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MAT 122 Fall 2011
Overview of Calculus
Homework #3 Solutions
Problems
1.6.10. Solve for t using natural logarithms if 10 = 6e0.5t.
Evaluating this numerically, t ≈ 1.022.
The function P = 3.2e0.03t represent exponential growth or decay. What is the initial quantity? What is the growth rate? State if the growth rate is continuous.
Solution: The initial quantity is 3.2, and the (continuous) growth rate is 0.03.
1.6.24. Write the function P = 2e−0.5t in the form P = P0at. Does the function represent
exponential growth or decay?
Solution: The continuous growth rate k is −0.5, so the growth factor a is e−0.5 ≈ 0.607.
Since k < 0, this function represents decay.
1.6.28. Put the function P = 10(1.7)t in the form P = P0ekt.
Solution: In this case, the growth factor a is 1.7, so k = ln(1.7) ≈ 0.531.
MAT 122 Fall 2011
Overview of Calculus
1.6.36. The gross world product is W = 32.4(1.036)t, where W is in trillions of dollars
and t is years since 2001. Find a formula for gross world product using a continuous
growth rate.
Solution: The growth factor is a = 1.036, so the continuous growth rate is k = ln(1.036) ≈0.0354. Hence, the new formula is The half-life of nicotine in the blood is 2 hours. A person absorbs 0.4 mg of nicotine by smoking a cigarette. Fill in the following table with the amount of nicotinein the blood after t hours. Estimate the length of time until the amount of nicotine isreduced to 0.04 mg.
A formula describing the amount of nicotine at time t is N(t) = 0.4(2)−t/2. Then 0.04 =0.4(2)−t/2, so 0.1 = 2−t/2, and −t/2 = log ( 2 0.1). Finally, t = −2 log2 0.1) ≈ 6.64 hours.
1.7.6. Suppose 1000 is invested in an account paying interest at a rate of 5.5% per year.
How much is in the account after 8 years if the interest is compounded
Solution (a): The growth reate is 0.055, so the amount is 1000(1.055)8 ≈ 1534.69.
Solution (b): The continuous growth rate is 0.055, so the amount is 1000e0.055(8) ≈ 1552.71.
1.7.16. The antidepressant fluoxetine (or Prozac) has a half-life of about 3 days. What
percentage of a dose remains in the body after one day? After one week?
Solution: We write a function P(t) that gives the proportion of Prozac remaining aftertime t. Since the half-life is 3 days, P(t) = 2−t/3. Then P(1) = 2−1/3 ≈ 0.794 = 79.4% isthe percentage remaining after 1 day, and P(7) = 2−7/3 ≈ 0.198 = 19.8% is the percentageremaining after 7 days.
MAT 122 Fall 2011
Overview of Calculus
1.7.20. The number of people living with HIV infections increased worldwide approx-
imately exponentially from 2.5 million in 1985 to 37.8 million in 2003.
(a) Give a formula for the number of HIV infections, H, (in millions) as a function of years, t, since 1985. Use the form H = H0ekt. Graph the function.
(b) What was the yearly continuous percent change in the number of HIV infections Solution (a): We have that H0 = 2.5. In 2003, t = 2003 − 1985 = 18, so we have 37.8 =2.5e18k. Then k = ln 37.8 /18 ≈ 0.151, so the formula is The graph of this function is as follows: Solution (b): The continuous percentage change is 15.1% per year.
1.8.2. If f (x) = x2 + 1, find and simplify:
Solution (a): f (t + 1) = (t + 1)2 + 1 = t2 + 2t + 1 + 1 = t2 + 2t + 2.
Solution (b): f (t2 + 1) = (t2 + 1)2 + 1 = t4 + 2t2 + 2.
MAT 122 Fall 2011
Overview of Calculus
Solution (d): 2 f (t) = 2(t2 + 1) = 2t2 + 2.
Solution (e): [ f (t)]2 + 1 = (t2 + 1)2 + 1 = t4 + 2t2 + 2.
1.8.8. Find the following if f (x) = 2x2 and g(x) = x + 3:
Solution (a): f (g(x)) = 2(x + 3)2 = 2x2 + 12x + 18.
Solution (c): f ( f (x)) = 2(2x2)2 = 8x4.
1.8.14. Use the variable u for the inside function to express each of the following as a
composite function:
Solution (a): Set u = 5t2 − 2, so y = u6.
Solution (b): Set u = −0.6t, so P = 12eu.
Solution (c): Set u = q3 + 1, so C = 12 ln u.
1.8.20. Estimate g( f (2)) from the graphs of f and g.
Solution: First, from the graph of f above x = 2, f (2) ≈ 0.4. Then from the graph of gabove x = 0.4, g(0.4) ≈ 1.2.
MAT 122 Fall 2011
Overview of Calculus
1.8.22. Using Table 1.36, create a table of values for f (g(x)) and for g( f (x)).
Solution: We calculate each value: for example, for f (g(−1)), we first look up g(−1) =−2. Then f (g(−1)) = f (−2), so we look that up to find f (−2) = 1. Hence, f (g(−1)) = 1.
1.8.28. The Heaviside step function, H, is graphed in Figure 1.79. Graph the following
functions:

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