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.

This problem demonstrates volume rate change for a cylinder with decreasing height. Given $V = \pi r^2 h$ with $r = 2$ cm (constant) and $dh/dt = -0.3$ cm/hr (negative for decreasing), taking the derivative: $dV/dt = \pi r^2 (dh/dt) = \pi(2)^2 (-0.3) = \pi(4) (-0.3) = -1.2 \pi \text{ cm}^3/\text{hr}$. The negative sign indicates volume is decreasing as the candle burns down, which matches physical expectation.

This problem uses constant area constraint for a rectangle. Given $A = lw = 144$ (constant), when $l = 9$ and $\frac{dl}{dt} = -2$ (negative for decreasing), we find $w = \frac{144}{9} = 16$. Taking the derivative: $l \frac{dw}{dt} + w \frac{dl}{dt} = 0$. Substituting: $9 \frac{dw}{dt} + 16(-2) = 0$, so $9 \frac{dw}{dt} = 32$. Therefore, $\frac{dw}{dt} = \frac{32}{9} \text{ ft/min}$. The positive result indicates width increases as length decreases to maintain constant area.

This problem connects volume and surface area rates for a cube. Given $V = s^3$ and $SA = 6s^2$, with $\frac{dV}{dt} = 150 , \text{cm}^3/\text{min}$ when $s = 5 , \text{cm}$. First, find $ds/dt$: $150 = 3s^2 (ds/dt) = 3(25)(ds/dt) = 75(ds/dt)$, so $ds/dt = 2 , \text{cm/min}$. Then $dSA/dt = 12s(ds/dt) = 12(5)(2) = 120 , \text{cm}^2/\text{min}$. This shows how volume changes translate to surface area changes through the common rate of side length change.

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 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 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 requires differentiating an exponential function. Given P = $200(1.5)^t$, taking the derivative: dP/dt = $200(1.5)^t$ × ln(1.5) using the chain rule for exponential functions. At t = 0, dP/dt = $200(1.5)^0$ × ln(1.5) = 200(1) × ln(1.5) = 200ln(1.5). Students might forget the natural logarithm factor or confuse the exponential differentiation rule.

This problem involves a constant area constraint where the area is unchanging. Given A = πr² = 400π (constant), taking the derivative: dA/dt = 2πr(dr/dt) = 0. Since area is constant, its rate of change must be zero regardless of how radius might be changing. This demonstrates that constraints can override individual parameter changes. Students might mistakenly calculate a non-zero rate based on the radius change.

This problem applies the basic formula for the rate of change of circular area. Given $A = \pi r^2$ and $\dfrac{dr}{dt} = 0.2$ mm/hr when $r = 10$ mm, taking the derivative: $\dfrac{dA}{dt} = 2\pi r (\dfrac{dr}{dt}) = 2\pi(10)(0.2) = 4\pi$ mm$^2$/hr. This is a direct application of the chain rule to the area formula. Students might forget the coefficient 2 from differentiating r$^2$ or miscalculate the substitution.