HM

Application Problem: Ball Inflation (copy)

  • Radius-based Volume Formula
    • Volume of a sphere: V = \frac{4}{3} \pi r^3
  • Problem Setup: Ball Inflation
    • Initial diameter = 6 cm, so initial radius = 3 cm
    • Radius increases by 1.5 cm per second
    • Let r be the radius and t be time in seconds
  • Part A: Model radius as a function of time
    • Radius grows linearly with time: r(t) = 1.5t + 3
    • Why r is used: the volume depends on radius, not diameter
    • Check some values:
    • After 1 second: r(1) = 1.5(1) + 3 = 4.5\text{ cm}
    • After 2 seconds: r(2) = 1.5(2) + 3 = 6\text{ cm}
  • Part B: Find the volume after four seconds
    • Step 1: Determine the radius at t = 4
    • r(4) = 1.5(4) + 3 = 9\text{ cm}
    • Step 2: Use the volume formula with this radius
    • V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (9)^3
    • Step 3: Compute the cube and constants
    • 9^3 = 729
    • \frac{4}{3} \cdot 729 = 972
    • Step 4: Include π for the final volume
    • V = 972\pi\ \text{cm}^3 \approx 3.05363 \times 10^3 \ \text{cm}^3
    • Note on the transcript: The calculation shown there omitted the factor of π and stated the result as 972\ \text{cm}^3, which is missing the π factor. The correct result is V = 972\pi\ \text{cm}^3 (approximately 3053.63\ \text{cm}^3).
  • Part C: Find f(g(t)) and interpret
    • Define the functions:
    • f(r) = \frac{4}{3} \pi r^3 (volume as a function of radius)
    • g(t) = 1.5t + 3 (radius as a function of time)
    • Composition: (f \circ g)(t) = f(g(t)) = \frac{4}{3} \pi \big(1.5t + 3\big)^3
    • Evaluation at t = 4:
    • (f \circ g)(4) = \frac{4}{3} \pi \big(9\big)^3 = 972\pi\ \text{cm}^3
    • Connection to Part B: The composition gives the same result as computing radius first and then volume; numerically, it matches the corrected value above (972π cm^3).
  • Discussion and insights
    • The radius grows linearly with time, while volume grows as the cube of the radius, illustrating why small radius changes produce large volume changes
    • Units: radius in cm, time in seconds, volume in cm^3
    • Distinguishing diameter vs radius is crucial: starting diameter 6 cm implies starting radius 3 cm
    • Common pitfall: forgetting the factor of π when calculating volume
    • If expanding further, you could also express the time-to-volume mapping directly via the composite function as above
  • Conceptual check and practical implications
    • This is a straightforward application of function composition and the sphere volume formula to a real-world process (inflating a ball)
    • The approach demonstrates how to model a physical process with two simple, separate rate laws and then combine them via composition
  • Summary of key formulas
    • Radius as a function of time: r(t) = 1.5t + 3
    • Volume as a function of radius: V(r) = \frac{4}{3} \pi r^3
    • Composite volume as a function of time: (f \circ g)(t) = \frac{4}{3} \pi \big(1.5t + 3\big)^3
    • Volume after 4 seconds (correct): V = 972\pi \text{ cm}^3 \approx 3053.63 \text{ cm}^3
  • Takeaways
    • Always verify whether you’re using radius or diameter in formulas
    • When computing with π, keep the π factor explicit to avoid missing terms
    • Function composition provides a compact way to express “volume directly from time” without stepwise intermediate results