1 = ln + C
1 = ln + C = 1.693147
Thus, f (x) = ln(sin x) + 1.693147.
At x = 1, we get f (1) = ln(sin1) + 1.693147 = 1.521 (rounded to three decimal places).
- B We solve this differential equation using separation of variables.
First, move the y to the left side and the dt to the right side, which gives us = k dt.
Now, integrate both sides.
ln y = kt + C
Next, it’s traditional to put the equation in terms of y. We do this by exponentiating both sides to
the base e. We get y = ekt + C.
Using the rules of exponents, we can rewrite this as: y = ekt eC. Finally, because eC is a
constant, we can rewrite the equation as: y = Cekt.
Now, we use the initial condition to solve for k. At time t = 60 (seconds), y = .
We are assuming a starting amount of y = 1, which will make C = 1. Actually, we could assume
any starting amount. The half-life tells us that there will be half that amount after 1 minute.
Therefore,
= e^60 k
Solve for k: k = ln.
This gives us k = −0.012 (rounded to three decimal places).
