#$&* course Mth 158 4/16 8pm 027. `* 27
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Given Solution: * * ** P = (x, y) is of the form (x, x^2 - 8). So the distance from P to (0, -1) is sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) = sqrt(x^2 + (-7-x^2)^2) = sqrt( x^2 + 49 - 14 x^2 + x^4) = sqrt( x^4 - 13 x^2 + 49). ** What are the values of d for x=0 and x = -1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x = 0 we have sqrt( x^4 - 13 x^2 + 49) sqrt(0^4 - 13 * 0 + 49) sqrt(49) = 7 If x = -1 we have sqrt( x^4 - 13 x^2 + 49) sqrt((-1)^4 - 13 * (-1)^2 + 49) sqrt( 63) = 7.94 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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)^2 + 49) = sqrt( 63). sqrt(64) = 8, so sqrt(63) is a little less than 8 (turns out that sqrt(63) is about 7.94). 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 verify that these distances make sense. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok Self-critique Rating:3 ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The circle has radius r The square has sides the length of 2r and an area of (2r)^2 = 4r^2 The area of the circle is pi*r^2 The area of the square not covered by the circls is 4r^2 - pi*r^2 = (4 - pi)r^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. ** What is the expression for perimeter p as a function of the radius r of the circle? The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: * 3.6.19 / 27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph Give your expression for the distance d between the cars as a function of time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Time is t At t = 0 the first car is 2mi south of the origin, and its distance from the origin is increasing at 30mph At time t it will have traveled the distance 30t from the 2 mile marker. It’s position will be at 2 + 30t from the origin along the negative y axis At t = 0 the second car is 3mi east, or to the right, of the origin and its distance from the origin is increasing at 30mph At time t it will have traveled the distance 30t from the 3 mile marker. It’s position will be 3 + 40t east of the origin, along the positive x axis The distance between the two cars is the hypotenuse of a triangle A^2 +b^2 = c^2 so distance = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) sqrt( 2500 t^2 + 360 t + 13) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** Assume a coordinate system with the y axis pointing north and south, the x axis east and west. At time t= 0 the position of one car is 2 miles south of the origin, and its distance from the origin is increasing at 30 mph. So at time t it will have traveled distance 30 t from the 2-mile position. Its position will therefore be at distance 2 + 30 t from the origin along the negative y axis. The position of the other is found by similar reasoning to be 3 + 40 t east of the origin, putting it at distance 3 + 40 t along the positive x axis. At clock time t, then, a straight line from the position of the first car to that of the second will therefore form a right triangle with the x and y axes. The legs of the triangle will be 2 + 30 t and 3 + 40 t. The distance between the cars is the hypotenuse of this triangle, so distance = 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!