This page consists of a table representing the main concepts and procedures in the VHCC Precalculus I course. Click on a link for an explanation of a statement or an example of the means by which the statement is implemented.
statement |
means |
result |
nature |
y = a t^2 + b t + c mathematical model |
exact for 3 data points |
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a = **, b = **, c = ** |
solution if indep vbl values distinct |
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ave. dev. = ** |
measures closeness, not appropriateness |
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random graph of pattern |
eyeball indication; statistical analysis possible |
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plot points or stretch and shift basic function or use quadratic formula |
symmetric graph with vertex |
parabola is a very specific shape; no asymptotes |
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t where at^2 + bt + c = 0 |
may be 1, 2 or 3 solutions (sign of b^2 - 4 * a * c) |
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extreme point (t, yVertex) |
may be anywhere |
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points (t +- 1, yVertex + a) |
determines 'spread' of parabola |
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a = vert. displacement at tVertex +- 1; yVertex = -b / (2a); then c easy |
model y = a t^2 + bt + c |
spread and horiz. shift + a, b; then c from vert. shift |
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y = f(t); plug in known t, solve for y; plug in known y, solve for t |
value of variable assoc with known value of other |
project from x or y axis to graph then to other axis |
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#11: f(x) notation |
when x is replaced in f(x), it's replaced in defining expression |
generalized notation |
allows us to talk and conceive of functions in general terms |
#12: basic functions |
y = x (linear); y = x^2 (quadratic); y = 2^x ( exponential); y = x^p (power) |
graphs and behavior not easily deformable into others |
classification of different behaviors |
#13: function families |
vertical, horizontal stretches and shifts |
recognizable patterns, behavior |
image of deformation |
increasing/decreasing at increasing/decreasing rate; periodicity; extrema; asymptotes;zeros |
intervals, classification |
modeing of characteristics of real-world systems |
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#15: rate of change |
dynamic behavior |
average or instantaneous |
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#16: slope |
rise / run |
rate of change |
average or instantaneous |
nested square brackets and commas |
vector to plot or to analyze |
graphical images, objects for analysis |
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#18: DERIVE plot |
plot window, set ranges, include axes if desirable |
graphical image |
observe characteristics, estimate rates, zeros, etc. |
#19: DERIVE fit |
fit( [vbl, function], data) |
function model |
analytical function |
#20: DERIVE equation |
expression1 = expression2 |
solvable equation |
equation with meaningful solution |
plot left side, plot right side |
intersection points |
two intersecting(?) graphs |
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plot (left side - right side), locate zeros |
zeros are solutions |
graph through horizontal axis |
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| #23: linear function model | fit best straight line to points; pick two points and plug into y = mx + b; solve system of 2 eqns. | m is slope, b is y intercept | use points on line, not data points; for accuracy don't choose points close together |
| #24: average rate of change of a function between points | determine coordinates of points; average rate is rise / run | average rate often denoted `dy / `dx | rate units are 'y units / x units', or 'y units per x unit' |
| #25: slope = slope equation of a straight line given two points | slope between points = slope between one point and (x,y) | (y - y1) / (x - x1) = (y2-y1)/(x2-x1) | can be solved for slope -intercept form y = mx + b |
| #26: equation of a straight line by simul. eqns. given two points | substitute coordinates into std. form y = mx + b, solve eqns. for m, b | eliminate b by subtracting equations | easiest if y-intercept is one of the points |
| #27: approx. equation of a linear fit to a data set | use two points on approx. best-fit line | use slope = slope or simul. eqns. | best-fit minimizes `sqrt(ave. squared deviation) |
| #28: DERIVE best fit of linear function | FIT( [x, mx+b], #1) if data set in line 1 | [x,mx+b] gives vbl, form | highlighted data set can also be entered with F3 |
| #29: basic points, linear fn. y = mx + b | y-int (0,b); (1, b+m) 1 unit to right | x-int. (-b/m,0) | run of 1 => rise of slope * run = m * 1 = m |
| #30: interpretation of rate | rate is change in dependent vbl. quantity / change in indep. vbl quantity | units of dep. vbl. / units of indep. vbl. | units of dep. vbl. per unit of indep. vbl. |
| #31: difference equation for linear function | a(n+1) = a(n) + c a(0) = b |
a(1) = a(0) + c a(2) = a(1) + c etc. |
