球型气球膨胀速率问题求解及相关微分公式应用咨询

Hey there! Let me break this down for you clearly—you’re already on the right track with your initial guess, great job getting that far!

First, let’s unpack the formula you were given:
$$\dfrac{dV}{dt} = \dfrac{dV}{dr} \cdot \dfrac{dr}{dt}$$
This is just the chain rule from calculus, a tool we use to link the rate of change of one quantity to another that depends on it. Here, the balloon’s volume ($V$) depends directly on its radius ($r$), and both are changing over time ($t$). Instead of trying to find how $r$ changes with $t$ directly, we use the relationship between $V$ and $r$ (that’s $\frac{dV}{dr}$) to connect the volume’s growth rate to the radius’s growth rate.

Now let’s walk through solving your problem step by step:

• Start with the sphere volume formula
  For a perfect sphere, volume is calculated as:
  $$V = \dfrac{4}{3} \pi r^3$$

• Calculate $\frac{dV}{dr}$ (volume change relative to radius)
  Differentiate the volume formula with respect to $r$, and we get:
  $$\dfrac{dV}{dr} = 4 \pi r^2$$
  This makes intuitive sense too—$4\pi r^2$ is the surface area of the sphere, think of it as the "speed" at which volume accumulates when you nudge the radius a tiny bit.

• Plug into the chain rule equation
  Substitute $\frac{dV}{dr}$ into the chain rule formula, and we end up with:
  $$\dfrac{dV}{dt} = 4 \pi r^2 \cdot \dfrac{dr}{dt}$$

• Plug in known values and solve for $\frac{dr}{dt}$
  We know two key values:

  • $\frac{dV}{dt} = 0.1$ L/s (the rate the balloon is being inflated)
  • $r = 20$ cm (the radius we’re focusing on)

  Rearrange the equation to isolate $\frac{dr}{dt}$:
  $$\dfrac{dr}{dt} = \dfrac{0.1}{4 \pi r^2}$$

  Now substitute $r=20$ cm:
  $$\dfrac{dr}{dt} = \dfrac{0.1}{4 \pi (20)^2} \approx 0.00000785 \text{ cm/s}$$
  This is the instantaneous rate at which the radius is increasing when the balloon’s radius hits 20 cm.

• Total radius increase over a specific time (if needed)
  If you want to find how much the radius grows over a set period, just multiply this rate by the time. For example, over 1 minute (60 seconds):
  $$\text{Total radius increase} = 0.00000785 \cdot 60 \approx 0.00469 \text{ cm}$$

Your initial understanding was totally correct—you already grasped the core idea of how this differential equation links the rates. The big takeaway here is that the chain rule is our go-to for these "related rates" problems, letting us connect changes in dependent quantities easily.

备注：内容来源于stack exchange，提问作者cricket900