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0901 Quiz

1.  We have looked at the basic graphs of linear, quadratic, power and exponential functions.  Among power functions some have positive powers greater than one, some have positive powers less than 1, and some have negative powers.  List the functions which have graphs with the following characteristics:

• The graph is a parabola.

The basic quadratic function y = x^2 has a parabolic graph, as do the generalized quadratic functions.

• The graph has a horizontal asymptote.

The graphs of the basic negative-power power functions have horizontal asymptotes, as does the graph of the basic exponential function.  Examples:    y = 1 / sqrt(x) = x^(-1/2) and y = 2^x.

• The graph has a vertical asymptote.

The graphs of the negative-power power functions have vertical asymptotes.  Example:  y = 1 / sqrt(x) = x^(-1/2).

• The graph has both vertical and horizontal asymptotes.

The graphs of negative-power power functions have vertical asymptotes.  Example:  y = 1 / sqrt(x) = x^(-1/2).

• The graph is a straight line.

The graphs of linear functions have straight-line graphs.

• The graph is increasing at a decreasing rate.

The graphs of power functions for powers between p = 0 and p = 1 are increasing at a decreasing rate.  Example:  y = sqrt(x) = x^(1/2).

2.  Which of the descriptions in #1 do you think would apply to the frequency of a pendulum vs. its length?

Frequency is the number of times the pendulum oscillates per minute.  The longer the pendulum, the fewer the number of oscillations per minute.  There is no length at which frequency is zero, so the graph has to be asymptotic to the horizontal axis.  It turns out that there is no limit to how many times a pendulum can oscillate per minute if you make it short enough, so there is also a horizontal asymptote at the vertical axis.

This description is consistent with a negative-power function.

Which of the descriptions do you think would apply to the period of a pendulum vs. its length?

The period function turns out to be T = .2 sqrt(L), which is a generalized p = 1/2 power function of form T = a * L^(1/2).

3.  Solve the system

2 a + 3 b = 12

3 a + 5 b = 19

by elimination.

If we wish to eliminate b we can multiply the first equation by -5 and the second by 3 to get

-10 a - 15 b = -60

9 a + 15 b = 57.

Adding these equations we get -a = - 3 so we know that a = 3.

Substituting a into the first equation we get

2 * 3 + 3 * b = 12, or

6 + 3 b = 12 so that

3 b = 6 and

b = 2.

We could substitute a and b into the second equation to verify this solution.

4.  Give the generalized form of each of the functions given below:

• A generalized power function.

y = A x ^ p

(most generalized is y = A ( x - h ) ^ p + k)

• A generalized quadratic function.

y = a x^2 + b x + c

(another generalized form is y = A ( x - h ) ^ 2 + k).

• A generalized exponential function.

y = A * 2^(k x)

(another important generalized form is y = A b^x)

(most generalized:  y = A * 2^(kx) + c or y = A * b^x + c or y = A e^(kx) + c).

• A generalized linear function.

y = m x + b.

5.  Sketch a graph of each of the following sets of points and tell which of the functions in #1 you think the general shape of the graph is more similar to:

• The points (6, 100), (10, 10) and (20, 2).

This graph appears to have a vertical asymptote at the y axis and a horizontal asymptote at the x axis, as does the graph of y = 1 / sqrt(x).

• The points (5, 100), (10, 40) and (20, 150).

This graph starts high, comes to a low point then goes high again, like the basic quadratic function y = x^2.

• The points (2, 8), (12, 20) and (20, 100)
• The points (3, 10), (10, 20) and (20, 25).
• The points (4, 10), (10, 13) and (16, 16)

6.  Write down the equations we would get in each of the following situations.  You don't need to solve the equations at this point, and to solve some of the equations will in fact require that you use methods you probably don't know yet:

• We have y values 3, 9 and 21, which correspond to x values 8, 12 and 17, and we want to construct a generalized quadratic model.
• We have y values .7, 1.8 and 9.7 corresponding to x values 5, 10 and 15 and we want to construct a linear model.
• We have y values 2, 12 and 73 corresponding to x values 3, 9 and 13 and we want to construct an exponential function model.

7.  Multiplying or dividing both sides by the same quantity, adding the same quantity to both sides, raising both sides to the same power, etc., how do you solve each of the following for x?  Which of these equations cannot be solved for x using just these procedures?  Be sure you specify each step in each of your solutions..