sequence is c, b+c, b+2c, . . . , b+n*c, . . .; graph of a(n) vs. n is linear |
| #32: rate equation for linear function | `dy / `dt = constant | constant slope | in limit as `dt -> 0 we write dy / dt |
| #33: sequence behavior of linear function | seq: eval. fn. at regular intervals -> 1st difference is constant | subtract prev. number from each number | if regular interval is `dx, then 1st diff. for y = mx + b is m * `dx |
| #34: geometric proportionalities | square area prop. x^2 cube volume prop. x^3 |
area = k x^2 vol. = k x^3 |
x is any linear measure |
| #35: proportionalities of geom. similar objects | area prop. x^2 volume prop. x^3 |
area = k x^2 vol. = k x^3 |
area covered by tiny squares; vol. filled by tiny cubes; similarity -> uniform inflation of squares or cubes |
| #36: determining prop. eqn. from data by finding k | solve y = k x^p for k, using data pts. (x,y) | if k same for all data pts, prop. is exact | if k nearly same for all data pts., prop. is good |
| #37: determining best prop. from data using DERIVE | fit([x,k*x^p], #1) for different values of p; determine best p | best p gives close fit with random residuals | use geometry or other insights to predict best p |
| #38: ratios implied by proportionality | if y = k x^p then (y1 / y2) = (x2 /
x1) ^ p; and x = k y^(1/p) so (x2 / x1) = (y2 / y1) ^ (1/p) |
(y2/y1) = (k*x2^p)/(k*x1^p) = (x2/x1)^p | k of y = k x^p different than k of x = k y^(1/p) (k value of one is 1/k^(1/p) of the other) |
| #39: area to volume and volume to area ratios for geom. sim. figs. | area = k * vol^(2/3) vol = k * area^(3/2) |
area = kA * x^2, vol = kV * x^3 -> given prop. | k for area prop. different than k for volume prop. |
Plug y vs. t data points (t1, y1), (t2, y2) and (t3, y3) into form y = a t^2 + b t + c to obtain 3 equations
## a + ## b + c = ##
## a + ## b + c = ##
## a + ## b + c = ##
to solve for the three parameters a, b, c.
Example: set up 3 simul. linear equations
Data points (2, 5), (5, 3), (9, 7) plugged into a t^2 + b t + c = y yield system
a * 2^2 + b * 2 + c = 5
a * 5^2 + b * 5 + c = 3
a * 9^2 + b * 9 + c = 7
which simplifies to
4 a + 2 b + c = 5
25a + 5 b + c = 3
81a + 9 b + c = 7
which can be solved for a, b and c.
A system of 3 simultaneous equations in 3 variables can be reduced by elimination to determine whether the system has no solution, a single unique solution, or an infinite number of solutions.
If the system is the result of plugging three data points with distinct t values into the form y = a t^2 + b t + c, we obtain a system of the form
## a + ## b + c = ##
## a + ## b + c = ##
## a + ## b + c = ##
which has a single unique solution for a, b and c.
Example: eliminate variables
The system
4 a + 2 b + c = 5
25a + 5 b + c = 3
81a + 9 b + c = 7
of 3 simultaneous linear equations in 3 unknowns can be reduced to a system of two equations in two unknowns by subtracting the first equation from the second, then from the third, to
21a + 3b = -2
77a + 7b = 2
from which b can be eliminated and the value of a determined by multiplying the first equation by -7 and the second by 3 then adding:
-147a - 21b = 14
241a + 21b = 6
94a = 20, so a = 20/94
The value of b can be obtained by substituting this value of a into one of the two-variable equations.
The value of c can be obtained by substituting the values of a and b into one of the original equations.
It is a good idea to check the solution by substituting into the other two equations to obtain identities.
The deviation of a mathematical model from a data point is defined as the absolute value of the difference between the observed value and the value predicted by the model. If for example we have a function model y(t) of water depth y vs. clock time t and an observed data point, say, (11.3, 7.0), and when we evaluate y(t) for t = 11.3 would find that the model predicts a depth y = 70.6, then the deviation is the difference .6 between the observed depth 70.0 and the predicted depth 70.6.
Had the model predicted depth y = 69.4, the deviation would still be .6. It doesn't matter which is higher, the observed value or the predicted value, since we're taking the absolute value of the difference. The deviation, in other words, will always be positive.
If we have a set of observed data, then we can calculate the average of the deviations to obtain some measure of how close our model is to the data. Often and even better indication is obtained by first squaring in the deviations, then averaging the squares then taking the square roots. This measure is called the standard deviation and is more closely related to standard statistical measures.
The average deviation (or more often the standard deviation) is one measure of the quality of our model. Another measure, often more important, is the pattern formed by the residuals.
Residuals are much like deviations, except that they can be positive or negative. We obtain the residual by subtracting the value of the observed the data point from the value predicted by the model. So for the case for the observed data value is 70 and the prediction is 70.6, the residual would be 70.6 - 70 = .6; had the predicted model been 69.4, the residual would be 69.4 - 70 = -.6.
