Introduction - The What and Why of Calculus

Calculus is the study of how functions [i]change[/i]. [br][br]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. [br][br]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 displacement (distance traveled in a given direction). So not only can we relate [i]change in displacement[/i] to [math]velocity[/math], we can reverse the process and find out about [math]displacement[/math] when we know [math]velocity[/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 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 displacement: an object moving at a [i]speed[/i] of 3 m/s is [i]changing its displacement[/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 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 velocity and displacement: displacement = velocity [math]\times[/math] time, and the area under the velocity graph is velocity [math]\times[/math] time! And as for the displacement graph, we have gradient = displacement [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]displacement[/i] - it can only give us the [i]change[/i] in displacement 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 displacement = 3 m. We would have to be told that the object [i]started[/i] at position 5 m, and its displacement [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 displacement 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 displacement 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 displacement during this time as "height" (0.6 m/s) times "width" ([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 displacement of the ball (the ball's initial position is zero). When the ball rises,both its displacement and its velocity are positive. When the ball falls, its displacement 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 displacement 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]