Assignment 27 36

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course Mth 158

3/31/13 Around 6:00 PM

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:

 d(x) = sqrt(x^4 - 13x^2 + 49)

 

 

confidence rating #$&*: 2

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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:

 d(0) = sqrt(0^4 - 13(0)^2 + 49) = sqrt(49) = 7

 

 d(-1) = sqrt((-1)^4 - 13(-1)^2 + 49) = sqrt(1 - 13 + 49) = sqrt(37) roughly = 6.08

 

 

confidence rating #$&*: 2

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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):

 I don't think the answer to d(-1) is sqrt(63). The part in the given solution: sqrt((-1)^4 - 13 * (-1) + 49), should read: sqrt((-1)^4 - 13 * (1) + 49), I think. -1 squared is positive 1 not negative.

 

 

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Self-critique Rating: 2

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You are correct. Good eye, and thanks for pointing this out.

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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 = 2r * 2r

 A(r) = 4r^2

 

 P = 2r + 2r + 2r + 2r

 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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Self-critique (if necessary): OK

 

 

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Self-critique Rating: OK

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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:

 I am at a loss for this problem, I don't know where to start. I am mostly confused on how to structure the equation, if I knew that, I might be able figure it out from there.

 

 

confidence rating #$&*: 1

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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):

 I did not know what to do with the information I was given. I see how, in the given solution, the answer was gotten, but just am not sure how I would know to do some of that. I have no idea how 2 + 30t and 3 + 40t give the ""position function"". I understand that it is the distance from the intersection plus their speed multiplied by time, but I don't know how I would know to format it that way. Also, if I am correct, you are using the intersection as the origin which gives us:

 Distance formula: sqrt(x - 0)^2 + (y - 0)^2) = sqrt(x^2 + y^2)

 And then you are replacing x and y with the two position functions to express the distance between the two as a function of time. If I knew why the 2 + 30t and 3 + 40t were formatted how they are, I think I would've been able to get it.

 

 

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Self-critique Rating: 1

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You would probably begin by sketching a picture of the positions of the two cars, beginning with the point 2 miles south and 3 miles east of the intersection. You would want to include the north-south line, along which the first car moves, and the east-west line, along which the second car moves. At the first point a line from the first car to the second would the hypotenuse of a right triangle with the two lines. The legs of the triangle would be 2 miles and 3 miles, so the distance would be sqrt(2^2 + 3^2) miles = sqrt(13) miles.

This would give you a start.

Then you might sketch additional positions of the cars. 6-minute intervals are convenient, since the car moves 3 miles south and 4 miles east every 6 minutes. After 6 minutes the first car will be 5 miles south of the intersection, the second 7 miles east of the intersection, and their separation will be sqrt(5^2 + 7^2) miles = sqrt(74) miles. After 12 minutes the distance will be sqrt(8^2 + 11^2) miles = sqrt(175) miles, as you can check.

Now to get the separation as a function of time, we would need to express the distance of each car from the intersection as a function of time. If t is the time in hours, and the cars are at the 2 mile and 3 mile positions at t = 0, then when the clock time is t the first car will have traveled 30 t miles and the second will have traveled 40 t miles, putting them as positions (2 + 30 t) miles south of the intersection and (3 + 40 t) miles east of the intersection. They will at this time be sqrt( (2 + 30 t) ^ 2 + (3 + 40 t) ^ 2 ) miles apart.

I've rewritten the solution to the problem in the qa document as follows:

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).

Let me know if this helps.

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Self-critique (if necessary):

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&#Good responses. See my notes and let me know if you have questions. &#