We often graph the residuals and simply look at the pattern they form. If the residuals seem to follow a definite pattern, for example steadily increasing or decreasing, or decreasing to a minimum value and then increasing, this is an indication that our model could probably be improved. Perhaps a different type of a function will be necessary, or perhaps the same type of function will be appropriate if the parameters are modified somewhat.
Ideally the residuals will be randomly distributed. A random distribution of residuals is often a good indication of the appropriateness of a model.
If the graph of the basic parabola y = x^2 is stretched uniformly in the x direction, then shifted to the right or left, or up or down, the graph will remain a parabola with its axis of symmetry parallel to the y axis. Any such parabola is the graph of some quadratic function y = a t^2 + bt + c, and any quadratic function y = a t^2 + b t + c has a graph which is such a parabola.
A parabola of this type has its lowest point (or its highest point, if it is 'upside down') on its axis of symmetry. This point is called its vertex. If the parabola and its function are a model of a real world y vs. t, then the vertex indicates the time t and the value y at which this maximum or minimum value occurs.
The quadratic formula is useful when we graph a parabola. It tells us whether or not the parabola passes through the horizontal axis and, if so, at what point or points. It also tells us the horizontal coordinate of the vertex of the parabola (see below for details).
The quadratic formula states that
if y is given by the function y = a t^2 + b t + c, then y = 0 for t = [-b +- `sqrt(b^2 - 4 * a * c] / (2 * a), and only for those values of t.
Depending on whether the quantity b^2 - 4 * a * c, called the discriminant, of which the square root is taken is positive, zero or negative, there will be either 2, 1, or 0 such values. The reason for this is fairly obvious. If the discriminant is negative then the square root is not defined, so the expression makes no sense. If the discriminant is positive then we have a square root to either add or subtract from the quantity -b in order to obtain the numerator of our solution, thereby obtaining to solutions. If the discriminant is 0 then its square root is 0 and the numerator is simply -b, giving us exactly one solution.
Examples
The quadratic function y = 2 t^2 - 8 t + 3 is of the form y = a t^2 + b t + c for a = 2, b = -8 and c = 3. We therefore have y = 0 when t = [-(-8) +- `sqrt( (-8)^2 - 4 * 2 * 3] / (2 * 2) = [8 +- `sqrt(40)] / 4 = (8 +- 6.3) / 4 (approx.), so we have the two zeros t = 1.7 / 4 or t = 14.3 / 4 (approx).
The exact solutions of the above are [8 +- 2 `sqrt(10)] / 4 = 2 +- `sqrt(10)/2, using the fact that `sqrt(40) = `sqrt(4 * 10) = `sqrt(4) * `sqrt(10) = 2 * `sqrt(10).
We conclude that the graph of the function y = 2 t^2 - 8 t + 3 will pass through the t axis at the two t values found as solutions of the equation 2 t^2 - 8 t + 3 = 0.
The equation 2 t^2 + 12 = 8t + 9 can be recognized as a rearranged form of the equation a t^2 + b t + c = 0. The original equation is therefore quadratic equation. We can put into the standard form, with 0 on the right-hand side, if we subtract the 8t + 9 from both sides. When we do we obtain the equation 2 t^2 - 8t + 3 = 0, which was solved in the previous example.
The vertex of the parabolaThe equation -2t = 4t^2 + 10 is a disguised quadratic equation. Rearranged the equation becomes 4 t^2 + 2t + 10 = 0, with solutions t = [-2 +- `sqrt(2^2 - 4 * 4 * 10] / (2 * 4). The quantity of which we are taking the square root is 2^2 - 4 * 4 * 10 = - 156; there is no real square root of this number so we have no real solution to the equation. (Note that there will be a complex solution, which will be dealt with later). The graph of y = 4 t^2 + 2t + 10 will therefore not pass through the t axis.
Since a parabola y = a t^2 + b t + c is symmetric a vertical line through its vertex, then if it has zeros its vertex must lie on a vertical line halfway between the zeros (more correctly, equidistant from the zeros). A little consideration shows that the value of t halfway between the two values t = b + - z is b. It follows that halfway between the two values t = [-b +- `sqrt(b^2 - 4 * a * c] / (2 * a) is tVertex = -b / (2 * a). This is therefore the t coordinate of the vertex. The y coordinate is easily found by substituting the t coordinate into y = a t^2 + b t + c.
Examples:
The quadratic function y = 2 t^2 - 8 t + 3 has zeros at t = (8 +- `sqrt(40) ) / 4, so its vertex has t coordinate halfway between these two values at t = 8 / 4 = 2. The y coordinate is found by substituting t = 2 into the original function, obtaining y = 2 * 2^2 - 8 * 2 + 3 = 8 - 16 + 3 = -5. The vertex therefore lies at the point (2, -5).
