Assignment 27

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course Mth 158

7/18/12 at 11:04 a.m.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

027. `* 27

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Question: * 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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Your solution: Answer 1: You plug in (0 - x)^2 + (-1 - (x^2 - 8)^2 and get d(x) = sqrt x^4 -13x^2 + 49. Answer 2: If you plug in 0 for x, you get sqrt 0^4 - 13(0)^2 + 49 = sqrt of 49, which is 7. Answer 3: If you plug in -1 for x, you get sqrt (-1)^4 - 13(-1)^2 + 49 which = approximately 6.08.

confidence rating #$&*: 3

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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?

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Your solution: See above for my shown work. When x = 0, your answer is 7. When x = -1, your answer is approximately 6.08.

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) + 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. **

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Self-critique (if necessary):

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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?

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Your solution: A(r) =( 4r - pi)^2. Perimeter is p(r)= 8r.

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. **

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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.

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Your solution: Car 1: 2 miles south with speed of 30 mph = 2 + 30t. Car 2: 3 miles east with speed of 40 mph = 3 + 40t. Distance = sqrt(x^2 + y^2). You plug in: sqrt (2 + 30t)^2 + (3 + 40t)^2 = sqrt(4 + 120 t + 900 + 9 +240t + 1600t) = sqrt(2500t^2 + 360t + 13)

confidence rating #$&*: 2

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Given Solution:

* * ** 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). **

qa college algebra part 2

030. * 30

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Self-critique (if necessary): It took me a little while to figure out 360t.

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