The Clearwater Chemical Company is developing a product that uses enzymes to safely break down the foamy pollution from certain chemical spills. In a test, the company used the new product on foam that initially had a height of 100millimeters. After one hour, the foam had shrunk to a height of 80millimeters.
Let t represent the number of hours since the start of the test. Assuming the foam's height decreased exponentially, which expression best models the foam's height over time?
If the foam's height continued to decrease exponentially, what was the height 5hours after the start of the test?
The foam's height can be modeled using exponential decay. The general form of an exponential decay function is given by:
y = a * e^(kt)
where y is the height of the foam at time t, a is the initial height of the foam, k is the decay constant, and e is Euler's number (approximately 2.71828).
In this case, the initial height of the foam was 100 millimeters and the height after 1 hour was 80 millimeters. We can substitute these values into the exponential decay equation to solve for the decay constant k:
80 = 100 * e^(k*1)
Dividing both sides by 100:
0.8 = e^k
Taking the natural logarithm (ln) of both sides:
ln(0.8) = ln(e^k)
Using the property of logarithms that ln(e^k) = k:
ln(0.8) = k
Now we can substitute this value of k back into the exponential decay equation to find the height of the foam after 5 hours:
y = 100 * e^(ln(0.8) * 5)
Simplifying:
y = 100 * 0.8^5
Calculating:
y ≈ 32
Therefore, the height of the foam after 5 hours would be approximately 32 millimeters.