This problem uses the constraint relationship y = 100/x for a hyperbolic function. Taking the derivative: dy/dt = -100/x² × dx/dt. When x = 20 and dx/dt = 1, we have dy/dt = -100/(20)² × 1 = -100/400 = -1/4. The negative sign indicates that as x increases, y decreases to maintain the constant product relationship. Students might forget the negative sign from the chain rule.

This problem demonstrates the product rule applied to lateral surface area of a cylinder. Given L = 2πrh with r = 2, h = 9, dr/dt = 1, and dh/dt = -2, taking the derivative: dL/dt = 2π[h(dr/dt) + r(dh/dt)] = 2π[9(1) + 2(-2)] = 2π[9 - 4] = 2π(5) = 10π. The positive result shows lateral area is increasing despite height decreasing, because radius increase dominates.

This problem applies area rate change using diameter for a circle. Given A = π(d/2)² = πd²/4 and dd/dt = 0.8 cm/s when d = 10 cm, taking the derivative: dA/dt = π(d/2)(dd/dt) = π(10/2)(0.8) = π(5)(0.8) = 4π cm²/s. This demonstrates how diameter-based formulations affect the derivative calculation compared to radius-based approaches.

This problem demonstrates interpreting rates in a linear business context. Given $R = 50p$, taking the derivative: $\dfrac{dR}{dt} = 50 \dfrac{dp}{dt}$. Since price decreases at 0.4 dollars/day, $\dfrac{dp}{dt} = -0.4$. Therefore, $\dfrac{dR}{dt} = 50(-0.4) = -20$ dollars/day. The negative sign indicates revenue is decreasing as price decreases. A common error would be forgetting the negative sign or confusing the direction of change.

This problem requires finding the instantaneous rate of change using exponential differentiation. Given $A = 80e^{-0.1t}$, taking the derivative: $dA/dt = 80(-0.1)e^{-0.1t} = -8e^{-0.1t}$. At $t = 0$, $dA/dt = -8e^0 = -8(1) = -8$ mg/hr. The negative sign indicates the medication amount is decreasing over time, which is expected for drug elimination. Students might forget the chain rule or the negative coefficient.

This problem demonstrates the relationship between diameter and radius rates. Given $d = 2r$, taking the derivative: $\dfrac{dd}{dt} = 2 \dfrac{dr}{dt}$. Therefore, $\dfrac{dr}{dt} = \frac{1}{2} (\dfrac{dd}{dt}) = \frac{1}{2} (6) = 3$ mm/hr. This is a direct application of the constant multiple rule. Students might forget the factor of 2 or confuse the direction of the relationship between diameter and radius changes.

This problem involves related rates with constant perimeter constraint. For a rectangle with perimeter P = 2l + 2w = 40, we have l + w = 20. Taking the derivative: dl/dt + dw/dt = 0. Given l = 12 and dl/dt = 1 ft/min, we substitute: 1 + dw/dt = 0. Solving: dw/dt = -1 ft/min. The negative sign correctly indicates that as length increases, width must decrease to maintain constant perimeter. A common error would be forgetting the constraint or the negative relationship.

This problem applies the relationship between square side length and diagonal. Given $d = s\sqrt{2}$, taking the derivative: $\dfrac{dd}{dt} = \sqrt{2} \dfrac{ds}{dt}$. With $\dfrac{ds}{dt} = 3$ cm/min, we have $\dfrac{dd}{dt} = \sqrt{2}(3) = 3\sqrt{2}$ cm/min. This demonstrates how geometric relationships create proportional rate relationships. Students might forget the $\sqrt{2}$ factor or confuse the direction of the relationship.

This problem demonstrates area rate change for a decreasing circular radius. Given A = πr² and dr/dt = -0.05 cm/s (negative for decreasing) when r = 20 cm, taking the derivative: dA/dt = 2πr(dr/dt) = 2π(20)(-0.05) = -2π cm²/s. The negative sign correctly indicates the area is decreasing as the bracelet contracts. Students might forget the negative sign or miscalculate the coefficient.

This problem demonstrates constraint relationships in perimeter. Given $P = 2x + 2y$ and $\dfrac{dP}{dt} = 0$ (constant perimeter), taking the derivative: $2\dfrac{dx}{dt} + 2\dfrac{dy}{dt} = 0$. With $\dfrac{dx}{dt} = 3$, we have $2(3) + 2\dfrac{dy}{dt} = 0$, so $6 + 2\dfrac{dy}{dt} = 0$. Therefore, $\dfrac{dy}{dt} = -3$. The negative sign indicates that as one dimension increases, the other must decrease to maintain constant perimeter.