Rates of Change and Tangent Lines to Curves

Finding Slope and Equation of a Line

• Slope of a line containing two points:
  • Given by m = \frac{\text{rise}}{\text{run}} = \frac{\Delta y}{\Delta x}
  • Example: Find the equation of the line passing through (6, -1) and (3, 5).
    • Slope: m = \frac{5 - (-1)}{3 - 6} = \frac{6}{-3} = -2
• Equation of a line (point-slope form):
  • For a line with slope m passing through point (a, b), the equation is y - b = m(x - a).
  • Using point (3, 5): y - 5 = -2(x - 3)

Average and Instantaneous Speeds

• Goals:
  • Find average and instantaneous speeds of an object.
  • Find the equation of a line tangent to a curve at a given point.
• Example: Driving on a highway, pass Mile Post 50 at noon and Mile Post 80 at 12:30.
  • Instantaneous speed at 12:10: Cannot determine without more information.
  • Average speed over the half-hour trip:
    • Average speed = \frac{\text{distance traveled}}{\text{time elapsed}} = \frac{30 \text{ miles}}{0.5 \text{ hours}} = 60 \text{ mph}
• Graphical representation:
  • Average speed is the slope of the secant line connecting two points on a position graph.

Calculating Instantaneous Speed

• Example: Position of a train is s(t) = 3t^2 for the first 20 seconds, where t is in seconds and s is in feet.
  • Find the instantaneous speed at t = 4 seconds.
  • Cannot use the average speed formula directly because \frac{0}{0} is undefined.
• Method: Use time intervals that get closer and closer to zero.
  • Calculate average speed for each interval.
  • Instantaneous speed is the value the average speed approaches as the interval length approaches zero.
• Table for calculations:
  • Fix one end of the interval at 4 seconds.
  • h = difference between the beginning point (4 seconds) and the ending point (4 + h) seconds.
  • \Delta t = h
  • \Delta s = s(4+h) - s(4)
  • Average speed = \frac{\Delta s}{\Delta t}

• Guess: Instantaneous speed at t = 4 seconds is 24 feet per second.
• Exact Calculation:
  • Use a generic time interval of length h from 4 seconds to (4 + h) seconds.
  • Average speed = \frac{s(4 + h) - s(4)}{(4 + h) - 4} = \frac{s(4 + h) - s(4)}{h}
  • s(t) = 3t^2 so s(4 + h) = 3(4 + h)^2 and s(4) = 3(4)^2 = 48
  • \frac{3(4 + h)^2 - 48}{h} = \frac{3(16 + 8h + h^2) - 48}{h} = \frac{48 + 24h + 3h^2 - 48}{h} = \frac{24h + 3h^2}{h}
  • Factor out h: \frac{h(24 + 3h)}{h}
  • Cancel h: 24 + 3h
  • Let h go to zero: 24 + 3(0) = 24 feet per second.
  • The instantaneous speed at t = 4 is 24 feet per second.

Graphical Interpretation

• Secant Lines:
  • Draw a line through the points on the curve.
  • The slope of the secant line gives the average speed over the time interval.
• Tangent Line:
  • As the time interval approaches zero, the secant line approaches the tangent line.
  • The tangent line touches the curve at a single point.
  • The slope of the tangent line equals the slope of the curve at that point.
  • The slope of the tangent line equals the instantaneous speed.

Summary

• Average speed approaches instantaneous speed as the length of the time interval approaches zero.
• Apply the same idea to the rate of change of any quantity with a formula.
• For any graph, the slope of the tangent line at a point equals the limit of the slopes of the secant lines fixed at that point.

General Example: Tangent Line to a Graph

• Find the equation of the tangent line to the graph of f(x) = 1 - x^3 at x = 2.
  • f(2) = 1 - (2)^3 = 1 - 8 = -7
  • Point-slope form: y - (-7) = m(x - 2), where m is the slope of the tangent line.
• Calculate the average rate of change over an interval of length h.
  • \text{Slope of secant line} = \frac{f(2 + h) - f(2)}{h}

\frac{[1 - (2 + h)^3] - (-7)}{h} = \frac{1 - (8 + 12h + 6h^2 + h^3) + 7}{h} = \frac{-12h - 6h^2 - h^3}{h}

• Factor and cancel h: \frac{h(-12 - 6h - h^2)}{h} = -12 - 6h - h^2
• To find the slope of the tangent line, let h = 0: -12 - 6(0) - (0)^2 = -12
• Equation of the tangent line: y + 7 = -12(x - 2)