Introduction - The What and Why of Calculus

Calculus is the study of how functions [i]change[/i]. In Physics, you learned that an object whose position is changing is moving, and therefore has velocity (speed with direction). If it moves farther in a given amount of time, it is going faster than if it moves less in the same amount of time. Conversely, we know that if an object is moving at a certain speed for a given amount of time, we can calculate the change in its position (distance traveled). So not only can we relate [i]change in position[/i] to [math]speed[/math], we can reverse the process and find out about [math]position[/math] when we know [math]speed[/math]. Calculus ties these two things together and gives us tools to analyze them in both "directions".[br][br]On this app, there are several scenarios to explore that will help illustrate the basics of what calculus is all about. The [color=#c51414][b]red[/b][/color] graph represents the object's position during the time interval from 0 to 10 seconds. The [b][color=#1551b5]blue[/color][/b] graph represents the velocity of the object. Remember that "velocity" is the same as the [i]rate of change[/i] of the position: an object moving at a [i]speed[/i] of 3 m/s is [i]changing its position[/i] at a [i]rate[/i] of 3 m every second.[br][br]You can change the time by sliding the slider manually, or you can animate it by clicking the "START" button.[br][br]The first position function is a constant, 3 m. If the object stays at a position of 3 m for all time, then it is not moving, and its velocity is zero. (Check the "Show Velocity" box to see this). Here is how the two graphs connect. If we look at the [i]slope[/i] of the position function, we get the [i]height[/i] (value) of the velocity function. Since the position is not changing, its slope (and thus its velocity) is zero. We can determine the object's position from the velocity graph by looking at the [i]area under[/i] the velocity graph. Why? Let's look at the basic formula relating speed and distance: distance = rate [math]\times[/math] time. If the height of the velocity graph is the rate, and the width is the time, then the area is rate [math]\times[/math] time! And as for the position graph, we have rate = distance [math]\div[/math] time, which is rise over run, which is slope.[br][br]Notice however that the velocity cannot give us the actual [i]position[/i] - it can only give us the [i]change[/i] in position since some point in time. Thus, when you look at the second function in the list (constant 5 m/s), the velocity function is the same as it was for position = 3 m. We would have to be told that the object [i]started[/i] at position 5 m, and its position [i]changed[/i] by 0 m during whatever period of time. Thus, when we find a quantity from its rate, we always need to know its [i]initial quantity[/i], otherwise all we can calculate is the amount of the quantity's [i]change[/i].[br][br]Move on to the third function, [math]0.6x[/math] ([math]x[/math] is really time "[math]t[/math]"). In this case, the object is moving steadily from position 0m to position 6m after 10 sec. Since this motion is a linear function, we can easily calculate the slope as 6m per 10 sec, or 0.6 m/s. The short blue "[math]m[/math]" slope segment on the position graph shows this. As time moves on from zero, the object's position changes, and we see that it has velocity. How much velocity? 6 m every 10 sec, or 0.6 m/s. The velocity is constant, so its graph is a horizontal line at a height of 0.6 m/s.[br][br]Now let's consider the area under the velocity graph. Its height is 0.6 m for all time [math]t[/math]. Thus, the amount of distance it moves in any time [math]\Delta t[/math] is 0.6 m/s [math]\times[/math] [math]\Delta t[/math] sec = 0.6[math]\Delta t[/math] m. This gives us the change in position during this time as "height" or "rate" (0.6 m/s) times "width" or "time" ([math]\Delta t[/math] sec), or area under the graph.[br][br]Now try the fourth function, the quadratic. Imagine that this is the path of a ball tossed into the air. You can see that at time [math]t=0[/math] that the object has an initial velocity (slope) of 2.4 m/s. This matches both the [i]height[/i] of the velocity graph and the [i]slope[/i] of the position graph. "Slope" is a bit more complicated now that we're trying to apply it to a curve instead of a line. We use a "tangent line" to the function at a point to tell us the "slope of the curve" at that point. As time goes on, the velocity slows as the ball rises. The position curve flattens out as the slope approaches zero. Then the velocity is zero for an instant, after which the ball begins to fall (with a negative velocity).[br][br]First, note how the slope of the position graph always matches the height of the velocity graph. Then notice how the area under the velocity graph always matches the position of the ball (the ball's initial position is zero). When the ball rises,both its position and its velocity are positive. When the ball falls, its position is still positive (still above ground), but its velocity (slope of position) is negative (since it is moving down).[br][br]Finally, play around with the last two functions. Notice how the velocity is positive whenever the position is increasing, and that position is decreasing whenever velocity is negative. Try to understand the connections between the two graphs. This will really help later on - as things become more technical and theoretical in class, it helps to keep these concrete examples in mind.[br][br]Here is calculus in a brief picture:[br][br][b][color=#c51414]Amount[/color] [math]\rightarrow[/math] slope [math]\rightarrow[/math] [color=#1551b5]Rate of Change[/color][br][color=#c51414]Amount[/color] [math]\leftarrow[/math] area [math]\leftarrow[/math] [color=#1551b5]Rate of Change[/color][/b]

Illustrates how Point-Slope and Slope-Intercept line formulas work.