The Effect of the parameter aThe quadratic function y = 4 t^2 + 2t + 10 has zeros at t = (-4 +- `sqrt(-156) ) / 8. Though we can't take the square root of the negative number, we can still think of the vertex as having t coordinate -4 / 8 = -.5. The white coordinate of the vertex is therefore found by substituting -.5 into the definition of the function, obtaining y = 10.
A parabola defined by a function y = a t^2 + b t + c at can be fat or thin, upright (opening upward) or inverted (opening downward). All these things are controlled by the value of the parameter a, and can be understood by considering the graph of the parabola defined by y = a t^2.
A table for the function y = a t^2 will contain the point (0,0), which is the vertex, and the points (-1,a) and (1,a), each one unit to the right or left of the vertex. If the value of a is large the parabola will go up or down from the vertex quickly, resulting in a 'thin' parabola, while a small value of a will result in a 'fat' parabola that goes up or down from the vertex slowly. Whether the parabola goes up or down depends on whether a is positive or negative.
The graph of the function y = a t ^ 2 can be shifted vertically or horizontally, in order to locate its vertex at any desired point. The vertical shift is accomplished by vertically raising every point of the y = a t ^ 2 graph by the same desired distance. If the distance is k, then every y coordinate is thereby increased by c. The vertex will be moved from (0,0) to (0, k), and the graph represents the new function y = a t ^ 2 + k.
A horizontal shift of h units is accomplished by displacing every point of the graph k units in the horizontal direction, thereby moving the vertex of the y = a t ^ 2 + k from (0, k) to (h,k).
We see that we can thereby control the location of the vertex by vertical and horizontal shifts, and the 'width' of the parabola by the vertical stretch factor a. The resulting parabola is to graph of the function y = a (t-h) ^ 2 + k.
If for example we wished to construct a parabola with vertex at (2, 3), and such that a horizontal displacement of 1 unit from the vertex result of vertical displacement of .2, we would first stretch the parabola by factor .2, obtaining the graph of y = .2 t ^ 2, then shift the parabola vertically 3 units to obtain the graph of y = .2 t ^ 2 + 3, and finally shift the parabola horizontally 2 units, obtaining the graph of y = .2 (t-2) ^ 2 + 3.
We note that this function can be put into y = a t^2 + b t + c form by expanding the square and simplifying. We obtained y = .2 (t-2) ^ 2 + 3 = .2 (t^2 - 4t + 4) + 3 = .2 t^2 - .8 t + 3.8, with a = .2, b = -.8 and c = 3.8.
The graph of any parabola can be modeled by a function y = a t^2 + b t + c.
The parameter a is the rise of the graph corresponding to a run of 1, starting at the vertex. The x coordinate of the vertex is -b / (2a), and c is the y intercept. Using this information we can construct a function model of any parabola.
Alternatively, the form is y = a ( t - h) ^ 2 + k, where a used as before and (h,k) are the coordinates of the vertex.
If we know the function y = f(t) which relates to values of y to the values of t, we can easily determine the value of y corresponding to any value of t, are we can find the value or values of t corresponding to a given value of y.
We imagine a graph of y = f(t). A given value of t can be located on the horizontal axis. From this point we project vertically to the graph then horizontally to the y axis, obtaining the corresponding value of y. For a given value of y, we locate the value on the vertical axis then project horizontally in both directions to find all points at which the projections intersect the graph. From each such point we project vertically to the horizontal axis to obtain our values of t.
y = f(t); plug in known t to get y; plug in known y, solve for t
If we know the function y = f(t), we easily find y from given t of plugging t into the formula for f(t). For example, if y = f(t) = a t ^ 2 + b t + c, then we can determined the value of y corresponding to a given value of t by substituting t and simplifying the result.
If we know the value of y, then substituting y into the expression y = f(t), we obtain an equation which can be solved for t, either algebraically or numerically. For example if we know y, the function y = a t ^ 2 + b t + c gives us a quadratic equation when we substitute the value of y. We easily solve this equation for t, using the quadratic formula.
f(x) notation allows us to talk and think about functions without referring to specific examples or resorting to cumbersome sentences and paragraphs. For example, we can make a statement like 'when we vertically stretch graph of a function by factor a, the resulting function is obtained by multiplying right-hand side of the expression for the y = ... function by a', which makes sense but requires a lot of reading, or we can say that 'the graph obtained by vertically stretching the graph of y = f(x) by factor a user graph of y = a f(x)', which is much more specific and concise. We could even abbreviate the statement, e.g. 'y = f(x) stretched vertically by factor a gives y = a f(x)'.
