course mth 158 äÛÈÿé¯DýÆËè”Lüëæ«™’¸‰‹šü€Á÷«assignment #027 027. `query 27 College Algebra 07-26-2008
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18:46:03 query 3.6.2 / 10. P = (x, y) on y = x^2 - 8. Give your expression for the distance d from P to (0, -1)
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RESPONSE --> d= sqrt x^2 = y ^2 d (x) = sqrt x^2 + (x^2 -8)^2
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18:53:15 ** P = (x, y) is of the form (x, x^2 - 8). So the distance from P to (0, -1) is sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) = sqrt(x^2 + (-7-x^2)^2) = sqrt( x^2 + 49 - 14 x^2 + x^4) = sqrt( x^4 - 13 x^2 + 49). **
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RESPONSE --> so sfter using the distance formula sqrt x^2 + y^2 I should have subsituted both the y coordinate and the formula for the Y^2 and then solve.
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18:56:58 What are the values of d for x=0 and x = -1?
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RESPONSE --> d(0) =sqrt of 0^4 -13 (0) + 49 d(1) = sqrt of (1)^4 - 13(1)^2 + 49
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19:00:06 ** If x = 0 we have sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7. If x = -1 we have sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8. Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. **
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RESPONSE --> After substituting the x values you solve the equation.
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19:07:28 query 3.6.9 / 18 (was and remains 3.6.18). Circle inscribed in square. What is the expression for area A as a function of the radius r of the circle?
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RESPONSE --> A(r) = (2r) (r) = 2r^2 P(r) = 2(2r) + 2 r
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19:09:23 ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **
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RESPONSE --> so for the area you use the regular formula for the area of a circle pi r^2
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19:15:16 What is the expression for perimeter p as a function of the radius r of the circle?
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RESPONSE --> P (r) = 2 pi r
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19:16:51 ** The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **
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RESPONSE --> I thought the question asks for the perimeter of a circle. So when using this I need to use the perimeter of the square 4 * 2r= 8 instead of the circle as the function of r.
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19:24:52 query 3.6.19 / 27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph Give your expression for the distance d between the cars as a function of time.
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RESPONSE --> distance 1 = 2-30t distance 2 = 3-30t d(t) - sqrt (2-30t) ^2 + (3- 40 t)^2
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19:28:48 ** At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t. The position function of the other is 3 + 40 t. If these are the x and the y coordinates of the position then the distance between the cars is distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **
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RESPONSE --> I am not sure how you get the 120t + 900t^2 and the 240 t + 1600t^2. Could you explain so I can see this more clearly.
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(2 + 30 t)^2 = 4 + 120 t + 900 t^2. (3 + 40t)^2 = 9 + 240 t + 1600 t^2 So (2 + 30 t)^2 + (3 + 40t)^2 = 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2, which simplified is 13 + 360 t + 1500 t^2.