Long-Term Investments Exercise based on the following table, which lists interest rates on long-term investments (based on 10-year government bonds) in several countries in 2008. HINT [See Example 4.]
Example 4:
a. The weight of carbon 14 that remains in a sample that originally contained A grams is given by
where t is time in years. Find the half-life, the time it takes half of the carbon 14 in a sample to decay.
b. Repeat part (a) using the following alternative form of the exponential model in part (a):
c. Another radioactive material has a half-life of 7,000 years. Find an exponential decay model in the form
for the amount of undecayed material remaining. (The constant k is called the decay constant.)
d. How long will it take for 99.95% of the substance in a sample of the material in part (c) to decay?
Solution
a. We want to find the value of t for which C(t) = the weight of undecayed carbon 14 left _ half the original weight = 0.5A. Substituting, we get
Dividing both sides by A gives
b. This is similar to part (a): We want to solve the equation
for t. Dividing both sides by A gives
Taking the natural logarithm of both sides gives
as we obtained in part (a).
c. This time we are given the half-life, which we can use to find the exponential model the amount of radioactive material is
Because half of the sample decays in 7,000 years, this sample will decay to 0.5A grams in 7,000 years (t = 7,000). Substituting this information gives
Canceling A and taking natural logarithms (again using Identity 3) gives
ln(0.5) = −7,000k
so the decay constant k is
Therefore, the model is
d. If 99.95% of the substance in a sample has decayed, then the amount of undecayed material left is 0.05% of the original amount, or 0.0005A. We have
If the interest on a long-term U.S. investment is compounded continuously, how long will it take the value of an investment to double? (Give the answer correct to two decimal places.)
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