061206
Problem Number 1
A graph of the range r(y) of a water stream coming from a hole in the side of a container, as a function of water depth y, is a power function r(y) = k `sqrt(y) with r-intercept (0,0) and graph point ( 80, 21). A graph of the depth y(t) as a function of clock time t, in seconds, is approximated by a parabola with its y-intercept at (0, 20) and vertex at ( 108 , 0 ).
Notes: A graph of y vs. t is a parabola which should be easy to sketch from the given information. A graph of r vs. y is a power function with exponent ½ (since sqrt(y) = y^ ½ ) so its shape will be similar to that of the p = ½ power function, which you should be able to sketch (remember that the basic points are at x = -1, 0, ½, 1 and 2; in this case the ½ power of -1 is undefined, which should remind you that the function is undefined for negative x). If the sketch is relabeled so it contains the point (80, 21) you will have the graph of r(y).
To get the graph of r(t), take a series of t values on the graph of y vs. t and estimate the corresponding values of y , then find these values of y on the graph of r vs. y and determine the corresponding values of r. Graph r values vs. t values.
Problem Number 2
If f(x) = x^ 4 and g(x) = .47 ln ( x ), sketch graphs of f(x) and g(x) for 0 < x <= 3, showing how you use the appropriate basic points to obtain the shape the graphs of f(x) and g(x). Show how you combine the graphs to obtain the graph of f(x) / g(x).
Notes: Use basic points to graph f(x), and graph g(x) by first graphing the e^x function (basic pts are (1, e), (0, 1) and asymptote at negative x axis), then graphing its inverse (basic pts are (e, 1), (1, 0) and the negative y axis), then vertically stretching this graph by factor .47.
The zero of g(x) is at x = 1; as you approach x = 1 the value of g(x) approaches 0 and the value of f(x) is approaches 1 so the magnitude of f / g approaches infinity, giving you a vertical asymptote at x = 1. To the left of x = 1 g is negative so f / g is negative, and f / g approaches –infinity. To the right of x = 1 g is positive and the value of f is positive so f / g approaches +infinity.
For x <= 0 the g(x) function is not defined so f / g is not defined.
Between x = 0 and x = 1 the value of f is positive and the value of g is negative so f / g is negative.
f(x) = 0 for x = 0 so as x approaches zero the numerator approaches 0. The denominator approaches –infinity. Dividing a number which approaches 0 by a number which gets very large gives you a number which approaches 0, since a very small number divided by a very large number yields a very small result.
Thus between x = 0 and x = 1 the graph of f / g starts just to the right of (0, 0) and curves downward to its asymptote at x = 1.
The g(x) function takes value 1 when .47 ln(x) = 1, which can be solved to give x = e^(1 / .47) = 8, approx.. You don’t need to solve the equation, though it would certainly be OK to do so. You can use the approximate graph of .47 ln(x) you sketched; if you sketched it reasonably well you should see that it intersects the line y = 1 reasonably near x = 8 (anything between x = 4 and x = 16 would be a reasonable graphical estimate). At this value of x, the f / g function will coincide with the f(x) function, and you should mark this point on your graph. Your f / g curve should fall from its vertical asymptote at x = 1 to this point.
Beyond this point it might not be clear to you what happens. However, since the graph of x^4 increases so rapidly, curving rapidly upward, and the value of ln(x) increases so much more slowly, increasing more and more slowly as x gets larger and larger, you should be able to conjecture that f / g gets larger and larger without bound, so the graph of f / g will get large as x approaches infinity. So at some point either before or after reaching the preceding point, the graph of f / g will begin curving upward and will continue to do so.
Problem Number 3
Show that for y = f(x) = 4 * 1.15^x the ratio f( x + h ) / f(x) does not depend on the value of x.
Note: f(x + h) = 4 * 1.15^(x + h), which by the laws of exponents is equal to 4* 1.15^x * 1.15^h, so
f(x+h) / f(x) = 4 * 1.15^(x+h) / (4 * 1.15^x) = 4 * 1.15^x * 1.15^h / (4 * 1.15^x) = 1.15^h.
This value does not involve x, so it does not depend on x.
