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Unformatted text preview: MATH 31A (Butler) Practice for Final (C) Try to answer the following questions without the use of book, notes or calculator; but you can use the equation sheet posted on the course website. Time yourself and try to finish the questions in less than three hours. 1. Find the area of the region bounded between the curves g ( t ) = sin( πt ) and f ( t ) = 8 t 2 2 t . (Hint: the best way to find the intersection points is to “guess” some easy values for t .) Since we are not given bounds we first need to figure out the bounds. This reduces to solving for t in the following equation. sin( πt ) = 8 t 2 2 t There is no method to solve this analytically, and since no calculators are allowed then it must be that we can guess solutions. For example t = 0 gives zero on both sides and so that is a point of intersection. The next “nice” value for sin( πt ) would occur at t = 1 / 2 and if we plug that in we see that we get 1 on both sides. It is not hard to see that these are the only two points of intersection (if in doubt do a rough sketch of these two functions). Now let us try to see which function is on top, if we pick t = 1 / 4 (a point in the middle) we see that g (1 / 4) = sin( π/ 4) = √ 2 / 2 while f (1 / 4) = 8(1 / 4) 2 2(1 / 4) = 0, so g ( t ) is the function on the top and f ( t ) is the function on the bottom (this last part is easy to see if you do a rough sketch of the functions). So we have Area = Z 1 / 2 ( g ( t ) f ( t ) ) dt = Z 1 / 2 ( sin( πt ) (8 t 2 2 t ) ) dt = Z 1 / 2 ( sin( πt ) 8 t 2 + 2 t ) dt = 1 π cos( πt ) 8 3 t 3 + t 2 t =1 / 2 t =0 = 1 π cos ( π 2 ) 8 3 ( 1 2 ) 3 + ( 1 2 ) 2 1 π cos(0) 0 + 0 = 1 π 1 12 . 2. (a) For a ≥ 1 2 find the point on the curve y = √ x closest to the point ( a, 0). First we note that the function y = √ x is only defined for x ≥ 0 and so throughout we will assume that x ≥ 0. Let ( x,y ) = ( x, √ x ) be a point on the curve, then the distance between this point and ( a, 0) is q ( x a ) 2 + ( √ x 0) 2 = √ x 2 2 ax + a 2 + x = p x 2 (2 a 1) x + a 2 . Since we are trying to minimize this expression, it is the same as trying to minimize the expression inside the square root, i.e., minimize g ( x ) = x 2 (2 a 1) x + a 2 . We have g ( x ) = 2 x (2 a 1) which is never undefined and is equal to 0 at 2 x = 2 a 1 or x = a 1 2 . (Note that a 1 2 ≥ 0 and so this is still a point on the curve.) By taking the second derivative we have g 00 ( a 1 / 2) = 2 > 0 showing that this is indeed a minimum. Therefore we can conclude that the point on the curve closest to ( a, 0) is at the point 1 a 2 , r 1 a 2 ....
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 Spring '10
 BUTLER
 Math, Derivative, Cos, Continuous function, dx

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