A certain radioactive material follows the law of exponential change and has a half life of 38 hours. Find how long it takes for 90% of the radioactivity to be dissipated.
Solution:
Use the formula:
S=Ce^{-kt}
First, find the constant of proportionality. In the problem, after 38 hours, half of the radioactivity has been dissipated and a half has been retained. So we can assume thatS = 0.5So when t = 38 hrs and C = So.
Now we can solve for the time(t). According to the problem, 90% of the radioactivity is dissipated, so 10% is retained. So we can assume that S = 0.1So and change C = So.
A tank contains 400 liters of brine. Twelve liters of brine, each containing 2.5 N of dissolved salt, enter the tank per minute, and the mixture, assumed uniform leaves at the rate of 8 liters per min. If the concentration is to be 2 N/litre at the end of one hour, how many newtons of salt should there be present in the tank originally?
Using the given working equation, solve for the value of S @ t=0
S\left(100+t\right)^2=10\left(100+t\right)^{^3}-8192000; t=0\\S\left(100+0\right)^2=10\left(100+0\right)^{^3}-8192000\\\frac{S\left(1000\right)^2}{1000}=\frac{1808000}{1000}\\S=180.8 N
A bacterial population follows the law of exponential growth. If between noon and 2 p.m. the population triples, at what time should the population become 100 times what it was at noon? At 10 a.m. what percentage was present?
SOLUTION:
First, we denote
P as the population of bacteria at anytime
Po as the original bacterial population
t = 0 (12 noon)
t = 2 (2 p.m.)
Let us determine the given and the required
GIVEN:
@12nn to 2p.m.; P= 3Po
REQUIRED:
what time should the population become 100 times
at noon
percentage at 10 a.m.
Using the formula of Applications of Ordinary First-Ordered Differential Equations under Exponential Growth or Decay
Find the equation of the curve so drawn that every point on it is equidistant from the origin and the intersection of the x-axis with the normal to the curve at the point.
Solution:
Plot points on the curve,
A(x_{1},y_{1})
We all know that a Slope of a Tangent corresponds to m, and its negative reciprocal is equal to the Slope of a Normal. Thus, we use Point-Slope Formula.
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