Geometric Similarity in Related Rates

---

Concept Explanation

1. The Multi-Variable Degrees of Freedom Problem

The volume of a right circular cone is given by:

If water is filling or draining from a conical tank, both the water surface radius $r(t)$ and depth $h(t)$ change continuously over time. If one naively applies the time differential operator $\frac{d}{dt}$ directly using the Product Rule: The resulting equation contains two unknown time derivatives ($\frac{dr}{dt}$ and $\frac{dh}{dt}$). Unless telemetry measures both rates simultaneously, the problem cannot be solved.

2. Cross-Sectional Similar Triangles

Because the physical boundary walls of a conical vessel or triangular prism maintain a fixed angle, any vertical cross-section passing through the central axis forms two nested right triangles:

1. The Container Triangle: Formed by the fixed top radius $R$ and total container height $H$.
2. The Fluid Triangle: Formed by the instantaneous fluid radius $r(t)$ and instantaneous fluid depth $h(t)$.

Because these triangles share the same apex angle and each contain a right angle, they are mathematically similar (AAA similarity):

3. Pre-Differentiation Variable Elimination

By isolating the variable that is not requested in the problem, we substitute it into the governing volume equation before taking the derivative:

• If solving for $\frac{dh}{dt}$ (rate of water level rise/fall):
  $$r = \left(\frac{R}{H}\right)h \implies V = \frac{1}{3}\pi \left(\frac{R}{H}h\right)^2 h = \frac{\pi R^2}{3H^2} h^3$$
  Differentiating with respect to $t$ now produces a clean single-variable rate:
  $$\frac{dV}{dt} = \frac{\pi R^2}{H^2} h^2 \frac{dh}{dt}$$
• If solving for $\frac{dr}{dt}$ (rate of surface radius expansion):
  $$h = \left(\frac{H}{R}\right)r \implies V = \frac{1}{3}\pi r^2 \left(\frac{H}{R}r\right) = \frac{\pi H}{3R} r^3$$

4. Dynamic Shadow Projections

A parallel application of geometric similarity occurs in optical shadow casting (a pedestrian moving away from a streetlight or toward a spotlight wall).

• Streetlight Scenario: Light post height $H$, person height $h_{\text{p}}$, person distance from pole $x(t)$, shadow length $s(t)$. By similar triangles:
  $$\frac{h_{\text{p}}}{s} = \frac{H}{x + s} \implies h_{\text{p}}(x + s) = H s \implies s(t) = \frac{h_{\text{p}}}{H - h_{\text{p}}} x(t)$$
  Notice that the rate of change of shadow length is directly proportional to walking speed:
  $$\frac{ds}{dt} = \frac{h_{\text{p}}}{H - h_{\text{p}}} \frac{dx}{dt}$$
• Rate of Shadow Length vs. Rate of Shadow Tip:
  • The rate at which the length of the shadow is growing is $\frac{ds}{dt}$.
  • The rate at which the tip of the shadow is moving along the ground relative to the light pole is the derivative of total distance from the pole:
    $$\frac{d}{dt}[x + s] = \frac{dx}{dt} + \frac{ds}{dt}$$

---

Engineering Applications

• Civil & Environmental Hydrology: Retention basins, settling funnels, and irrigation channels are designed with sloping trapezoidal or conical walls. Environmental engineers apply similar triangles to relate ultrasonic water depth gauge readings directly to volumetric storage and discharge rates.
• Biomedical & Pharmaceutical Manufacturing: Conical feed hoppers deliver powdered excipients and active pharmaceutical ingredients into tablet-pressing machinery. Engineers model mass discharge by tracking the rate of descending powder level $\frac{dh}{dt}$.
• Computer Vision & Optical Tracking: Automated security and motion-capture systems track moving objects using structured shadow casting and optical camera projections. Similar triangle ratios establish linear coordinate mappings from 2D projected shadows to true 3D spatial velocity vectors.

---

Example Problems

Example 1: Inverted Conical Hopper

A conical grain hopper has a top radius of $R = 2\text{ m}$ and a total vertical height of $H = 6\text{ m}$. Corn grain pours out of an orifice at the bottom at a steady volumetric flow rate of $\frac{dV}{dt} = -0.8\text{ m}^3/\text{min}$. At what instantaneous rate is the grain height $h$ dropping when the grain level is $h = 3\text{ meters}$ deep?

Solution

1. Governing Formula: Volume of a cone and similar triangle reduction:
   $$V = \frac{1}{3}\pi r^2 h, \qquad \frac{r}{h} = \frac{2}{6} = \frac{1}{3} \implies r = \frac{1}{3}h$$
2. Express volume in terms of $h$:
   $$V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{\pi}{27} h^3$$
3. Differentiate implicitly with respect to time ($t$):
   $$\frac{dV}{dt} = \frac{\pi}{9} h^2 \frac{dh}{dt}$$
4. Substitute known rates and snapshot values ($\frac{dV}{dt} = -0.8\text{ m}^3/\text{min}$, $h = 3\text{ m}$):
   $$-0.8 = \frac{\pi}{9}(3)^2 \frac{dh}{dt} \implies -0.8 = \pi \frac{dh}{dt}$$
   $$\frac{dh}{dt} = -\frac{0.8}{\pi} = -\frac{4}{5\pi}\text{ m/min} \approx -0.255\text{ m/min}$$
5. Interpretation: The grain height is falling at a rate of $\frac{4}{5\pi}\text{ m/min}$ (approx $25.5\text{ cm/min}$).

---

Connections & References

• Parent Concepts:
  • Related Rates
  • Implicit Differentiation
• Related Geometry:
  • Similar Triangles
  • Pythagorean and Trigonometric Related Rates
• Course Applications:
  • Calc-1 Discussion 05
  • Calc 1 Session 11 - Related Rates
  • Hydraulic Reservoir Surge Control (Civil Engineering)