Light is absorbed when it passes through a glass window. If r% of light is absorbed by a glass with thickness w , then the percent of light that is absorbed by a piece of glass with a thickness nw of for any natural number n is modeled by the following function:
A(n) = 100 [1 - (1 - r/100)^n], f or 0 _< r _< 100
a. Show that is an increasing function of n if 0 < r < 100.
b. Sketch the graph of A when r = 10.
c. Evaluate limn� oo A(n) and interpret the results.
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v=1-r/100 <1, so it's a fraction
as n increases, z=v^n thus decreases
so, 100[1-z] increases
use your favorite graphing calculator or online grapher to sketch thebgraph
As n gets large, A(n) approaches 100, since z approaches zero.
Hello Steve. Thank you for your help. Can you please explain the steps and in details, so i will know where my problem was. Once again thanks.
To show that A(n) is an increasing function when 0<r<100, we need to show that A(n+1) > A(n) for any n.
Let's start by calculating A(n+1) and A(n):
A(n+1) = 100 [1 - (1 - r/100)^(n+1)]
A(n) = 100 [1 - (1 - r/100)^n]
Now, we need to compare A(n+1) and A(n):
A(n+1) - A(n) = 100 [1 - (1 - r/100)^(n+1)] - 100 [1 - (1 - r/100)^n]
= 100 - 100(1 - r/100)^(n+1) - 100 + 100(1 - r/100)^n
= 100(1 - r/100)^n - 100(1 - r/100)^(n+1)
To prove that A(n+1) > A(n), we should prove that A(n+1) - A(n) > 0.
A(n+1) - A(n) > 0
100(1 - r/100)^n - 100(1 - r/100)^(n+1) > 0
(1 - r/100)^n - (1 - r/100)^(n+1) > 0
To simplify this inequality, let's multiply through by (1 - r/100)^n:
(1 - r/100)^n[(1 - r/100)^n - (1 - r/100)^(n+1)] > 0
[(1 - r/100)^n - (1 - r/100)^(n+1)][(1 - r/100)^n] > 0
Now, we can see that both factors are positive:
(1 - r/100)^n > 0 (since 1 - r/100 is between 0 and 1, raised to any power, it remains positive)
(1 - r/100)^n - (1 - r/100)^(n+1) > 0 (since (1 - r/100)^(n+1) is smaller than (1 - r/100)^n)
Therefore, A(n+1) - A(n) > 0, which means A(n+1) > A(n). Hence, A(n) is an increasing function for 0<r<100.
To sketch the graph of A when r = 10, we can substitute r = 10 into the function A(n) = 100 [1 - (1 - r/100)^n]:
A(n) = 100 [1 - (1 - 10/100)^n]
= 100 [1 - (1 - 0.1)^n]
= 100 [1 - 0.9^n]
Now, we can evaluate A(n) for different values of n to plot the graph. For example, when n = 0, A(0) = 100 [1 - 0.9^0] = 100 [1 - 1] = 0. When n = 1, A(1) = 100 [1 - 0.9^1] = 100 [1 - 0.9] = 10.
As n increases, A(n) will approach 100. The graph will start at 0 when n = 0 and gradually increase towards 100 as n increases.
Finally, to evaluate lim(n->∞) A(n) and interpret the results:
lim(n->∞) A(n) = lim(n->∞) 100 [1 - (1 - r/100)^n]
As n approaches infinity, the term (1 - r/100)^n becomes very close to zero. Therefore, the limit becomes:
lim(n->∞) A(n) = 100 [1 - (1 - r/100)^∞]
= 100 [1 - 0]
= 100
This means that as the thickness of the glass increases infinitely, the percentage of light absorbed approaches 100%. In other words, if the glass becomes infinitely thick, no light will pass through it, and all of it will be absorbed.