Introduction to Instantaneous Velocity and Tangent Lines

Calculated Motion and the Study of Change

• Calculus is defined as the study of great communication and change over time.

• A fundamental question in calculus is determining how fast an object is moving at a particular instance.

• While calculating average speed over a duration is straightforward, finding the speed at an exact instant is more complex and requires looking at progressively smaller intervals to get closer to the actual value.

The Rock Motion Example: Data and Observations

• Suppose a rock is thrown out of a window. Its height ($h$) in feet is measured against time ($t$) in seconds. The following data points represent the rock's trajectory:

  • At $t = 0$, $h = 48$ feet.

  • At $t = 1$ , $h = 60$ feet.

  • At $t = 2$, $h = 64$ feet.

  • At $t = 3$, $h = 60$ feet.

  • At $t = 4$, $h = 48$ feet.

  • At $t = 5$, $h = 28$ feet.

  • At $t = 6$, $h = 0$ feet.

• Physical Description of Motion: The rock follows an arc. It travels upward, reaches a maximum peak, and then falls down toward the ground.

Average Velocity and Secant Lines

• Average Velocity Formula: $\text{Average Velocity} = \frac{\text{Change in Position}}{\text{Change in Time}}$

• Connection to Algebra: If time is represented by $x$ (or $t$) and height/position is represented by $y$ (or $h$), the change in position over the change in time is identical to the formula for slope: $\frac{\Delta y}{\Delta x}$.

• Secant Line: A secant line is defined as a straight line drawn between two distinct points on a curve. The slope of a secant line between two points is equivalent to the average velocity over the interval between those two points.

Calculations of Average Velocity Over Intervals

• Interval $[0, 2]$:

  • $h(2) = 48$, $h(0) = 48$.

  • $\frac{48 - 48}{2 - 0} = \frac{0}{2} = 0$.

  • Because the heights at the start and end are the same, the average velocity appears to be zero, even though the rock was moving in between.

• Interval $[1, 2]$:

  • $h(2) = 48$, $h(1) = 64$.

  • $\frac{48 - 64}{2 - 1} = \frac{-16}{1} = -16$.

  • The negative value indicates that the rock is falling.

• Interval $[2, 2.5]$: Average velocity is $-24$.

• Interval $[1, 3]$: Average velocity is $-32$.

• Interval $[2, 3]$: Average velocity is $-48$.

• Interval $[3, 4]$: Average velocity is $-64$.

Instantaneous Velocity and Tangent Lines

• Instantaneous Velocity: This is the velocity of an object at one specific instant in time ($t$).

• Tangent Line: A tangent line is a line that "kisses" the curve, touching exactly at one point and then continuing straight in either direction. The slope of the tangent line represents the instantaneous velocity at that point.

• The Calculation Difficulty: Finding the slope of a tangent line using the standard slope formula is difficult because using only one point results in $\frac{0}{0}$, which is undefined. To solve this, intervals must be made smaller and smaller to approach the point of interest.

Refining Velocity Calculations using Functions

• To get closer to the instantaneous velocity at $t = 2$, a functional definition for height is used:     $h(t) = 64 - 16(t - 1)^2$

• By plugging in values extremely close to $t = 2$ into the calculator (such as Desmos), the average rate of change over very small intervals can be computed:

  • Interval $[1.5, 2]$: $-24$

  • Interval $[1.9, 2]$: $-30.4$

  • Interval $[1.99, 2]$: $-31.84$

  • Interval $[1.999, 2]$: $-31.984$

  • Interval $[2, 2.5]$: $-40$

  • Interval $[2, 2.1]$: $-33.6$

  • Interval $[2, 2.01]$: $-32.16$

  • Interval $[2, 2.001]$: $-32.016$

• Conclusion: As the intervals shrink and get closer to $t = 2$ from both below and above, the velocity values converge toward $-32$. Therefore, the instantaneous velocity at $t = 2$ is estimated to be $-32\,ft/s$.

Deriving the Equation of the Tangent Line

• To write the equation of the tangent line at $t = 2$, two pieces of information are needed:

  1. Slope ($m$): $-32$ (as determined by the limit of the average velocities).

  2. Point ($x_1, y_1$): When $t = 2$, the height $h(2) = 48$. This gives the point $(2, 48)$.

Method 1: Slope-Intercept Form ($y = mx + b$)

• $48 = -32(2) + b$

• $48 = -64 + b$

• $112 = b$

• Final Equation: $y = -32x + 112$

Method 2: Point-Slope Form ($y - y_1 = m(x - x_1)$)

• $y - 48 = -32(x - 2)$

• Distributing the slope: $y - 48 = -32x + 64$

• Adding 48 to both sides: $y = -32x + 112$

Questions & Discussion

• Question (Audience): Why is the velocity negative?

• Response: Velocity can be positive, negative, or zero. It is negative in this case because the rock is falling. Positive velocity would indicate the rock is moving upward.

• Question (Audience): Why can't we just use the points from $2$ to $2$ to get the answer?

• Response: You cannot divide by zero. Using the same point twice in the slope formula results in a zero in the denominator, which is not mathematically possible.

• Pedagogical Note: To help students be successful, the pace of the lecture can be adjusted. Students are encouraged to request repetitions or slower explanations if the material is being covered too quickly.

Practice Problems

• Problem 1: Let $s(t) = 100 + 20t + t^2$. Estimate the velocity at $t = 2$ by computing the average velocity over the following intervals:

  • $[2, 2.2]$

  • $[2, 2.1]$

  • $[2, 2.01]$

  • $[2, 2.001]$

• Note on Homework: The first half of the current homework assignment is similar to these practice problems. The remaining portion of the homework will be covered in the next session. The homework is due next Wednesday at midnight.