In calculus, we often use the [i]tangent line[/i] to the function for analysis. As you'll see, usually we can find the slope from the function, and we know the one point where we want to place the tangent line. Thus, the point-slope method of finding the line equation will be used much more than the more-familiar slope-intercept form. This app illustrates how the point-slope formula works and how to change it to slope-intercept if need be.[br][br]We start by noting that the slope of a line passing through any two points [math]P_1=(x_1, y_1)[/math] and [math]P_2=(x_2, y_2)[/math] can be found as [math]m=\frac{y_2-y_1}{x_2-x_1}[/math]. If we multiply this equation on both sides by [math]x_2-x_1[/math], and switch the sides, we get [math]y_2-y_1=m(x_2-x_1)[/math]. But let's allow [math](x_2, y_2)[/math] to be any point [math](x,y)[/math]. This gives us the [b]point-slope formula[/b] [math]y-y_1=m(x-x_1)[/math]. We can then change this to the slope-intercept form by solving for [math]y[/math]: [math]y=mx+(y_1-mx_1)[/math]. Note that the quantity in the parentheses is "[math]b[/math]", the [math]y[/math]-intercept.[br][br]To begin, leave the three check boxes checked. One or more can be turned off to hide parts of the display for clarity.[br][br]The two purple dots are the points [math](x_1, y_1)[/math] and [math](x_2, y_2)[/math]. You can drag them anywhere on the graph. As you do, you can see how the slope is determined, by the difference in the points' [math]y[/math]-values divided by the difference in their [math]x[/math]-values ([math]m=\frac{\Delta y}{\Delta x}[/math]). This is illustrated by the dashed brown lines and the black numbers next to them. In purple, the value of the ratio is shown.[br][br]When we convert to the slope-intercept form, we saw above that [math]y=mx+(y_1-mx_1)[/math]. The part in parentheses gives us [math]b[/math], the [math]y[/math]-intercept value. It says that the [math]y[/math]-intercept will be at the [math]y[/math]-value of [math]P_1[/math], minus a distance given by the slope [math]m[/math] times the [math]x[/math]-coordinate of [math]P_1[/math]. But [math]x_1[/math] is just the "run" distance [math]\Delta x[/math] from the [math]y[/math]-axis to [math]P_1[/math]. Since [math]m=\frac{\Delta y}{\Delta x}[/math], we can solve for [math]\Delta y=m \Delta x[/math]. So we would expect the [math]y[/math]-intercept to be [math]m \Delta x[/math] up or down from [math]P_1[/math]'s [math]y[/math]-value: [math]b=y_1-mx_1[/math]. This is illustrated by the blue and red dotted arrows.

If velocity is the slope of a position graph, how can we find the velocity at a [i]single [/i]point in time, when it takes [i]two [/i]points to make a line?