For example, consider the function y = f(x) = a x^2 + b x + c. In the expression f(2) we have substituted 2 for the x of the original definition, so we retain the substitution throughout the expression defining the function, obtaining y = f(2) = a (2^2) + b(2) + c. We can also substitute symbolic expressions for x, obtaining, e.g., f(x + `dx) = a (x + `dx) ^ 2 + b ( x + `dx) + c.
The expressions we get when we substitute can of course be simplified and manipulated algebraically.
The four basic functions studied in this course are the linear, quadratic, exponential and power functions y = x, y = x^2, y = 2 ^ x and y = x ^ p for various powers p. By stretching and shifting transformations these functions can be manipulated into the members of the following families: linear functions y = m x + b; quadratic functions y = a t ^ 2 + b t + c or y = a(t-h)^2 + k; y = A b^x + c or y = A * 2 ^ (kx ) + c; and power functions y = A (x-h) ^ 2 + k.
The linear function family y = mx + b can be obtained by applying vertical stretches and vertical shifts to the basic y = x function. The vertical stretch factor is m and the vertical shift is b. For constant b the graph of the resulting function family consists of all possible straight lines passing through the y intercept (0, b). For constant m the family consists of all possible straight lines with slope m; these lines will all be parallel.
The quadratic function family y = a (x-h) ^ 2 + k is obtained by applying vertical stretches and both horizontal and vertical shifts to the basic y = x ^ 2 function. If h and k are held constant the graph of the family consists of all possible parabolas with vertex at (h, k), using all possible vertical stretch factors a. The parabolas will be fat and skinny, upright and inverted. If a and h are held constant, we will end up with a stack of identical parabolas with vertices running up then down the vertical line x = h, and with the 'spread' of the parabola determined by the vertical stretch factor a. If a and k are held constant we end up with a role of identical parabolas with vertices running across the horizontal line y = k.
The exponential family y = A * 2 ^ (kx) + c is obtained by vertically stretching the basic y = 2 ^ x function by factor A, then stretching the resulting graph horizontally by factor 1/k, and finally by vertically shifting the graph c units. The y intercept of the basic exponential graph at (0, 1) is vertically stretched to (0, A); the horizontal stretch by factor 1/k either increases or decreases the steepness of the graph at every vertical point by factor k; the asymptote y = 0 is shifted vertically c units so that y = c is the horizontal asymptote and the y intercept shifts to (0, A+c).
If k and c are held constant, we obtain to family of exponential graphs with common horizontal asymptotes at the line y = c, with steepness determined by k, and with y intercepts varying over all points of the y axis. If A and k are held constant, we obtain a 'stack' of exponential graphs with asymptotes at every possible line y = c, each with y intercept A unit above the asymptote, and with steepness determined by k. If A and c are held constant, we obtain all possible parabolas with horizontal asymptote y = c and y intercept A units above the horizontal asymptote, with all possible steepness factors k.
The power function families y = A (x - h) ^ p + k have different characteristics depending upon the power p.
For positive even powers p, the graphs resemble parabolas which for p > 2 are progressively flattened near the axis of symmetry and steeper a way from the axis of symmetry. The vertical stretch factor A acts as it does with a parabola, and the 'vertex' (for p > 2 we don't call the point on the axis of symmetry a vertex) is shifted to the point (h,k).
For positive odd powers p, the graphs have an axis about which they are antisymmetric, and a point of inflection on this axis. The graphs are progressively flattened near the point inflection for increasing values of p and steeper away from the point inflection. The vertical stretch A acts in the usual manner, and the horizontal and vertical shifts plays the point of inflection at (h, k).
For negative powers p, the graphs will have vertical asymptotes on either side of the line x = h, and horizontal asymptotes y = k. For even negative powers p, the graphs will be symmetric about the line x = h; for odd negative powers p the graphs will be antisymmetric about this line.
The graph of function can be characterized in many ways. Some of the primary characteristics are the increasing and decreasing behaviors of the graph, the increasing in decreasing behaviors of the rate of increase or decrease, and asymptotes.
The graph is increasing if it rises as we move from left to right, decreasing if it falls as we move from left to right.
An increasing graph can increase faster and faster, slower and slower, or at a constant rate. We can thus say that a graph can be increasing at an increasing, a decreasing or a constant rate. An increasing graph that increases at increasing rate is said to be concave upward; if it increases added decreasing rate it is said to be concave downward.
The decreasing graph can decrease faster and faster, slower and slower or at a constant rate. We can the say that a graph can be decreasing added increasing, a decreasing or constant rate. A decreasing graph that decreases at an increasing rate is said to be concave downward; if it decreases added decreasing rate it is said to be concave upward.