There are statements of this problem that say ‘Show that for y = f(x) = 4 * 1.15^x the ratio f( x + h ) / f(x) does not depend on the value of h’. This would be a false statement, since 1.15^h does depend on h. You would be expected to simplify the expression to get 1.15^h. Ideally you would then say that 1.15^h does indeed depend on h, but if after obtaining 1.15^h you get confused by the erroneous problem statement you would still get credit.
Problem Number 4
Find tDoub if 7 * 1.05^(x + tDoub) = 2[ 7 * 1.05^x ].
Note: Solve the equation for tDoub. Using the laws of exponents 1.05^(x + tDoub) = 1.05 ^x * 1.05^tDoub, so you get
7 * 1.05^x * 1.05^ tDoub = 2 [ 7 * 1.05 ^ x ]. Dividing both sides by 7 * 1.05^x gives you
1.05^tDoub = 2 so that
tDoub = ln(2) / ln(1.05).
Problem Number 5
Sketch a graph of y = (x + 6) ^ 3 (x – 6) (x^2 + 2 x + 3).
Note: Check x^2 + 2 x + 3 to see if it factors. This will be the case if there are values of x for which x^2 + 2 x + 3 = 0; the discriminant of this quadratic equation is 2^2 – 4 * 1 * 3 = -8, so there are no solutions.
Thus the only zeros are at x = -6 and x = 6.
As x -> infinity all factors are positive so y will approach infinity very rapidly and the graph will curve rapidly upward as we move beyond the rightmost zero.
As x ->- infinity the factors x + 6 and x – 6 will both be large negative numbers while the quadratic factor is dominated by the x^2 and will be positive, so y be a produce of four large negative numbers and one large positive number. Thus y will approach infinity very rapidly and the graph will curve rapidly upward to the left beyond the leftmost zero.
The y intercept is obtained by substituting x = 0 to get y = 6^3 * (-6) * (6^2 + 2 * x + 3) = 60,000 or so. Be sure to label the y axis to an appropriate scale so this value appears.
You should by this point know that you can get your calculator to display this graph. You should also be aware that a copy of calculator output, without specifying that the above details to justify it, will not receive any credit.
Problem Number 6
The population of a certain organism is governed by the recurrence relation a(0) = 4, a(1) = 4, a(2) = 3 and a(n) = a(n-1) + 3 a(n-2) + a(n-3), where n is the number of the population transition.
You will get
a(3)= a(2) + 3 a(1) + a(0) = 3 + 3 * 4 + 4 = 19, then
a(4)= a(3) + 3 a(2) + a(1) = 19 + 3 * 3 + 4 = 29 then
a(5)= a(4) + 3 a(3) + a(2) = 29 + 3 * 19 + 4 = 90 then
a(6)= a(5) + 3 a(4) + a(3) = 90 + 3 * 29 + 19 = 196,
etc..
The sequence is thus 4, 4, 3, 19, 29, 90, 196, … and the ratios are 4/4 = 1, ¾ = .75, 19/3 = 6.33, 29/19 = 1.45, 90/29 = 3.1, 196 / 90 = 2.2, etc. Starting with .75, the ratios go up, down, up, etc., with all ratios following any two consecutive ratios lying between these ratios. The ratios will approach a limiting value, which by computing the ratios through n = 10 you will be able to estimate fairly accurately. This limiting ratio will be the growth factor for the exponential.
Problem Number 7
Linearize the data set consisting of the weight (pounds) vs. time (years) points ( 13, 6), ( 19, 9.120001), ( 25, 13.862) and ( 31, 21.07).
To linearize a data set you make a table for x, y, log(x) and log(y), then see if any of the graphs y vs. x, log(y) vs. x, y vs. log(x), log(y) vs. log(x) gives a good straight line, and if so which gives the best straight line. After determining the slope and vertical intercept of that line, you obtain an equation which can be solved for y using laws of exponents, logarithms, and common operations on equations.
Problem Number 8
If a patient starts with no drug in her body takes a 400 mg dose every 6 hours, losing 30 % of the after-dose amount during that time, then how much drug will remain 6 after the initial dose?
The amounts are:
The just-before dose amounts are 0, 280, 476, 613.2 mg.
We will look at sequence of ratios, then if necessary at sequence of ratios of differences to see that the function is exponential.