• y = a x^2

Divide both sides by a to get y / a = x^2.

Take + - square root of both sides to get x = +- sqrt(y/a).

• y = a sqrt(x)

Divide both sides by a then square both sides and you get x = (y/a)^2.

• y = a ( x - h ) ^ 2 + k

Starting with

y = a ( x - h ) ^ 2 + k we first subtract k from both sides to get

y - k = a ( x - h ) ^ 2.  Then divide both sides by a to get

(y - k) / a = (x - h)^2.  Then take +- square root to get

+-sqrt(( y-k)/ a) = x - h and finally add h to both sides ...

• y = a x^2 + b x + c

Use the quadratic formula.

• y = a / x^7

Multiply both sides by x^7 to get

x^7 y = a  then divide both sides by y to get

x^7 = a / y.  Then take the 1/7 power of both sides to get

(x^7)^(1/7) = (a / y)^(1/7) or

x = (a / y ) ^ (1/7).

8.  For pendulum period vs. length, if we obtain periods of .69, 1.08 and 1.41 seconds at lengths 12, 30 and 50 cm then the model T = a L^2 + b L + c gives us the three equations

• .69 = a * 12^2 + b * 12 + c
• 1.08 = a * 30^2 + b * 30 + c
• 1.41 = a * 50^2 + b * 50 + c

or

• .69 = a * 144 + b * 12 + c
• 1.08 = a * 900 + b * 30 + c
• 1.41 = a * 2500 + b * 50 + c

If we subtract the second equation from the first we eliminate c from the two equations, and if we subtract the third from the first we eliminate c from those two equations.  This gives us two equations

• -.39 = -756 a- 18 b  and
• -.72 = -2356 a - 38 b.

Solve these two equations for a and b, using the process of elimination.  Suggestion:  Start by eliminating b from the two equations.

If we multiply the first equation by -38 and the second by 18 the coefficients of b will match:

-38( -.39) = -38 ( -756 a) - 38 ( -18 b) gives us 14.82 = 28728 a + 684 b.

18 ( -.72) = 18 ( -2356 a) + 18 ( -38) b gives us -12.96 = - 42408 a - 684·b

Adding

14.82 = 28728 a + 684 b to

-12.96 = - 42408 a - 684·b we get

1.86 = -13780 a

so that

a = 1.86 / -13780

 = -.000136.

Substituting this into the first equation we have

-.39 = 756 * -.000136 - 18 b so that

b = .0274.

Substituting these values of a and b into the first equation we get

• .69 = -.000136 * 12^2 + .0274 * 12 + c

We easily solve for c to get c = .381.

Substitute your values of a, b and c into the form T = a L^2 + b L + c to get a quadratic model of T vs. L. 

Our final model T = a L^2 + b L + c is

• T = -.000136 L^2 + .0274 L + .381.

• What period does your function predict if L = 40 cm?

If L = 40 then we obtain T = -.000136 * 40^2 + .09274 * 40 + .391 = 1.26.

• What period does your function predict of L = 80 cm?

If L = 80 then we obtain T = -.000136 * 80^2 + .09274 * 80 + .381 = 1.70..

9.  Sketch a graph of T vs. L for the preceding data.

The correct T vs. L function is T = .2 sqrt(L). 

• What is the period T if L = 40 cm? 

If L = 40 then we obtain .2 sqrt(40) = 1.26

• What is the period if L = 80 cm?

If L = 80 then we obtain T = .2 sqrt(80) = 1.79.

How does each of these period compare with the predictions of the quadratic model? 

• By how much does each differ from the quadratic model of #7?

For L = 40 cm the actual period is T = 1.26 sec and the quadratic model gives us T = 1.26 sec.  So the quadratic model works very well for this length, with 0 error when periods are calculated to the nearest .01 sec.

For L = 80 cm the actual period is T = 1.79 sec and the quadratic model gives us T = 1.70 sec.  So the quadratic model is a bit low for this length, with an error of .09 when periods are calculated to the nearest .01 sec. 

• What percent of the correct result is each difference?

The above figures indicate no error for the L = 40 cm length, when calculations are rounded to the nearest .01 sec.

An error of .09 is .09 / 1.79 = .05 of the ideal result 1.79.  So we say that the percent error in this case is 5%.