Related Rates – Changing Base of a Triangle

Problem Overview

  • Objective: Determine the rate of change of a triangle’s base, \frac{db}{dt}, given that both the height and the area are changing with respect to time.
  • Context: Classic “related‐rates” application from differential calculus; links geometry (area of a triangle) with implicit differentiation.

Diagram & Variable Definitions

  • Right or generic triangle sketched to visualize:
    • Vertical segment = altitude/height (h).
    • Horizontal segment = base (b).
  • Variables & rates:
    • h = height (cm).
    • b = base (cm).
    • A = area (\text{cm}^2).
    • \frac{dh}{dt} = rate of change of height (cm/min).
    • \frac{dA}{dt} = rate of change of area (\text{cm}^2/min).
    • \frac{db}{dt} = desired rate of change of base (cm/min).

Given Information (Instantaneous Snapshot)

  • \frac{dh}{dt} = 2.5\,\text{cm/min} (height increasing ⇒ positive).
  • \frac{dA}{dt} = 3\,\text{cm}^2!/\text{min} (area increasing ⇒ positive).
  • Specific “moment” to evaluate rate:
    • h = 12\,\text{cm}.
    • A = 84\,\text{cm}^2.
  • b is not supplied directly → must compute from area formula.

Fundamental Formula (Triangle Area)

  • A = \frac12 b h.
  • Relates all three instantaneous quantities and underlies the differentiation step.

Step 1: Determine the Base at the Given Instant

  • Substitute A = 84 and h = 12 into A = \frac12 b h:
    84 = \frac12 b (12)
    84 = 6b
    b = 14\,\text{cm}.
  • Record for later substitution in the derivative equation.

Step 2: Differentiate Implicitly w.r.t. Time t

  • Differentiate A = \frac12 b h using the product rule (both b and h depend on t):
    \frac{dA}{dt} = \frac12 \big( b\,\frac{dh}{dt} + h\,\frac{db}{dt} \big).
    • Product rule: firstderivative of second + secondderivative of first.

Step 3: Substitute Known Values & Solve for \frac{db}{dt}

  • Plug in numbers:
    3 = \tfrac12 \big( 14 \times 2.5 + 12 \times \frac{db}{dt} \big).
  • Clear the \tfrac12 by distributing:
    • \tfrac12 (14 \times 2.5) = 17.5.
    • \tfrac12 (12) = 6, so term becomes 6\,\frac{db}{dt}.
  • Equation becomes:
    3 = 17.5 + 6\,\frac{db}{dt}.
  • Isolate \frac{db}{dt}:
    3 - 17.5 = 6\,\frac{db}{dt}
    -14.5 = 6\,\frac{db}{dt}
    \frac{db}{dt} = \frac{-14.5}{6} = -\frac{29}{12}\,\text{cm/min}.
  • Decimal form (optional): \frac{db}{dt} \approx -2.417\,\text{cm/min}.

Interpretation of the Result

  • Negative sign: the base decreases while both height and area increase.
  • Physical meaning: To keep the area increase relatively small (3 \text{cm}^2!/\text{min}) while the height shoots up at 2.5 cm/min, the base must shorten.
  • Valid only at the instant where h = 12\,\text{cm},\ A = 84\,\text{cm}^2,\ b = 14\,\text{cm}.

Connections & Broader Significance

  • Demonstrates the power of implicit differentiation in real‐time geometry problems.
  • Mirrors real‐world scenarios: structures changing shape, fluid levels in variable‐width containers, etc.
  • Ethical/practical angle: accurate rate calculations prevent engineering misjudgments (e.g., changing load paths in expanding trusses).
  • Reinforces: product rule, related‐rates workflow (draw, list quantities/units, connect via formula, differentiate, substitute, solve, interpret).

Numerical & Symbolic Summary

  • Formula used: A = \tfrac12 b h.
  • Differentiated form: \frac{dA}{dt} = \tfrac12 \big( b\,\frac{dh}{dt} + h\,\frac{db}{dt} \big).
  • Final rate: \boxed{\frac{db}{dt} = -\frac{29}{12}\,\text{cm/min} \; (\approx -2.417)}.
  • Positive \frac{dh}{dt}, positive \frac{dA}{dt}, negative \frac{db}{dt} → highlights the interplay between variables when constrained by a geometric relation.