Conical Pile Related Rates Problem

Conveyor belt pours sand at a constant volumetric rate.
- Given rate: $\dfrac{dV}{dt}=25\text{ ft}^3/\text{min}$
Sand forms a pile shaped as a right circular cone.
- Visual cue: classic conical heap where base always perfectly circular.
Geometric constraint: height equals diameter at every instant.
- Expressed as $h=d$ .
- Since diameter $d=2r$ , the relation becomes $h=2r\;\Rightarrow\;r=\dfrac{h}{2}$ (radius always half the height).
Instant of interest: pile is $h=18\text{ ft}$ tall.
Target: find $\dfrac{dh}{dt}$ (rate of change of height) at that moment.

Volume of a right circular cone:
$V=\dfrac{1}{3}\pi r^2 h$
Substitute radius–height relation to reduce variables (optional but conceptually useful): $V=\dfrac{1}{3}\pi \left(\dfrac{h}{2}\right)^2 h=\dfrac{1}{12}\pi h^3$
- Many instructors keep $r$ in place for implicit differentiation; both paths are valid.

Keep variables symbolic, differentiate volume formula with respect to time $t$ using implicit differentiation.
Use product rule for $r^2 h$ term if radius is not eliminated beforehand.
Inject known numeric values ((h, r, \dfrac{dV}{dt})) and algebraically isolate $\dfrac{dh}{dt}$ .
Leverage the permanent proportion $\dfrac{dr}{dt}=\dfrac{1}{2}\dfrac{dh}{dt}$ derived from $r=\dfrac{h}{2}$ to eliminate the extra rate.

Start: $V=\dfrac{1}{3}\pi r^2 h$
Differentiate both sides w.r.t. $t$ : $\dfrac{dV}{dt}=\dfrac{1}{3}\pi \big( r^2\,\dfrac{dh}{dt}+h\,(2r)\dfrac{dr}{dt} \big)$
- Product rule applied to $r^2 \cdot h$ (first (r^2), second (h)).
- Chain rule invoked for $r^2$ (derivative $2r\dfrac{dr}{dt}$ ).

Compute instantaneous radius when $h=18\text{ ft}$ :
$r=\dfrac{h}{2}=\dfrac{18}{2}=9\text{ ft}$
Express $\dfrac{dr}{dt}$ via height rate:
$\dfrac{dr}{dt}=\dfrac{1}{2}\dfrac{dh}{dt}$
Plug into differentiated equation:
$25 = \dfrac{1}{3}\pi \Big[ (9)^2 \dfrac{dh}{dt} + 18\cdot 2\cdot 9 \big(\dfrac{1}{2}\dfrac{dh}{dt}\big) \Big]$
Simplify inside brackets:
- $9^2=81$
- Factor $18\cdot 2 \cdot 9 = 324$ then times $\dfrac{1}{2} \rightarrow 162$
- Combined expression: $81\dfrac{dh}{dt}+162\dfrac{dh}{dt}=243\dfrac{dh}{dt}$
Isolate $\dfrac{dh}{dt}$ :
$25 = \dfrac{1}{3}\pi (243)\dfrac{dh}{dt} \;\Longrightarrow\; 75 = 243\pi\dfrac{dh}{dt}$
$\dfrac{dh}{dt}=\dfrac{75}{243\pi}\approx 0.0982\text{ ft/min}$

Height grows roughly $0.098\text{ ft}$ (E1.18 in) each minute at the specified instant.
Units validate: (ft³/min)/(ft²) yields ft/min.
Consistency: Positive sign shows pile is growing taller.
Sensitivity: Any change in pour rate or cone proportion would alter result proportionally.

Using $V=\dfrac{1}{12}\pi h^3$ eliminates $r$ and $\dfrac{dr}{dt}$ entirely.
- Differentiate: $\dfrac{dV}{dt}=\dfrac{1}{4}\pi h^2 \dfrac{dh}{dt}$
- Insert $h=18\text{ ft}$ , $\dfrac{dV}{dt}=25$ :
  $25=\dfrac{1}{4}\pi (18)^2\dfrac{dh}{dt} \Rightarrow\dfrac{dh}{dt}=\dfrac{25}{81\pi}\approx0.0982\text{ ft/min}$
Confirms original answer, showcasing elegance of variable reduction.

Related-rates technique extends to:
- Liquid draining/filling, spread of light/shadows, motion in physics.
Geometric similarity (constant ratio between variables) often simplifies analysis.
Understanding derivative chains crucial for multivariable systems in engineering.

Accurate rate calculations can prevent overflows or structural failures in industrial silos and hoppers.
Misjudging pour rates may lead to environmental spills or economic loss.

Always identify all rates and relations before differentiating.
Decide whether to substitute constraints early (reduces algebra) or keep variables (eases conceptual tracking); both must yield same result.
Keep units visible; they serve as an algebraic error-check.
For cones with proportional dimensions, volume often simplifies to a single-variable power, yielding quick differentiation.