AP Calculus AB Applet Collection
A collection of interactive apps to illustrate important concepts taught in AP Calculus AB.
Table of Contents
- Introduction to Calculus
- Calculus - The Big Picture
- Basic Calculus Connections
- Calculus - Connecting How Fast with How Much
- Review: Lines
- Line Equations
- Slopes of Perpendicular Lines
- Rate of Change of a Function
- Rate of Change of a Function
- The Tangent Line Problem
- Local Linearity
- Limits and Continuity
- Why We Use Limits
- Understanding "x Approaches c" Notation
- Continuity
- Limits & Continuity
- The Intermediate Value Theorem
- Continuity and Differentiability with Parameters
- Vertical Asymptotes as Infinite Limits
- Rational Function End Behavior
- Dominance and Limits to Infinity
- Derivatives
- An Intuitive Approach to the Derivative
- Derivative Intuition 2
- Tangent and Normal Line Equations
- Applications of Derivatives
- Related Rates - Circular Ripple
- Rate of Change - Bubblegum
- Related Rates - The Sliding Ladder
- Particle Motion
- Connecting f, f', and f''
- Connecting Position, Velocity, and Acceleration
- Integrals
- Rectangular Sums
- Connecting Derivatives and Integrals
- The Fundamental Theorem of Calculus
- Fundamental Theorem of Calculus
- The Accumulation Function and the FTC
- Average Value of a Function
- Area and Volume
- Area of Planar Regions
- Volumes of Solids by Cross-Section
- Volumes of Solids - Disk & Washer Method
- Differential Equations
- Initial Value Problems
- Slope Fields
- Separable Differential Equations
Calculus - The Big Picture
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]
Line Equations
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.
Rate of Change of a Function
The Rate of Change of a function is how much the y-value changes as the x-value changes by some amount. For a linear function, we would call that "slope". On a position vs time graph, it measures change in position per change in time, which we call velocity. If we measure this between two distinct points (with two distinct x-values), we call it the Average Rate of Change (AROC). In calculus, we will use the AROC to find the Instantaneous Rate of Change (IROC) at a single point (single x-value).
The app begins with "Show f" and "Show Secant Line" both unchecked. There are two points in the plane, (a, f(a)) and (b, f(b)). Looking at just these two points, we can easily find the slope of a line beween them using the familiar slope formula (with function notation instead of y notation). Check the "Show Secant Line" box to display the segment between the points.[br][br]If these two points existed in isolation, then all we would know is the change in x and the change in y between them, from which we calculate a slope. But if we think of these points as being points on a graph (check the "Show f" box), we now see that the slope calculation actually finds the AROC of the function between the two points.[br][br]Drag the x values (a and b) to various places, and observe the behavior with various functions. Notice that the AROC of a linear function is constant everywhere, equal to the slope of the line.
Why We Use Limits
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].
An Intuitive Approach to the Derivative
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!
Related Rates - Circular Ripple
Drop a pebble in a calm pond. A circular ripple moves away at a constant speed; that is, its radius increases at a constant rate. How does the circle's area change in relation to the change in radius?
Rectangular Sums
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.
Fundamental Theorem of Calculus
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.
Area of Planar Regions
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.
Initial Value Problems
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].