The asymptote is the before-dose maintenence-level concentration such that the 30% lost is equal to the dose taken.
Let L be the after-dose maintenance level. Then
so
We easily solve to get L = 1333 1/3 mg.
This is the after-dose maintenance level. The before-dose level is therefore 400 mg less:
The before-dose concentrations 280, 476, 613.4 mg approach 933 1/3 mg as an asymptote.
See notes from the preceding class for more detail on how to obtain the exponential function
Problem Number 9
A graph of the power P(v), in horsepower, of an automobile as a function of its velocity v, in mph, is linear with vertical intercept (0, 4) and slope 4.5. A graph of the velocity v(t) in mph vs. clock time t in seconds is approximated by a parabola with its y-intercept at (0, 0) and vertex at ( 17 , 52 ).
Note: You should be able to construct the parabolic graph of v vs. t from the given info (see also Problem Number 1 above), as well as the linear graph of P vs. v. By choosing a series of t values you should be able to estimate the corresponding v values, then use these v values to estimate the corresponding P values, to obtain a graph of P vs. t.
You could obtain a reasonable estimate of the slope of the v vs. t graph by estimating v for t = 0 and for t = 13, then dividing the change in v (obtained from your estimates) by the change in t (if you use t = 9 and t = 13 this change will be 4). Note that t = 11 is halfway between the chosen t values. You could use other pairs of t values, e.g., t = 10 and t = 12. The values should be chosen with 11 in the middle, far enough apart that you can make a reasonably accurate estimate of the change in , but not so far the the average slope between the points differs too much from the actual slope at t = 11. Alternatively you could sketch the tangent line at t = 11 and estimate the rise and run between two points on this line, then calculate slope = rise / run.
If you then estimate the value of v for t = 11 and locate this value on the P vs. v graph, you can use a similar procedure to estimate the slope of the P vs. v graph.
The slope of your P vs. t graph should be the product of the two slopes.
These two slopes represent the rate at which v changes with respect to t, and the rate at which P changes with respect to v. The product represents the rate at which P changes with respect to t.
Problem Number 10
What dosage is required to maintain a minimum level of 390 mg if 30 % of the drug is removed between doses?
If the minimum level is 390 mg and the (unknown) dosage is D, then after the dose the amount will be 390 mg + D.
When 30% is then removed before the next dose, 70% will remain, and the remaining amount will be 70% ( 390 mg + D).
This remaining amount will be equal to the 390 mg minimum level, so we have
70% ( 390 mg + D) = 390 mg, or
.70 ( 390 mg + D) = 390 mg.
This equation can easily be solved for D.
Problem Number 11
Which of the following are valid and which are not? Be sure to state why for each case.
Very similar problems were asked on daily quizzes. Be sure you know the laws of exponents well enough to justify your answer to each. As on all problems, you must justify your answers.
Problem Number 12
The population of a certain organism is governed by the recurrence relation a(0) = 2, a(1) = 4, a(n) = a(n-1) + 4 a(n-2), where n is the number of the population transition.
This is identical in form to Problem 6, and the solution proceeds by similar methods.
Problem Number 13
Determine whether the following altitude vs. clock time data are best modeled by an exponential or a y = x^2 power function: Data points are (t, T) = ( 5, 7.36974), ( 6, 8.327806), ( 8, 10.63378), ( 9, 12.01617).
Note: Follow the procedure for linearizing the data. If the function is an exponential function then log(y) vs. x will be linear. If it is a power-2 function then log(y) vs. log(x) will be linear and will have slope 2.
Problem Number 14
If f(x) = .41 x + 4 and g(x) = -6.761 x + 2.112, sketch graphs of f(x) and g(x). Show how you combine the graphs to obtain the graph of f(x) * g(x).
Note: Follow the procedures used in class to find the points where f(x) = 0, g(x) = 0,| f(x) | = 1, | g(x) | = 1, the intervals where | f(x)| < 1, where | g(x) | < 1, where each function is positive and where negative, etc., and use to construct the graph.
Where f(x) = 0 the product function is 0.
Where gf(x) = 0 the product function is 0.
Where one function has value 1 the product function has the same value as the other function.
Where one function has value -1 the product function has a value equal and opposite to the value of the other function, i.e., the point would be on a graph of the reflection of the other function through the horizontal axis.
