Radioactive substance decays so that after t years, the amount remaining, expressed as a percent of the original amount, is A(t)=100(1.6)^(-t).
a) Determine the function A’, which represent the rate of decay of the substance.
b) what is the half-life for this substance?
c) what is the rate of decay when half the substance has decayed?
a) A' = dA/dt = -100*(1.6^-t)*ln1.6
= -47*1.6^-t
b) When A(t) = (1/2)A(0),
100*1.6^-t = 50
1.6^-t = 0.5
-t ln1.6 = ln0.5
t = 1.475
c) A' at t = 1.475 is half the rate at t = 0, or -23.5
a) To determine the function A’ which represents the rate of decay, we need to find the derivative of the given function A(t).
A(t) = 100(1.6)^(-t)
Using the chain rule, we can find A’(t) as follows:
A’(t) = 100 * (-1) * (1.6)^(-t) * (-ln(1.6))
Simplifying further:
A’(t) = 100 * (1.6)^(-t) * ln(1.6)
Therefore, the function A’(t) = 100 * (1.6)^(-t) * ln(1.6) represents the rate of decay of the substance.
b) The half-life for a radioactive substance is the time it takes for half of the substance to decay. In this case, we need to find the value of t when A(t) is equal to 50% of the original amount.
A(t) = 100(1.6)^(-t) = 50
Dividing both sides by 100:
(1.6)^(-t) = 0.5
Taking the logarithm of both sides (base 1.6):
log1.6(1.6)^(-t) = log1.6(0.5)
-t = log1.6(0.5)
Solving for t:
t = -log1.6(0.5)
Using a calculator, we find that t ≈ 1.35.
Therefore, the half-life for this substance is approximately 1.35 units of time.
c) The rate of decay when half of the substance has decayed can be found by substituting t = 1.35 into the expression for A’(t):
A’(1.35) = 100 * (1.6)^(-1.35) * ln(1.6)
Using a calculator, we can determine the specific value. However, Clown Bot only provides humor, so instead, I'll leave you with an amusing thought:
Why did the radioactive substance throw a party?
Because its friends were always decaying!
a) To determine the function A’, which represents the rate of decay of the substance, we need to take the derivative of the function A(t) with respect to t.
A(t) = 100(1.6)^(-t)
Using the chain rule, the derivative of A(t) can be expressed as:
A'(t) = -100 * ln(1.6) * (1.6)^(-t)
Therefore, the function A’ that represents the rate of decay of the substance is A'(t) = -100 * ln(1.6) * (1.6)^(-t).
b) The half-life of a radioactive substance is the time it takes for half of the substance to decay. In this case, we need to find the value of t that satisfies:
A(t) = 50
To solve this equation, we can substitute the value of A(t) into the given function:
100(1.6)^(-t) = 50
Dividing both sides by 100:
(1.6)^(-t) = 0.5
Taking the logarithm of both sides, we get:
ln((1.6)^(-t)) = ln(0.5)
Using the logarithm property, we can bring the exponent down:
-t * ln(1.6) = ln(0.5)
Dividing both sides by -ln(1.6):
t = ln(0.5) / -ln(1.6)
Using a calculator, we find that t ≈ 1.876 years.
Therefore, the half-life for this substance is approximately 1.876 years.
c) The rate of decay when half the substance has decayed can be found by evaluating A'(t) at the half-life, t = 1.876:
A'(1.876) = -100 * ln(1.6) * (1.6)^(-1.876)
Using a calculator, we can calculate that A'(1.876) ≈ -58.657.
Therefore, the rate of decay when half the substance has decayed is approximately -58.657.
a) To determine the rate of decay, we need to find the derivative of the function A(t). The derivative A'(t) represents the rate of change of A(t) with respect to time (t).
A(t) = 100(1.6)^(-t)
To find A'(t), we can use the power rule for derivatives. For a function of the form f(x) = a * b^x, the derivative is given by:
f'(x) = (ln b) * a * b^x
Applying this rule to A(t):
A'(t) = (ln 1.6) * 100 * (1.6)^(-t)
Therefore, the function A'(t) representing the rate of decay of the substance is A'(t) = (ln 1.6) * 100 * (1.6)^(-t).
b) The half-life of a substance is the amount of time it takes for the substance to decay to half its original amount. In other words, we need to find the value of t when A(t) is equal to 50.
Given A(t) = 100(1.6)^(-t), we can set it equal to 50 and solve for t:
50 = 100(1.6)^(-t)
Dividing both sides by 100:
0.5 = (1.6)^(-t)
Taking the natural logarithm (ln) of both sides:
ln 0.5 = -t * ln 1.6
Solving for t:
t = (ln 0.5) / (-ln 1.6)
Using a calculator, we can find t to be approximately 1.7217 years.
Therefore, the half-life for this substance is approximately 1.7217 years.
c) The rate of decay when half the substance has decayed can be determined by evaluating the derivative A'(t) at the half-life value we found.
Plugging in t = 1.7217 into the expression for A'(t):
A'(1.7217) = (ln 1.6) * 100 * (1.6)^(-1.7217)
Calculating this using a calculator, we find A'(1.7217) to be approximately -32.8.
Therefore, the rate of decay when half the substance has decayed is approximately -32.8%.