I’m revising my fundamentals on rates of change and got stuck on something that feels simple but keeps tripping me up. Suppose I have a square whose area A(t) is increasing at a constant rate, like dA/dt = 3 cm²/s. What does that mean for how fast the side length s(t) is changing over time?
My attempt: since A = s², I wrote s = √A and then (using the chain rule) ds/dt = (1 / (2√A)) · dA/dt. If dA/dt is constant, that makes ds/dt = constant / (2√A), which gets smaller as A gets larger. So it seems like the side length speeds up at first but then its rate actually slows down as the square gets bigger. That feels backwards to me because the area is steadily increasing. Am I mixing up which thing depends on which, or is this actually right?
Analogy that might be wrong: spreading pizza dough. If I add dough at a steady “area” rate, the radius seems to grow more slowly as the pizza gets bigger, because new dough is spread over a longer edge. Is that the right intuition for a square’s side length too, or am I overthinking it?
Could someone explain-preferably in a plain, conceptual way-why the side length’s rate would decrease even though the area’s rate is constant? And is my chain-rule step a legitimate way to justify it, or is there a trap there?
Thanks! I’m trying to strengthen my basics on how one quantity’s steady change translates into another’s change.
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3 Responses
Your chain-rule step is right: since A = s², dA/dt = 2s·ds/dt, so ds/dt = (dA/dt)/(2s), which decreases as s grows because the same added area is spread along a longer edge. Example: if dA/dt = 3 cm²/s, then when s = 2 cm, ds/dt = 3/(2·2) = 0.75 cm/s, but when s = 6 cm, ds/dt = 3/(2·6) = 0.25 cm/s.
You’ve got it: A = s² gives dA/dt = 2s·ds/dt, so with dA/dt = 3 we get ds/dt = 3/(2s), which falls as s grows-same area per second spread over a longer edge means a thinner added “strip.”
Want to plug in a starting side s(0) and solve for s(t), or compare it to the circle/pizza case?
You’ve got it right! If A = s^2 and dA/dt is constant (say 3 cm²/s), then differentiating gives dA/dt = 2 s · ds/dt, so ds/dt = (dA/dt)/(2 s) = 3/(2 s). Since s grows over time, that fraction gets smaller, so the side length’s rate slows down. Concretely, A(t) = A0 + 3t, so s(t) = sqrt(A0 + 3t), and then ds/dt = 3/(2 sqrt(A0 + 3t)), which clearly decreases in t. Conceptually: a tiny outward nudge of the side sweeps out an area about 2 s · ds (bigger s means the same ds creates more area), so to keep the area gain fixed, ds has to shrink as s gets larger. Your pizza-dough intuition is on point too: for a circle A = π r^2, constant dA/dt gives dr/dt ∝ 1/r, the same “slows as it grows” behavior. Small sanity check numbers: at s = 1 cm, ds/dt = 1.5 cm/s; at s = 10 cm, ds/dt = 0.15 cm/s. I’m pretty sure that’s the whole story, and your chain-rule step is a perfectly legitimate justification. If you want a quick refresher on related rates in this style, Khan Academy’s intro is handy: https://www.khanacademy.org/math/differential-calculus/dc-related-rates-new/dc-related-rates/v/related-rates-introduction.