Name
MATH 116
PRACTICE MIDTERM 1 Please work out each of the given problems.
Credit will be based on the steps that you show towards the final answer.
Show your work. PROBLEM 1 Find the derivative of the following functions A.
f(x) = ln(ln x)
Solution
u = ln x u' = 1/x f(u) = ln u f '(u) = 1/u We get
1
1
1
Solution
and use the chain and product rules. We have u = ex ln x u' = ex ln x + ex / x f(u) = eu f '(u) = eu so that
Solution
now the derivative is simply f '(x) = 1
PROBLEM 2
Find the following anti-derivatives. A.
Solution
Now we use the power rule for integration to get 2/3 31/2 x3/2 + 1/2 ln|x| + 1/2 x8 + 2x1/2 + C
Solution For this integral, we use u-substitution. We let u = 4 - 3x5 du = -15x4 dx We get
Solution u = x + ex du = 1 + ex We get
PROBLEM 3
During an experiment with a deadly new virus, Tom
drops a flask of the virus on the floor. Tom
and his four lab aids are immediately infected.
During the next six hours they infect an additional forty unsuspecting
people. Assume the rate of
spreading of the virus is proportional to the number of people who have been
infected. A. Write a differential equation that models
this situation. Be sure to label
your variables. Solution We let t = the time in hours after the flask is dropped P(t) = the number of people infected after t hours Then
dP
B. How long will it be until one million
people have been infected? Solution Since this is the standard exponential growth model, the solution to the differential equation is P = P0 ekt Since P(0) = 5, we have P0 = 5 Hence P = 5 ekt Now use the fact that P(6) = 45 to get 45 = 5 e6k 9 = e6k ln 9 = 6k
ln 9 Hence P = 5 e0.3662t We want the time when P = 1,000,000. We write 1,000,000 = 5 e0.3662t 200,000 = e0.3662t ln 200,000 = 0.3662 t
ln 200,000 We can conclude that there will be 1,000,000 people infected in less than 34 hours.
PROBLEM 4 A population of bacteria is growing at the rate of
dP
3000
where t is the time in
days. When t
= 0, the population is 1000.
A. Write an equation that models the population P
in terms of the time t. Solution We need to solve the integral
We let u = 1 + 0.25t du = 0.25 dt We have
We have P(0) = 1000. Hence 1000 = 12,000 ln|1 + 0.25(0)| + C 1000 = 12,000 ln|1| + C = 12,000 (0) + C C = 1000 We can conclude that P(t) = 12,000 ln|1 + 0.25 t| + 1000
B. What is the population after 3
days? Solution We plug in 3 for t to get P(3) = 12,000 ln|1 + (0.25)(3)| + 1000 = 7715 We can conclude that the population will be 7715 after 3 days.
C. After how many days will the population be 12,000? Solution We set P = 12,000 and solve for t. 12,000 = 12,000 ln|1 + 0.25 t| + 1000 11,000 = 12,000 ln|1 + 0.25 t| 11/12 = ln|1 + 0.25 t| e11/12 = 1 + 0.25 t e11/12 - 1 = 0.25 t t = 4(e11/12 - 1) = 6 We can conclude that after 6 days, the population will be 12,000.
Extra Credit:
Write
down one thing that your instructor can do to make the class better. |