A graph that increases or decreases at a constant rate is said to be linear.
Any straight line approached by a graph more and more closely, coming as close as we could wish but never touching the line, is said to be a horizontal asymptote for that graph.
When a quantity y changes in response to changes in a quantity x, then the average rate at which y changes with respect to x, between any two values of x, is obtained by dividing the change in y by the change in x. We denote the change in y by `dy (standing for 'delta' y), and a change in x by `dx, so that the rate is rate = `dy / `dx.
Change of dependent vbl / change of independent vbl
When y changes in response to changes in x, we say that y is the dependent and x the independent variable. So the average rate is the change in the dependent divided by the change in the independent variable.
The basic example of a rate of change is the depth function, where y = f(t) = a t^2 + b t + c is the quadratic function that represents depth y as a function of the independent variable t, the clock time. The change `dy in y is understood to be the change in depth, usually in cm, while the change `dt in t, usually measured in seconds, is the time interval during which the depth makes this change. When we divide the depth change by the time interval, we obtain the average rate, in cm / sec, which the depth changes.
The slope of a graph of y vs. x, between any two graph points, is the ratio rise / run between the points. The rise is the change `dy in the y coordinate and the run is the change `dx in the x coordinate. The slope is therefore slope = rise / run = `dy / `dx. It is clear from this definition, and from the definition of the rate of change, that this slope is the average rate of change of y with respect to x, between the x values represented by the two points.
In the Algebra window (not the one that says Algebra on the command line -- that's the plot window -- but the one that says Author on the command line), choose Author, type and opening bracket, type in each data pair separated by commas within brackets, separating the data pairs by commas, and type the closing bracket.
An example: [ [3, 2], [4, 7], [8,11] ]
DERIVE will plot whatever expression is highlighted in the Algebra window. Choose Plot, which takes you to the Plot window, set ranges if necessary (left-right, bottom-top; tab from choice to choice, ALT-Tab to go backwards), use the Plot command when ready.
To fit a quadratic function to data in, for example, line number 17, the command would be fit( [x, ax^2 + bx + c], #17). The first x tells DERIVE what the variable is going to be; otherwise it wouldn't know whether a, b, c or x was the variable. You have to use brackets and not parentheses when giving DERIVE the variable and the function. If the data you want is highlighted you can use the F3 key instead of typing '#17'.
You can fit any function as long as the parameters of the function don't appear within exponents, square roots, or other functions.
Before plotting the fit function you have to simplify or approximate it. Otherwise you'll get an error saying 'too many variables' or something like that.
An equation consists of two possibly variable expressions separated by an equal sign. You can solve the equation by choosing soLve; if there is more than one variable you can choose which one to solve for (DERIVE will prompt you). You can solve in the Exact or the Approximate mode. If you solve in the Exact mode you will get every possible solution. If you solve in the Approximate mode you will get only the first solution which is greater than the lower bound of the interval over which you choose to search for solutions.
A DERIVE equation can be solved approximately by graphical means. You can highlight the left-hand side of the equation and plot it, then highlight the right-hand side and plot it. Any point where the two plots cross provides a solution, since the values of the two sides will clearly be the same data point where the graphs cross.
Solving equation by finding zerosThe equation (left side) = (right side) can be rearranged to give the equation (left side) - (right side) = 0. This is for example what we typically do when we encounter the quadratic equation. To solve this equation graphically we simply plot the expression corresponding to (left side) - (right side) and see where it crosses the horizontal axis, since the horizontal axis corresponds to a value of zero.
When a set of data seems to indicate a linear pattern, we attempt to fit it with the straight line that comes as close is possible on the average to the data points (actually we attempt to minimize the average of the squares of the vertical distances, but that's a topic for another course). We then take two points, not to close together so the small errors in estimation don't get magnified, and plug them into the form y = mx + b. We obtain two equations in the variables m and b; these equations are easily solved. We plug the values found for m and b back into the for y = mx + b and we have our model.
To find the average rate of change of a given function between two values of its independent variable (e.g., between two values of clock time t for a depth vs. clock time model), determine the corresponding values of the dependent variable (e.g., the values of y(t) = depth for a depth vs. clock time model). Plot the points and find the rise from the first point to the second (corresponding to the change in the dependent variable, `dy in the case of the depth vs. clock time model) and the run from the first point to the second (corresponding to the change in the independent variable, `dt in the case of the depth vs. clock time model). The average rate is the change in the dependent variable divided by the change in the independent variable (change in depth divided by changing clock time in the case of the depth vs. clock time model). This average rate will be in units of the dependent variaable divided by units of the independent variable (e.g., cm / sec for the depth vs. clock time model).