Where the functions have opposite signs the product function will be negative.
Etc.
Problem Number 15
Explain why the following statement must be true: If the quadratic polynomial f(x) = a x^2 + bx + c has no zeros, then that polynomial is not the product of two linear polynomials.
Note: Any linear polynomial will pass through the x axis at some point, and will therefore be zero at some value of x.
If the function was the product of two linear polynomials, then each of those linear polynomials would be zero at some value of x. At any such value of x, the product of the two polynomials would be zero.
So if f(x) is the product of two linear polynomials, it must have at least one zero.
If f(x) has no zeros, it is therefore not the product of two linear polynomials.
Problem Number 16
Find the equation of a line through ( 7, 5) and ( 4, 9) by each of the following methods:
Substitute x and y coordinates into y = m x + b to get two equations in unknowns m and b, and solve for m and b. Put those values back into the form y = m x + b to get the equation.
Then find the slope between the two given points, and the slope between either of the given points and the general point (x, y). Set the two expressions equal and solve for y. A sketch is also useful.
If your graph represents the velocity of a coasting automobile, in mph, vs. clock time in seconds:
Instead of y = m x + b, you might want to write the equation you obtained in the form v = m t + b, using the same values you obtained previously for m and b.
On a graph of velocity vs. clock time, rise represents change in velocity and run represents change in clock time so rise / run represents change in velocity / change in clock time, or average rate of change of velocity with respect to clock time.
To find the velocity when t = 12 sec plug in 12 for t in your equation v = m t + b and simplify to find y.
To find the clock time when velocity is 15 mph substitute v = 15 into your equation and solve for t.
Problem Number 17
Express the function y = 7 * 2 ^ ( .97 x) in the form y = A b ^ x, and also into the form y = A e ^ ( kx ).
If 7 * 2^(.97 x) = A b^x, then A = 7 and b^x = 2^( .97 x) = (2^.97)^x= 1.9588 ^ x, so b = 1.9588.
If 7 * 2^(.97 x) = A e^(kx), then A = 7 and e^(kx) = 2^( .97 x) = (2^.97)^x= 1.9588 ^ x, so e^(kx) = 1.9588 ^ x. Since e^(kx) = (e^k)^x, it follows that e^k = 1.9588 so that k = ln(1.9888) = .6724.
Problem Number 18
For the function y = f(t) = .93 ^ t construct a table of y vs. t for t running from t = 2.7 to t = 2.9 in four equal increments. Using appropriate transformation(s) on the y column, the t column, or both, linearize this data set and demonstrate that the data set has in fact been linearized.
If t runs from 2.7 to 2.9 in four equal increments, then the four increments span an interval of 2.9 - 2.7 = .2. So each increment has length .2 / 4 = .05.
The t values will therefore be 2.7, 2.75, 2.8, 2.85 and 2.9.
Evalute the function at these values to get the corresponding y values and make a table of y vs. t. Extend this table to also include log(t) and log(y). Then follow the procedure for linearization.
Problem Number 19
Make a table for y = x^4 for 0 <= x <= 5 and use it to obtain a table for its inverse function. Sketch graphs of both functions and show how the graphs are related. What is the formula for the inverse function? Are there any numbers that the inverse function cannot act on?
Make the table then reverse the columns. Graph the reversed columns. That is how you graph the inverse function.
Alternatively graph the function and ‘reflect’ it through the line y = x, which reverses the coordinates of the graph and gives the same result as the preceding. You should understand this both ways.
The inverse function is the function y = x^(1/4). The inverse of the function y = x^p is y = x^(1/p). The ¼ power ‘undoes’ the 4th power: take any number and raise it to the power 4, then raise the result to the power ¼. You will get the original number back. This also works if you raise the original number to the ¼ power then raise the result to the 4th power.
Problem Number 20
Evaluate f( 4.7), f( 8.1), f( 11.5) and f( 14.9) for the function y = f(t) = 3 * 1.13^t and show that successive ratios of the sequence formed by these values appear constant.
Find the values. The successive ratios are
f(8.1) / f(4.7), then f(11.5) / f(8.1), then f(14.9) / f(11.5). All these ratios should be the same, within roundoff error.