You may have learned in Physics class that the [i]slope [/i]of a position versus time graph gives us the [i]velocity[/i] of the object. So, if the object moves [math]\Delta d=10[/math] meters in [math]\Delta t=2[/math] seconds, we calculate its velocity as [math]v=\frac{10 m}{2 sec}=5m/s[/math]. But suppose this object is not moving at a constant speed - perhaps it is falling and therefore constantly speeding up. In this case, it is not going 5 m/s for the entire 2 seconds. We could try to get closer to its velocity at a single point in time by measuring [math]\Delta d[/math] for a shorter [math]\Delta t[/math], since we expect less variation in the velocity in a shorter time interval. As we make [math]\Delta t[/math] smaller, our estimate of the instantaneous velocity improves. But as long as [math]\Delta t \neq 0[/math], any such calculation is still an approximation over a time [i]interval[/i], and not the instantaneous velocity at a [i]single point[/i] in time![br][br]At first the answer seems simple - just set [math]\Delta t=0[/math]! But then [math]\Delta d=0[/math] too - the object moves zero meters in zero time. We cannot calculate [math]\frac{\Delta d}{\Delta t}=\frac{0}{0}[/math], because division by zero is undefined.[br][br]In this demonstration, the object's [i]position [/i](its distance from the origin) varies with time according to [math]d(t)=t^3[/math], which is graphed above. We want to know what its velocity is at time [math]t=2[/math] seconds by using the slope at that point. To do this, we'll construct a line (in green) between the point at our target (in blue) and a nearby point along the curve (in red). We'll use the line's slope as our estimate of the velocity. We can adjust the movable point's location with the slider. As we do, we see that [math]\Delta d[/math] and [math]\Delta t[/math] both change, and that the slope [math]\frac{\Delta d}{\Delta t}[/math] of the green line - our estimate of the velocity - changes with it. The closer we bring the movable point to the target point, the closer we get to the velocity at [math]t=2[/math]. But if we set the movable point to exactly the same position as the target, we have a problem.[br][br]But notice that as you position the movable point closer to the target, the slope (velocity) seems to be approaching a certain number. If we approach from the left, the velocity estimate is increasing, and if we approach from the right, it is decreasing. Is it possible that the "real" instantaneous velocity is some number between these values? The answer is "Yes"![br][br]We can write a formula for the velocity estimate [math]\frac{\Delta d}{\Delta t}[/math] and use our calculator to get very close estimates. From this exercise, perhaps we could deduce the "true" instantaneous velocity. We have two points for our line: [math](2,2^3)[/math] for the target point, and [math](x,x^3)[/math] for the movable point, for which time [math]t=x[/math]. Thus the slope (velocity) is [math]\frac{(x)^3-\left(2\right)^3}{(x)-(2)}[/math], using the slope formula. Now, plug in values of [math]x[/math] slightly larger than and slightly smaller than 2 into [math]\frac{x^3-8}{x-2}[/math] and see that "12" seems to be a reasonable result.[br][br]We call this number the [i]limit[/i] of [math]\frac{x^3-8}{x-2}[/math] as [math]x[/math] approaches [math]2[/math], and we write it as [math]\lim_{x \to 2}\frac{x^3-8}{x-2}=12[/math].

Let's look at the Derivative simply for what it is - a function that gives us the slope of another function.

The basic idea of the derivative is actually pretty simple - it's the function [color=#0000ff][math]f'[/math][/color] that gives the [i]slope[/i], rather than the [i]height[/i] ([math]y[/math]-value), of a function [color=#009900][math]f[/math][/color] at each value of [math]x[/math]. One way to think about this is that if the point on the graph of [color=#009900][math]f[/math][/color] at [color=#ff0000][math]x[/math][/color] is [math]([/math][color=#ff0000][math]x[/math][/color], [color=#009900][math]y[/math]-value of [math]f[/math] at [/color][color=#ff0000][math]x[/math][/color][math])[/math], then the point on the graph of [color=#0000ff][math]f'[/math][/color] at [color=#ff0000][math]x[/math][/color] would be [math]([/math][color=#ff0000][math]x[/math][/color], [color=#0000ff]slope of [/color][color=#009900][math]f[/math][/color][color=#0000ff] at [/color][color=#ff0000][math]x[/math][/color][math])[/math].[br][br]As you work with this app, check out Khan Academy's Derivative Intuition Module at [url]https://www.khanacademy.org/math/differential-calculus/taking-derivatives/visualizing-derivatives-tutorial/v/derivative-intuition-module[/url]. This app is based on the same concept, but it allows [b]YOU[/b] to move the dots, and also lets you try many different functions.[br][br]To start with, be sure the "[color=#000000]Show Derivative[/color]" checkbox is cleared ([i]un[/i]checked). The graph of [color=#009900][math]f[/math][/color] is shown in green. Different functions can be investigated by selecting one from the drop-down list at the top right. You can even enter your own "user" function in the "[color=#000000]f(x) =[/color]" box; it will then appear at the bottom of the drop-down list. Now the task is to slide the [color=#ff0000]red dots[/color] up or down so that the [color=#ff0000]red line segment[/color] at that value of [math]x[/math] is [i]tangent to[/i] (matches the slope of) [color=#009900][math]f[/math][/color] at that point. The [math]y[/math]-value of the [color=#ff0000]red dot[/color] is the slope of the corresponding [color=#ff0000]red tangent line segment[/color]. Since the derivative function [color=#0000ff][math]f'[/math][/color] gives us the slope of the original function [color=#009900][math]f[/math][/color], the red dot should therefore lie on the graph of [color=#0000ff][math]f'[/math][/color]. Adjust all the dots so that all the segments are tangent to the green [color=#009900][math]f[/math][/color] graph.[br][br]Now check the "[color=#000000]Show Derivative[/color]" box. The graph of [color=#0000ff][math]f'[/math][/color], the derivative of [color=#009900][math]f[/math][/color], will appear in blue. If you were successful in adjusting the tangent line slopes, your [color=#ff0000]red dots[/color] should lie on or very close to the graph of [color=#0000ff][math]f'[/math][/color].[br][br]A special function is [math]f(x)=e^x[/math]. This is the second-to-last function in the drop-down box. You can drag the graph down and/or use the Zoom buttons to see more points. Make note of the result you get for this function - you'll see this many more times in this course!