The slope = slope equation of the straight line between two given points is obtained by setting the slope between the two points equal to the slope between the one of the points and a general point (x,y). If the two given points are (x1, y1) and (x2, y2), then the slope between them is (y2 - y1) / (x2 - x1). The slope between the first of the given points and the general point is (y - y1) / (x-x1). Setting these slopes equal we obtain the equation (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1). If we first multiply both sides of this equation by (x-x1), then add y1 to both sides we will obtain the slope-intercept form of the equation.
We can obtain the equation of the straight line from two points on its graph by substituting the coordinates of each point into the linear form y = mx + b. We obtain a system of simultaneous equations. For example, if the points are (x1, y1) and (x2, y2), we obtain the two equations y1 = m x1 + b and y2 = m x2 + b. It should be clear that when the second equation is subtracted from the first, we eliminate b and obtain an equation we can then solve for m. Having obtained m, we can substitute it back into either of the equations and easily solve for b.
If we have a set of data which can reasonably be fit by a straight line, we can find the approximate equation of the best fit by first sketching a straight line which comes as close, on the average, to the data points. We then find coordinates of two points on this straight line, not too close together, and from these coordinates determine the approximate equation of the line.
The actual best-fit linear function, which is the one given by DERIVE, is the one whose graph minimizes the square root of the average of the deviations between the function and the data point. In this course we will not concern ourselves with techniques for finding this equation, but will simply think of the best-fit linear function as the one which comes as close as possible, on the average, to the points.
The DERIVE command fit( [x, mx+b], #1) tells the computer to return the equation of the best-fit line, specified by the slope-intercept expression mx + b, of the linear function whose graph comes as close as possible to the graphs of the data points in line #1. Closeness here is measured by the square root of the average of the squared deviations between the graph of the linear function an the data points.
The basic points of the linear function y = mx + b are taken to be the y intercept (0, b) (which is the point where x = 0), and the point one unit to its right. Moving one unit to the right, the run is 1; since the slope is rise / run = m, it follows that the rise must be m. Thus the y coordinate of the point will be b + m and, since the x coordinate is clearly 1, the point will be (1, b + m). We note that this point is higher than the y intercept when m is positive, and below the y intercept when m is negative.
We note also that the x intercept of the function y = mx + b occurs when y = 0, and is therefore found by solving 0 = mx + b for x. We obtain x = - b/m.
To interpret the average rate of change between two points on a graph, we note that the rate is the change in the quantity represented by the dependent variable (the rise) divided by the change in the quantity represented by the independent variable (the run). When these quantities are divided we obtain the number of units of change in the dependent variable per unit of change in the independent variable. In our first flow model, for example, we obtained the number of centimeters of depth change per second. A second is the unit of change of the independent variable, and centimeters represent the units of change of the dependent variable.
The difference equation a(n+1) = a(n) + c, a(0) = b, is evaluated for n = 0, 1, 2, 3, . . .. Starting with n = 0, we obtain a(1) = a(0) + c; since we know that a(0) = b, we obtain a(1) = b + c. Continuing with n = 1, we obtain a(2) = a(1) + c; since we now know that a(1) = b+c, we see that a(2) = (b+c) + c = b + 2c. Then n = 2 tells us that a(3) = a(2) + c; having found that a(2) = b + 2c we see that a(3) = b + 3c.
The clear pattern that emerges is a(0) = c, a(1) = b+c, a(2) = b + 2c, a(3) = b + 3c, . . . , a(n) = b + n * c.
It is clear that this pattern is linear, changing by the same amount for each subsequent value of n.
The average rate at which a function changes is given by `dy / `dx. This rate is represented by the average slope between two points of the graph of the function. If the function is linear, then the slope is always the same, so we have `dy / `dx = constant.
If we speak of the instantaneous rate which a function changes, we mean the rate at which it changes at a point of its graph, as the run of a series of slope triangles originating at that point shrinks down toward 0. We then no longer speak of the average slope or average rate, but the instantaneous rate. This rate is designated dy / dx. It should be clear that for a linear function, this rate is again always the same, so we right dy / dx = constant.
These equations, `dy / `dx = constant and dy / dx = constant, are called rate equations. The first is an average rate equation and the second is an instantaneous rate equation. Rate equations (later called differential equations) play an extremely important role in mathematics, since they allow us to express the ways in which rate change, and since these equations can often be solved to find important function models. Solving these equations requires the use of calculus, and we will not solve them in this course. We will, however, look at how to write certain important rate equations.
If a linear function is evaluated at regular intervals, the resulting sequence will have the property that its first difference is constant. (Recall that the first difference of sequence is obtained by subtracting each member the sequence from the member following it). If the regular interval at which the function y = mx + b is evaluated is `dx, then this difference is always m(x+`dx) + b - (mx + b) = m `dx.
