Practice Final

Please work out each of the problems below.  Credit will be based on the steps towards the final answer.  Show your work.

Problem 1

Sketch the following.

A.  The point (3,4,1).

Solution

We draw the xyz-axes, the shadow at the point (3,4) in the xy-plane and move it up 1 unit.

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.  The surface z  =  x2 + y2 

Solution

This is a paraboloid with the z-axis as its central axis.  The graph is shown below.

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C.  Five level curves to the surface z  =  x - y2 

Solution

Draw a table as follows

k -2 -1 0 1 2
k  =  x - y2 x  =  y2 - 2 x  =  y2 - 1 x  =  y2  x  =  y2 +1 x  =  y2 + 2

Notice that these are all sideways parabolas with distinct x-intercepts that correspond to the value of k.  The graphs below show these level curves.

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 2

Find fxz for

        f(x,y,z)  =  x2z + y2 - y cos(xz)

Solution

We first find the derivative with respect to x by letting y and z be constant.

        fx  =  2xz + yz sin(xz)  Note the chain rule, the derivative of xz is z

Now find the derivative of this with respect to z,

        fxz  =  2x + y sin(xz) + xyz cos(xz)  We used the product rule for the second term

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 3

A swimwear store sells both men's and women's swimsuits and has determined that the profit, P,  it can make in stocking x men's suits and y women's suits is 

        P(x,y)  =  2x2 + y2 + xy - 900x - 1100y  + 400,000

Determine the number of men's and the number of women's suits that should be stocked in order to maximize profit.  Then determine the maximum profit.

Solution

We find the partial derivatives and set them equal to 0.

        Px  =  4x + y - 900        Py  =  2y + x - 1100

This gives us the two equations

        4x + y - 900  =  0        x + 2y - 1100  =  0

The first equation can be written as

        y   =  900 - 4x

Substituting into the second equation gives

        x + 2(900 - 4x) - 1100  =  0

        x + 1800 - 8x - 1100  =  0

        7x  =  700

        x  =  100

Hence

        y  =  900 - 4(100)  =  500

The store should stock 100 men's suits and 500 women's suits.

The profit will be

        P(100,500)  =  2(100)2 + (500)2 + (100)(500) - 900(100) - 1100(500)  + 400,000

        =  80,000

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 4

A medical researcher is studying the effect that exercise has on a women's percent body fat.  The table below shows the results of the corresponding study done.  Determine the least squares regression line for this data and use it to estimate the mean percent body fat for a women that exercises 3 hours per week.

Exercise Hours 0 1 2 4 4 5
% Body Fat 31 28 26 20 18 15

 

Solution

We use a calculator to find the equation of the regression line.

        y  =  -3.155x +31.414

We plug in x  =  3 to get

        y(3)  =  -3.155(3) + 31.414  =  21.95

We predict that the expected percent body fat for a women who exercises three hours per week is 21.95%.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 5

Switch the order of integration to evaluate the following

       

 

Solution

To change the order we first draw the region shown below.

       

Now we switch the order using dydx.  Notice that y goes from 0 to x and x goes from 0 to 5.  We have

       

This integral can be solved using substitution with

        u  =  1 + x2          du  =  2xdx

This gives

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 6

Consider the following game.  Pick a card.  If you select a face card (Jack, Queen, or King) you win $3.  Otherwise, you lose $1.  Find the expected value and comment on how this number shows whether playing this game many times is a good idea.

 

Solution

We write down the probability distribution table

x 3 -1
P(x) 3/13 10/13

The expected value is just

             3                      10                1
        3          +      (-1)             =  -        
            13                     13               13

Since this is a negative value, we would expect to lose money on the average.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 7

Consider the following probability density function

        f(x)  =  x + 0.5        0  <  x  <  1

A.  Find P(0.5 < x < 1).

Solution

We just find the integral

       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.  Find the standard deviation.

Solution

First find the mean.

Now calculate the integral using a calculator.

Finally take a square root to get the standard deviation

        s  =  0.276

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C.  Find the median.

To find the median we set the following integral equal to 1/2.