Interpreting one-dimensional motion on a two-dimensional graph can be confusing. Let's try to connect a graph with the actual movement of a particle.

When considering [i]rectilinear[/i] (straight-line, one-dimensional) motion, we must always think of the moving object (often a sizeless, massless "particle") traveling along a straight line. It is either at rest (not moving), moving right, or moving left. (Up/down, backward/forward, and other opposite direction descriptors could be used as well). Yet when we look at the various motion graphs of position, velocity, and acceleration, what we see may not connect in our minds with the motion.[br][br][i]Position[/i] refers to the location of the object relative to the origin. If it is located 5 units to the left of the origin, we say its position is [math]-5[/math]. We often use [math]s(t)[/math] or [math]x(t)[/math] to represent the [i]position function[/i]; that is, the function that gives the object's position as a function of time [math]t[/math]. (Geogebra only allows "[math]x[/math]" as a variable when defining functions, but think of it as "[math]t[/math]" in the app above). The blue graph shows the object's position [i]along the [math]y[/math]-axis[/i] as a function of time. Notice that the blue graph is [i]not [/i]a trace of the particle's path in the [math]x-y[/math] plane! If you click RESET and then START, you'll see that the particle moves along the [math]y[/math]-axis, its position at any time [math]t[/math] given by the [math]y[/math]-value of the point on the graph indicated with the cursor.[br][br][i]Velocity[/i] is the particle's speed, with the [i]direction[/i] of travel given by the [i]sign[/i] of velocity. Velocity tells us how far a particle moves in a time period - that is, it tells us the rate of change of the particle's position. As such, velocity is the derivative of position: [math]v(t)=s'(t)[/math]. When looking at the green velocity graph, you must connect the particle's [i]speed[/i], not its [i]position[/i], with the [math]y[/math]-value of the graph. Thus, the particle could be located above the origin, but its velocity could be negative. This only means that the particle's [i]position [/i]is above but its [i]movement [/i]is in the down direction. When the velocity is positive, the particle moves up; when it's negative, the particle moves down. The farther away from the [math]t[/math]-axis the [math]y[/math]-value is, the faster the particle is moving. Thus, speed is determined by how far away from the [math]t[/math]-axis the velocity is: [math]speed=|v(t)|[/math]. The object is not moving whenever [math]v=0[/math] (when the velocity graph meets the [math]t[/math]-axis), and the particle [i]changes direction [/i]when [math]v[/math] [i]changes sign[/i] (when the graph [i]crosses[/i] the [math]t[/math]-axis).[br][br][i]Acceleration[/i] is the particle's change in velocity over time - the rate of change of velocity. Therefore acceleration is the derivative of velocity [math]a(t)=v'(t)[/math], and so it is also the second derivative of position [math]a(t)=s''(t)[/math]. Since we know that [math]\overrightarrow{F}=m\overrightarrow{a}[/math], imagine acceleration as the external force experienced by the particle. Acceleration is used in connection with velocity to determine when a particle is speeding up or slowing down. This is determined by comparing the signs of [math]a[/math] and [math]v[/math]: if they have the same sign, the particle is speeding up; otherwise it's slowing down. Graphically you can compare the sign of the red acceleration graph with the green velocity graph at any time [math]t[/math].

Estimate the area under the curve using rectangles or trapezoids.

Choose the number of subintervals on [math][a, b][/math] by sliding the "[math]n[/math]" slider. Check the "Trapezoidal Sum" box for a trapezoidal approximation; otherwise clear it for a rectangular sum. For a rectangular sum, choose the position of the [math]y[/math]-value within the interval using the "[math]p[/math]" slider.

