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.