Multiplying by 2 and bringing all the terms to the left hand side gives

        m2 + m - 1  =  0

Using the quadratic formula gives

        m  =  0.618

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 8

A pair of baby rabbits is introduced to an island.  Assume that rabbits take two months to reproduce when they have 2 pairs of new baby rabbits. Each month thereafter, the adult rabbits produce two additional pairs of rabbits. Assume that the rabbits never die.

A.  Write down the number of pairs of rabbits on the island during the first 6 months.

Solution

We have

        1, 1, 3, 5, 11, 21

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.  Find constants u and v such that

        an  =  u an-1 + v an-2 

Where an represents the number of rabbits after the nth month.

Solution

We just notice that to find the current generation, add the newborns to the prior population.  The newborns are coming from rabbits that were alive two months previously.  Hence

        u  =  1        and        v  =  2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 9

A very long river that runs along a vast area of farming communities has 3000 trout in its first mile.  Because of pollutants, each mile contains only 80% of the fish that the previous mile had.  so that the second mile has 2400 trout and the third has 1820 trout.  Considering the river as running an infinite distance, how many trout live in the river?

Solution

We have a geometric sequence

        3000, 2400, 1820,  ...

with

        r  =  0.8        and        a  =  3000

Hence the sum is

                       3000
         S  =                       =  15,000
                     1 - 0.8

There are 15,000 trout in the river.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 10

Determine whether the following series converge or diverge.  Be sure to cite the test that you are using.

A

Solution

We use the nth term test.

       

Since the limit does not go to 0, by nth term test, the series diverges.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B. 


Solution

We can rewrite this as

       

This is a P-series with p  =  3 > 1.  Hence by the P-Series Test the series converges.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 11

Find a power series representation for 

       

using the fact that

       

then find the radius of convergence for your solution

 

Solution

We substitute -x2 in for x to get

       

Now integrate to get

        

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 12

Find the third degree Taylor polynomial centered at x  =  0 for

        f(x)  =  x ex 

 

Solution

We find the derivative of f and evaluate them at x  =  0.

        f '(x)  =  ex + x ex  =  ex(1 + x)        f '(0)  =  1

        f ''(x)  =  ex + (1+ x) ex  =  ex(2 + x)        f '(0)  =  2

        f '''(x)  =  ex + (2+ x) ex  =  ex(3 + x)        f '(0)  =  3

We now plug this into the Taylor polynomial formula to get

                         0                  1                 2                  3
        P3(x)  =           x0   +            x1   +          x2    +          x3                    
                         0!                1!                2!                 3!

                            1
        =  x + x2 +         x3 
                            2

        

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 13

Use Newton's method  with initial guess of 1 to estimate 

       

within three decimal places of accuracy by finding the solution to 

        x2  -  2  =  0

Solution

With 

        f(x)  =  x2 - 2

we have

        f '(x)  =  2x 

so that

                                   xn2 - 2 
        xn+1  =  xn -                         
                               
      2xn

With our initial guess of x0  =  1, the table below gives the rest of the values

        

x0 x1 x2 x3 x4
1 1.5 1.4167 1.4142 1.4142

Hence 1.414 approximates the square root of 2 accurate up to three decimal places.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 14

Solve the following differential equation

                    1
        y ' -             y  =  2x2 + 1      
                    x

Solution

We first find the integrating factor

       

now multiply through by the integrating factor to get

        1                1                       1
              y ' -             y  =  2x +               
        x                x2                      x

so that

       

Now integrate both sides to get

        1        
             y  =  x2 + ln|x| + C
   
     x

and finally multiply both sides by x to get

        y  =  x3 + x ln|x| + Cx

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Problem 15

The rate of growth of a bacteria is equal to the quotient of the number of bacteria present and one more than the number of hour that the bacteria has been growing.  If initially there were 5 grams of bacteria, how many grams of bacteria will there be in 8 hours?

Solution

We have the differential equation

        dx             x
                =           
        dt           1 + t

Now separate the variables to get

        dx            dt
                =           
         x           1 + t

Now integrate to get

        ln x  =  ln(1 + t)  + C1 

Exponentiate both sides to get

        x  =  eln(1 + t) + C1  =  eln(1 + t) ec1  =  C(1 + t)

We have that when t  =  0,  x  =  5 so that

        5  =  C(1 + 0)          C  =  5

so that

        x(8)  =  5(1 + 8)  =  45

There will be 45 grams of bacteria after 8 hours.