| First, let's consider the line on a "normal", linear graph. Instead
of x and y, let's call the axes a and b. The equation describing
a line would be b = ma+n.
The relationship between a and x and b and y is simple: x = 10^a and y
= 10^b. Changed to the new graph, (.5, 2*10^7) becomes (-log 2,
7 + log 2) and (25, 10^3) becomes (2 - 2 log 2, 3).
This gives two linear equations, 7 + log 2 = m(-log 2) + n and
3 = m(2 - 2 log 2) + n. Solving these gives m = -2.532 (all decimals
are approximate) and n = 6.539.
Now we can say that y = 10^6.539 * x^-2.532.
To find the parts per cubic meter of parts greater than a given size
(s), integrate y from the given size to the upper limit of sizes. Are
the largest particles really only 25, or does the size go up
indefinitely with only the graph stopping at 25?
In the first case, the number of particles per cubic meter is:
(25^-1.532 - s^-1.532) * (10^6.539/-2.532).
In the second case, the number of particles per cubic meter is:
(- s^-1.532) * (10^6.539/-2.532).
-- edp
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| First, the equation of the line is:
log(y) = a*log(x) + b
Plugging in the two points and solving for a and b:
7.3 = a(-.3) + b --\ a = -2.5
3.0 = a(1.4) + b --/ b = 6.5
So,
log(y) = -2.5 log(x) + 6.5
or
y = 10 ^(-2.5 log(x) + 6.5)
= 10^(6.5) * 10^(-2.5 log(x)) = 10^(6.5) * (10^log(x))^(-2.5)
= 3.2*10^6 * x^(-2.5)
Now, I believe what's wanted is the 'area under the curve'. So, the density
of particles greater than size s should be given by the definite integral:
infinity
-
/ (3.2*10^6 * x^(-2.5)) dx
-
x=s
Here we note that the indefinite integral of x^k dx is x^(k+1)/(k+1),
when k is not equal to -1.
So our definite integral evaluates to:
(3.2*10^6 * infinity^(-1.4)/(-1.4)) - (3.2*10^6 * s^(-1.4)/(-1.4)))
or
(0) - (-2.3*10^6 * s^(-1.4)) = 2.3*10^6 * s^(-1.4)
Does this sound about right?
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