The Fundamental Theorem of Calculus Part I says [math]f'(x)=\frac{d}{dx}\left [\int_{a}^{x} {f'(t)}{d}t\right ][/math]. This can be a little difficult to navigate, with the [math]x[/math] in the limits of integration and the [math]t[/math] as the variable of integration. This app is intended to help make this concept more concrete using a real-life example.[br][br]On the left is a graph of a [i]rate of change[/i] (derivative). It represents the [b][i]rate[/i] [/b], [math]g'[/math], at which a pump moves water. In this example, the pump starts pumping at a rate of 5 gallons per minute, but continuously slows down as it runs, dropping to 2 gallons per minute somewhere around 9 minutes later. Time for a new pump, I guess![br][br]The amount of water that it pumps in a short time interval [math]\Delta t[/math] is found as the flow rate [math]g'[/math] times [math]\Delta t[/math]. (Since [math]g'[/math] is changing during [math]\Delta t[/math], we would hold constant some value of [math]g'[/math] within [math]\Delta t[/math]). You can think of these as the height ([math]g'[/math]) and the width ([math]\Delta t[/math]) of a rectangle on the left graph, and the amount pumped during [math]\Delta t[/math] as the area of this rectangle. So, the total amount pumped between times [math]t=a[/math] and [math]t=x[/math] is the sum of the areas of many of these narrow rectangles. As the number [math]n[/math] of rectangles increases, they get narrower, and the approximation of the area gets better. Thus we have a Riemann sum, [math]\lim_{n\rightarrow \infty} \sum_{i=1}^{n} {g'(t)\Delta}t=\int_{a}^{x}g'(t)dt[/math].[br][br]On the right is plotted a graph of an [i]accumulated quantity[/i] (integral). It represents the [b][i]amount[/i][/b] of water [math]g[/math] pumped at the rate [math]g'[/math] since time [math]t=a[/math]. Let's fix [math]t=a[/math] at some point in time that we want to start measuring the amount of water pumped. Now, we let [math]x[/math] be the (arbitrary) point in time at which we want to know the total amount of water pumped since [math]t=a[/math]. Thus, the total amount of water [math]g[/math] pumped between times [math]a[/math] and [math]x[/math] is really a function [math]g(x)-g(a)[/math], calculated by [math]{g(x)-g(a)}=\int_{a}^{x}{g'(t)}{d}t[/math]. This is what's graphed at the right - it's really [math]\Delta g[/math].[br][br]So we've established the relationship between [math]g'[/math] (the derivative) and [math]g[/math] (the integral). Now let's see how this relates to the Fundamental Theorem.[br][br]If you actually solve the "definite integral" [math]\int_{a}^{x}{g'(t)}{d}t[/math], you would first find the antiderivative of [math]g'[/math] (which is [math]g[/math]), then plug in [math]x[/math] and [math]a[/math] and take the difference: [math]\int_{a}^{x}{g'(t)}{d}t=g(x)-g(a)[/math]. Notice that the app's default is [math]g(a)=0[/math] so that what's graphed is just [math]g(x)[/math]. You can use the [math]g(a)[/math] slider to change the value of [math]g(a)[/math], which has the effect of shifting the graph vertically. Now let's take the derivative of [math]g(x)-g(a)[/math]. Since [math]g(a)[/math] is a constant, its derivative is zero. And the derivative of [math]g(x)[/math] is just [math]g'(x)[/math]. Thus, [math]g'(x)=\frac{d}{dx}\left [g(x)-g(a)\right ]=\frac{d}{dx}\left [\int_{a}^{x} {g'(t)}{d}t\right ][/math].[br][br]You can drag the "[math]a[/math]" and "[math]x[/math]" points on the [math]t[/math]-axis at the left to see the effects on [math]g(x)[/math]. You can also type a new [i]rate[/i] function for the [math]g'[/math] graph at the left, and see the effects on the [i]amount[/i] function [math]g[/math] on the right. (The software requires using "x" as the variable in the input box). Click the circular arrows at the top right of the left graph to reset the app.

Set the functions and the limits of integration to establish one or more regions between [math]f[/math] and [math]g[/math]. Use the checkboxes to examine the area under [math]f[/math], the area under [math]g[/math], and how the area between them is found.

The Initial Value Problems (IVPs) that we will study are essentially just antiderivatives with an initial condition applied, which allows us to obtain the value of "[math]C[/math]", the constant of integration.[br][br]Until we are given an initial condition (a point on the solution curve), we cannot determine the value of [math]C[/math], and so the general solution is really an infinitely-large family of curves, one curve for each value of [math]C[/math].

We write the equation as a Differential Equation (using a derivative):[br][br][math]y'=f(x)[/math][br][br]We then take the antiderivative of both sides, combining the two constants into one on the right side:[br][br][math]\int y'dx=\int f\left(x\right)dx[/math][br][math]y=F\left(x\right)+C[/math][br][br]This gives us the [b][i]general solution[/i][/b]. Finally, we apply the initial condition to obtain "[math]C[/math]", giving us the [b][i]particular solution[/i][/b].