If the length of the side of the square is doubled, we obtain a square large enough to contain 4 of the original squares (two rows of two squares each). If the side of cube is doubled, we obtain a cube large enough to contain 8 of the original cubes (two layers each containing to rows of two cubes each).
As the side x of a square is changed, its area is A = x^2. As the side x of a cube is changed, its volume is V = x^3.
If x is any linear dimension of a square, that is, any dimension that can be measured by a ruler (such as its perimeter or its diagonal), then the area of the square is A = k x^2, where k is a constant that depends on which linear dimension is being measured. We say that the area of the square is then proportional to the square of the linear dimension.
If x is any linear dimension of cube, then the volume of the cube is V = k x^3, were k is a constant depends on which linear dimension is being measured. We say that the volume of the cube is then proportional to the cube of the linear dimension.
If two objects are geometrically similar, then if x is any linear dimension of the corresponding class of objects (e.g., the diameter of the circle, the circumference of a sphere, the radius of the base of a cone whose height is equal to its diameter), the surface of the second object can be 'tiled' as accurately as desired by tiny squares in exactly the same way as the first, with the sides of the squares in the same proportion is the linear dimensions of the two objects. Furthermore, the volume of the second object can be filled as accurately as desired by tiny cubes in exactly the same way of the first, with the sides of the cubes in the same proportion is linear dimensions of the two objects.
It follows that for any class of geometrically similar objects, area is proportional to the square of a given linear dimension and volume is proportional to the cube of this dimension. That is, area = k x^2 and volume = k x^3.
If we know that two quantities y and x are exactly related by proportionality y = k x^p (e.g., A = k x^2 or V = k x^3), then one data point (x1,y1) suffices to find the value of k. We merely plug in the values x1 and y1, obtaining y1 = k x1^p, and solve for k. (The result is k = y1 / x1^p). If we then substitute this value of k into the form y = k x^p, we obtain a specific function relating y to x.
If the proportionality y = k x^p is not exact, we can find a reasonably good approximation to the best value of k by substituting every data point into the form to obtain a value of k for that data point. We then average the values of k for all the data points. (A better alternative is to use DERIVE or another computer algebra system to find the best y = k x^p fit to the data).
To determine the best y = k x^p proportionality for the data set by using DERIVE, we use the syntax FIT([ x, k x^p], #1) for different values of p. Using geometry or other insights to predict the best value of p, we try various values in the neighborhood of the predicted best value until the graph of the resulting fit function best matches the graph of our data. An optimal fit will lie close to the curve and will give random residuals.
If y = k x^p, then y2 / y1 = (k x2^p) / (k x1^p) = (k/k)(x2^p / x1^p) = x2 ^ p / x1 ^ p = (x2 / x1) ^ p.
If y = k x^p, then x^p = y / k and x = (y/k) ^ (1/p) = (1/k)^(1/p) * y^(1/p). Since k is constant, (1/k) ^ (1/p) is also constant so we might as well call it k. This k has a different value than the k of y = k x^p, but it is still constant. We can thus say that the proportionality of x to y is x = k y^(1/p).
Since we have x = k y ^ (1/p), we have x2 / x1 = (k y2^(1/p)) / (k y1^(1/p)) = ... = (y2 / y1) ^ (1/p), where the intermediate steps are identical to those used previously.
Thus, given the ratio of two x values we can obtain the ratio of the corresponding two y values, and given the ratio to y values we can obtain the ratio of the corresponding x values.
For most real objects the surface areas of geometrically similar objects are proportional to the square of linear dimension, and volumes are proportional to the cube of linear dimension. We could write this as area = kA x^2, and volume = kV x^3, where kA and kV are understood to be different proportionality constants.
To get the proportionality of area to volume, we can solve the volume proportionality for x, obtaining x = (volume / kV) ^ (1/3), and then substitute this into the area proportionality. We will obtain area = kA * [ (volume / kV) ^ (1/3) ] ^ 2 = kA * (volume / kV) ^ (2/3) = (kA / kV^(2/3) ) * volume ^ (2/3). The expression (kA / kV ^ (2/3) ), which since kA and kV are constant must also be a constant, can simply be called k, and we have the proportionality area = k (volume) ^ (2/3).
We can follow similar procedure, this time solving the area proportionality for x and substituting into the volume proportionality, to obtain volume = k (area) ^ (3/2), where the value of k is different than in the previous proportionality.
It will follow that for two geometrically similar figures, the ratio areas for two given volumes will be area 2 / area1 = (volume2 / volume1) ^ (2/3), while the proportionality of volumes for two given areas will be volume2 / volume1 = (area2 / area1) ^ (3/2).