin x + c 1 x + c 2 . As there are no other conditions, we’re done. (c) f ( x ) = x 4 +3 √ x x 2 . Solution: The right hand side looks quite complicated and it is not easy to guess an antiderivative. So we simplify first: x 4 + 3 √ x x 2 = x 4 + 3 x 1 / 2 x 2 = x 2 + 3 x 3 / 2 . Now, f ( x ) = x 3 3 + 3 x 1 / 2 1 / 2 + C = x 3 3 6 √ x + C. 2. Suppose a stone is thrown straight upward with an initial velocity of 10 ft/sec from a height of 4 feet. Determine the h ( t ) be the height of the stone above the ground at any time. Determine the time when the stone is at maximum height. (hint: you may use the fact that the the acceleration on the stone is a constant 32 ft/secmaximum height....
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This note was uploaded on 10/03/2011 for the course MATH 20 taught by Professor Moore during the Winter '02 term at Stanford.
 Winter '02
 Moore
 Math, Calculus

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