Problem Number 21
What specific function y = A b^x fits the data points ( 7, 11.38) and ( 12, 16.12)?
Plug the coordinates in to the form y = A b^x. You will get two equations in unknowns A and b.
Divide one equation by the other to eliminate A.
Solve the resulting equation for b, using the laws of exponents carefully and correctly.
Then plug your value of b back into either equation and solve for A.
Check your values of A and b by plugging them back into the other equation.
Problem Number 22
If a sound measures 100 decibels, then what is the intensity of the sound, as a multiple of the hearing threshold intensity?
dB = 10 log( I / I0 ), where I0 is threshold intensity.
So if dB = 100, we have
100 = 10 log(I / I0) so
log(I / I0) = 10. Rearranging to exponential form (raise 10 to the power of each side) we get
I / I0 = 10^10 so that
I = 10^10 * I0. This is a sufficient answer, but note that it means
I = 10,000,000,000 * I0.
Problem Number 23
What equation would you have to solve to find the doubling time, starting at t = 3, of a population that starts at 700 organisms and grows at annual rate 7.1%?
The growth rate is .071 so the growth factor is 1.071 and the function is
P(t) = P0 * 1.071 ^ t.
t = 0 population is P(0) = P0 * 1.071^0 = P0, and P(0) is given as 700, so P0 = 700 and the function is
P(t) = 700 * 1.071^t.
The population will have doubled when P(t) = 2 * 700 = 1400 so the equation would be
1400 = 700 * 1.071^t.
The question didn’t ask for the solution, but here it is:
Divide both sides by 700 to get
2 = 1.071^t. Take log of both sides to get
log(2) = log(1.071^t). Apply laws of logarithms to get
log(2) = t log(1.071) so that
t = log(2) / log(1.071). Plug this into your calculator; you’ll get t = 10, approximately.
Problem Number 24
What are the basic points of the exponential function y = f(x) = 7 * e^( 1.52 x)? Graph the function using these points.
The basic points of e^(1.52 x) are the x = 0 and x = 1 points (0, 1) and (1, e^(1.52)) = (1, 4.57), and the asymptote at the negative x axis.
When vertically stretched by factor 7 the points become (0, 1), (1, 32), approx., and the asymptote remains the negative x axis.
From these two points and the asymptote, and your knowledge of the shape of the graph of an exponential function, you should be able to sketch a reasonable graph the function without reference to your calculator or to a more extensive table.
Problem Number 25
What quadratic function describes the behavior of the graph of y = p(x) = (x- 3)(x- 3)(x+ 3.3) near the point ( 3,0)?
Near x = 3 the value of x – 3 is near 0 while the value of x + 3.3 is near 3 + 4.3 = 6.3. So the quadratic function is
y = (x – 3)(x - 3) * 7.3 or
y = 6.3 (x-3.3)^2.
At x = 3 – 1 = 2, p(x) = (2 – 3) (2 – 3) (2 + 3.3) = 5.3, while the quadratic approximation is y = 6.3 ( 2 – 3)^2 = 6.3.
At x = 3 – .5 = 2.5, p(x) = (2.5 – 3) (2.5 – 3) (2.5 + 3.3) = 1.45, while the quadratic approximation is y = 6.3 ( 2.5 - 3)^2 = 1.575.
At x = 3 – .1 = 2.9, p(x) = (2.9 – 3) (2.9 – 3) (2.9 + 3.3) = .063, while the quadratic approximation is y = 6.3 ( 2.9 - 3)^2 = .0063.
It is clear that the values of the approximation get much closer to the values of the actual polynomial as the value of x approaches 3.
Problem Number 27
A sandpile 5 cm in diameter has a mass of 150 grams.
Note: Be sure to use the appropriate proportionalities. For most problems of this nature the proportionalities are governed by the scaling of tiny squares or cubes.
For volume, which can be regarded as filled with tiny cubes, the proportionality is y = k x^3, and since mass occupies volume the same proportionality applies to mass. Since surface area can be regarded as being covered by tiny squares, the proportionality is y = k x^2.
If ratios are used, the ratio of volumes is the cube of the ratio of linear dimension: (y2 / y1) = (x2 / x1)^3. A ratio of areas is the square of the ratio of linear dimensions: (y2 / y1) = (x2 / x1)^2.