Difficulty: Introductory to Intermediate
Prerequisites: Sections 1.1 and 1.2 (velocity, distance, differences and sums).
This is where calculus properly begins. Sections 1.1 and 1.2 dealt with straight lines and constant velocities. Now the distance graph is a curve, and the velocity changes from moment to moment. The question becomes: what is the speed right now, at a single instant? The answer comes from taking the average speed over a shrinking interval and letting that interval approach zero. This limiting process produces the instantaneous velocity, which is the slope of the curve at a point.
To find the instantaneous velocity at time t, compute the average velocity over an interval from t to t + h, then let h shrink to zero. The result is the slope of the tangent line to the distance curve. For f(t) = t^2, this gives v(t) = 2t. For f(t) = 4t^2, this gives v(t) = 8t.
Average velocity
The total change in distance divided by the total change in time over an interval: v_avg = [f(t + h) - f(t)] / h. Think of it as the slope of the straight line connecting two points on the distance curve.
Instantaneous velocity
The velocity at a single moment in time. Found by taking the limit of the average velocity as the interval h shrinks to zero. This is the slope of the curve at that one point.
Limit (informal)
The value that a quantity approaches as some variable (here h) gets closer and closer to a target (here 0). You do not plug in h = 0 directly (that gives 0/0). Instead, you simplify the expression first, then let h go to zero.
Secant line
The straight line connecting two points on a curve. Its slope is the average velocity between those two points.
Tangent line
The straight line that touches the curve at exactly one point and has the same slope as the curve at that point. The slope of the tangent line is the instantaneous velocity.
Acceleration
The rate of change of velocity. If f(t) = (1/2)at^2, then v(t) = at, and the acceleration is the constant a.
Step 1: Compute the average velocity over an interval from t to t + h. This means calculating [f(t + h) - f(t)] / h.
Step 2: Simplify the expression algebraically. Cancel h from the numerator and denominator (after expanding and collecting terms).
Step 3: Let h approach zero. The simplified expression gives the instantaneous velocity v(t).
Average speed from t to t + h: [(t+h)^2 - (t+h) - (t^2 - t)] / h = [h^2 + 2ht - h] / h = h + 2t - 1.
Let h approach 0: v(t) = 2t - 1.
Check: at t = 1, v(1) = 2(1) - 1 = 1. At t = 2, the average speed from t = 1 to t = 2 is [f(2) - f(1)] / 1 = [2 - 0] / 1 = 2. The instantaneous speed (1) is less than this average (2), which makes sense since the function is accelerating.
Average slope from t to t + h: [4(t+h)^2 - 4t^2] / h = [4t^2 + 8th + 4h^2 - 4t^2] / h = 8t + 4h.
Let h approach 0: v(t) = 8t.
The instantaneous velocity is 8t. At t = 1, the speed is 8. At t = 0, the speed is 0 (the object starts from rest).
Given v(t) = 6 - 2t, the distance is the area under the v-graph from 0 to t.
For 0 <= t <= 3, the area is a trapezoid: f(t) = (1/2)(6 + 6 - 2t)(t) = 6t - t^2.
At t = 3, f(3) = 18 - 9 = 9 (the maximum). After t = 3, velocity is negative, so the area (and distance) decreases.
The graph of f(t) = 6t - t^2 is a downward-opening parabola with its peak at t = 3.
On the graph of f(t), draw the secant line from point A = (t, f(t)) to point B = (t+h, f(t+h)).
The slope of this secant line is the average velocity.
As h shrinks to 0, point B slides toward A, and the secant line rotates into the tangent line.
The slope of the tangent line at A is the instantaneous velocity: the derivative.
v_{\text{avg}} = \frac{f(t+h) - f(t)}{h}Average velocity over the interval from t to t + h. This is the slope of the secant line.
v(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}Instantaneous velocity. The limit of the average velocity as h approaches zero. This is the derivative.
f(t) = \frac{1}{2}at^2 \implies v(t) = atFor constant acceleration a, the distance is a parabola and the velocity is a straight line through the origin.
This is how speedometers work at a conceptual level. The car's computer measures distance over very short time intervals and divides to get speed. The shorter the interval, the closer the reading is to the true instantaneous speed. In physics, this process gives the velocity of any moving object, from a falling ball (f(t) = (1/2)gt^2, v(t) = gt) to a planet in orbit.
Students often try to compute the limit by directly substituting h = 0 into [f(t+h) - f(t)] / h. This gives 0/0, which is meaningless. You must simplify the algebra first, cancel h, and then set h = 0.
The average velocity and instantaneous velocity are different quantities. They are equal only when velocity is constant (straight-line distance graph). On a curve, the average over any interval is generally different from the instantaneous value at any point in that interval.
A common error is confusing "speed" and "velocity." Speed is the magnitude (absolute value) of velocity. A car going backward at 5 mph has velocity -5 and speed 5.
⚠️ The three-step process (write the difference quotient, simplify, take the limit) is the definition of the derivative. You will use it repeatedly in Chapter 2.
⚠️ Standard exam question: given f(t), find v(t) using the limit definition. Practise this for f(t) = t^2, t^3, 4t^2, and t^2 - t.
⚠️ The relationship between secant lines and the tangent line is a frequent conceptual question. Know that the tangent line is the limit of the secant lines as the two points merge.
Fill in the blank: The average velocity from t = 1 to t = 1 + h for f(t) = t^2 is ______. 2 + h.
True or False: As h approaches 0, the average velocity 2 + h approaches 2. True.
Fill in the blank: If f(t) = 4t^2, the instantaneous velocity at time t is ______. 8t.
True or False: The secant line and the tangent line are the same when h = 0. True (in the limit).
Q: For f(t) = t^2 - t, compute the average speed from t = 1 to t = 2.
A: f(2) = 4 - 2 = 2, f(1) = 1 - 1 = 0. Average speed = (2 - 0)/(2 - 1) = 2.
Q: For f(t) = t^2 - t, find the instantaneous velocity at t = 1 using the limit definition.
A: Average from t to t + h is h + 2t - 1. At t = 1: h + 1. As h approaches 0: v(1) = 1.
Q: If f(t) = (1/2)at^2 and v(t) = at, what is the acceleration?
A: The acceleration is a (constant). It is the rate at which velocity is increasing.
Q: Two cars start at the same place and end at the same place. Car F travels at constant speed. Car E goes faster, then slower, then backward. At what point is Car E's speed equal to Car F's speed?
A: At the instant where the slope of E's distance graph equals the slope of F's distance graph. This is the point where the two curves have the same tangent slope.
Q: For f(t) = 6t - t^2, at what time is the velocity zero? What is the maximum distance?
A: v(t) = 6 - 2t = 0 when t = 3. The maximum distance is f(3) = 18 - 9 = 9.
The limit process here is the formal definition of the derivative, developed fully in Chapter 2. The secant-to-tangent idea returns in Section 2.3 (tangent lines), Section 3.1 (linear approximation), and Section 3.7 (Newton's method). The area-under-the-curve idea for going from v(t) back to f(t) is the integral, covered in Chapter 5.
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