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
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)
Your solution:
Confidence Assessment:
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?
Your solution:
Confidence Assessment:
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):
Self-critique Rating:
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?
Your solution:
Confidence Assessment:
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):
Self-critique Rating:
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.
Your solution:
Confidence Assessment:
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):
Self-critique Rating: