Describe change at one instant

A ball thrown up

The curve shows the height of a ball, in feet, \(t\) seconds after it is thrown up. It is green where the ball rises and red where it falls.

The ball is at 5 ft at t = 0 and again at t = 4, so the faint line that joins those points is flat. The average rate over the whole flight is 0 ft/s.

But the ball moves at every moment in between. Drag the orange dot to t = 1.

The orange line is the tangent line, which touches the curve only at the dot and points the way the curve goes there.

At t = 1, the tangent rises 32 ft for each second, so the ball moves up at 32 ft/s at that moment.

At the top, at t = 2, the tangent is flat. After the top it slopes down, because the ball falls.

Any curve

The curve can show any quantity against an input. Drag the dot along the curve.

The tangent line follows the dot, and its slope is the rate of change at that one input.

The tangent slopes up where the curve is green and the quantity rises. It slopes down where the curve is red and the quantity falls.

Where the tangent is steep, the quantity changes fast. Where it is flat, the rate is 0 at that moment.

Notation

1. The instantaneous rate of change at \(a\) is how fast the quantity changes at the single input \(a\), not an average over an interval.

2. It equals the slope of the tangent line to the curve at the point \((a, f(a))\).

   \[\text{rate at } a = \cy{\text{tangent slope at } a}\]

   The curve needs one tangent line there, so it must not have a corner or a jump at that point.

3. For a moving object, the rate of change of its position is its velocity. Velocity has a sign: positive means up or forward, and negative means down or back.

4. Speed is the size of the velocity, its absolute value.

   \[\text{speed} = |\text{velocity}|\]

When it fails

The faint line through two dots gives the average rate between them. When the second dot moves away from t = 1, this line no longer follows the tangent.

From t = 1 to t = 2 the average is 16 ft/s, but the rate at t = 1 is 32 ft/s. An average over an interval does not give the rate at one input.

Example 1: reading a tangent

A ball has height \(h(t) = 5 + 64t - 16t^2\) feet. Its tangent line at \(t = 0.5\) passes through \((0, 9)\) and \((2, 105)\). Find the velocity and the speed at \(t = 0.5\).

1. Read two points on the tangent line. They are \((0, 9)\) and \((2, 105)\).

2. Between these points, the tangent rises 96 ft over a run of 2 s.

   \[105 - 9 = 96, \quad 2 - 0 = 2\]

3. The slope is the velocity at t = 0.5. It is positive, so the ball moves up, and the speed is the same 48 ft/s.

   \[\dfrac{96}{2} = \cy{48 \text{ ft/s}}\]

Example 2: after the top

Use the same ball. Find the velocity and the speed at \(t = 3\).

1. The ball is at 53 ft at t = 3. It was also at 53 ft at t = 1.

   \[h(1) = h(3) = 53\]

2. A tempting wrong move: The two heights are equal, so the rate is 0 ft/s.

   \[\dfrac{53 - 53}{3 - 1} = 0\]

   That is the slope of the faint flat line from t = 1 to t = 3, an average over the interval. At t = 3 the curve is red and the tangent slopes down, so the ball falls at that moment.

3. The tangent line at \(t = 3\) passes through \((2, 85)\) and \((4, 21)\), so its slope is negative.

   \[\dfrac{21 - 85}{4 - 2} = \cr{-32 \text{ ft/s}}\]

4. The velocity is −32 ft/s, so the ball moves down. The speed is the size of the velocity, 32 ft/s.

Practice

Use the same ball. At \(t = 2\) it is at its highest point, 69 ft. Find its velocity and speed at that moment.

Move the dot to the top of the curve. What is the slope of the tangent line there?

Questions

The graph shows the position \(g(t)\) of a cart on a track, in meters, \(t\) seconds after the start.

1. At \(t = 3\), is the velocity of the cart positive, negative or zero?

   • Negative
   • Positive
   • Zero
   Answer

   Negative

2. At which input is the velocity of the cart zero?

   • \(t = 2\)
   • \(t = 0\)
   • \(t = 5\)
   Answer

   \(t = 2\)

3. The graph of the height of a drone has a tangent line at \(t = 5\) through \((3, 40)\) and \((7, 12)\), in meters and seconds. Find the velocity and the speed at \(t = 5\).

   • Velocity −7 m/s, speed 7 m/s
   • Velocity 7 m/s, speed 7 m/s
   • Velocity −1/7 m/s, speed 1/7 m/s
   Answer

   Velocity −7 m/s, speed 7 m/s

Summary

You read the rate at one input as the slope of the tangent line, with its sign and its units.

This works because the tangent line shows the direction of the curve at that single point, while an average only joins two points.

Use the tangent slope when you need the rate at a moment, and the average for a stretch of time.

This is one of four skills in topic 1.1 of Which & Why: AP Calculus.