Consider the region in Quadrant 1 totally bounded by the 4 lines: x = 3, x = 9, y = 0, and y = mx (where m is positive). Determine the value of c such that the vertical line x = c bisects the area of that totally bounded region.
Needless to say, your first task should be to draw a representative diagram for the problem that has been described. Also, after determining the value of c, be sure to comment on the (probably) surprising non-contributing factor in this problem.
Bonus: What if the problem were modified such that a horizontal line, y = k was to be the area bisector of the totally bounded region. Determine what the value of k would be in that case.
To solve the problem, let's first draw a representative diagram that depicts the region described.
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m | \
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___________|______\_________
0 | c | 9
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In this diagram, we have the following lines:
- x = 3: This is a vertical line passing through x = 3.
- x = 9: This is a vertical line passing through x = 9.
- y = 0: This is the x-axis.
- y = mx: This is a line with a positive slope (m) passing through the origin.
Now, to determine the value of c such that the vertical line x = c bisects the area of the region, we need to find the point where the area on each side of x = c is equal.
To do this, we need to find the equation of the line y = mx. Since it passes through the origin, the equation will be y = mx.
Next, we can find the points where the line y = mx intersects the two vertical lines x = 3 and x = 9. By substituting the x-values into the equation y = mx, we can find the corresponding y-values:
For x = 3:
y = m(3) = 3m
For x = 9:
y = m(9) = 9m
Now, let's find the equation of the line segment that connects these two intersection points. The equation of this line segment will be the same as the equation of the line y = mx between the values of x = 3 and x = 9.
The equation of this line segment is given by:
y = mx
To find the area of the region bounded by the lines x = 3, x = 9, y = 0, and y = mx, we need to find the integral of y with respect to x between the limits of x = 3 and x = 9.
The area of the region is given by:
A = ∫[3, 9] y dx
To bisect the area, we need to find the value of c such that the integral of y with respect to x between the limits of x = 3 and x = c is equal to the integral of y with respect to x between the limits of x = c and x = 9.
Mathematically, this can be represented as follows:
∫[3, c] y dx = ∫[c, 9] y dx
Now, solving the integral equations would give us the expression for c. However, there is a surprising non-contributing factor in this problem.
Since the region is symmetric with respect to the line x = 6 (midway between 3 and 9), the line x = 6 will automatically bisect the area of the region. This means that the value of c is 6, regardless of the value of m.
Bonus:
If a horizontal line, y = k, is to be the area bisector of the totally bounded region, the value of k would be the y-coordinate of the centroid of the region. The centroid of a region is given by the formula:
(x̄, ȳ) = (1/A) ∫∫ (x, y) dA
where (x̄, ȳ) represents the coordinates of the centroid and A represents the area of the region.
In this case, A is the area of the region bounded by the lines x = 3, x = 9, y = 0, and y = mx.
By using the centroid formula, we can find the value of k (the y-coordinate of the centroid), which would be the value that divides the area of the region into equal halves along the line y = k.