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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

We use the chain rule.  We write

        u  =  ln x        u'  =  1/x

        f(u)  =  ln u    f '(u)  =  1/u

We get

                                1        1                    1
        f '(x)  =                           =                      
                          x        ln x              x ln x

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.     

Solution

We write

       

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

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C.  

 

 Solution

This problem is much easier if we use the properties of logarithms first.  We simplify to 

       

now the derivative is simply

        f '(x)  =  1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

PROBLEM 2  Find the following anti-derivatives.

A.    

Solution

We first simplify the exponents using algebra to get

       

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

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.     

Solution

For this integral, we use u-substitution.  We let

        u  =  4 - 3x5         du  =  -15x4 dx

We get

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C.  

 

 Solution

We use u-substitution again

        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       
                  =   kP        P(0)  =  5        P(6)  =  45
         dt  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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
        k  =                  =  0.3662
                     6

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
        t  =                              =  33.3
                     0.3662

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
                =                            
        dt            1 + 0.25t

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.