This problem involves solving related rates by relating coordinates on a circle. Given $x^2 + y^2 = 25$, differentiating implicitly with respect to time $t$ gives $2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0$, which simplifies to $\dfrac{dy}{dt} = -\dfrac{x}{y} \dfrac{dx}{dt}$. When $x=3$ and $y=4$, $\dfrac{dx}{dt}=4$, substituting yields $\dfrac{dy}{dt} = -\dfrac{3}{4} \times 4 = -3$. This assumes y is positive, consistent with the point (3,4) on the circle. A tempting distractor like $-\dfrac{16}{3}$ fails because it might arise from inverting the ratio incorrectly or misapplying signs. In general, for related rates problems, identify the variables and their relationship, differentiate with respect to time, plug in known values, and solve for the desired rate.

Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The shadow length s relates to the person's distance x by s = (2/3) x from similar triangles with heights 15 ft and 6 ft. Differentiating implicitly with respect to time gives ds/dt = (2/3) dx/dt. When dx/dt=5, this yields ds/dt = (2/3)*5 = 10/3 ft/s, independent of x=10. A tempting distractor is 25/3, which might come from finding the tip's speed d(x+s)/dt instead of the lengthening rate. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.

This problem involves solving related rates by relating the area of a circle to its radius. The area A = πr², and differentiating implicitly with respect to time t gives dA/dt = 2πr dr/dt. When r=6 cm and dA/dt=12π cm²/min, substituting yields 12π = 2π(6) dr/dt, so dr/dt = 1 cm/min. This uses the chain rule to connect the area change to the radius change accurately. A tempting distractor like 1/π cm/min fails because it might come from incorrectly dividing by 2πr instead of solving properly. In general, for related rates problems, identify the variables and their relationship, differentiate with respect to time, plug in known values, and solve for the desired rate.

Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The area A of the circle is A = π r². Differentiating implicitly with respect to time gives dA/dt = 2π r dr/dt. When r=10 and dr/dt=-3, substituting yields dA/dt = 2π10(-3) = -60π cm²/s. A tempting distractor is -30π, perhaps from forgetting the factor of 2 in the derivative. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.

This problem involves solving related rates by differentiating a geometric formula with respect to time. The separation is $s = \sqrt{x^2 + y^2}$, so implicit differentiation gives $\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s}$. After 2 hours, x = 24 mi, y = 10 mi, $\frac{dx}{dt}$ = 12 mph, $\frac{dy}{dt}$ = 5 mph, s = 26 mi, substituting yields $\frac{ds}{dt} = \frac{24 \times 12 + 10 \times 5}{26} = \frac{338}{26} = 13$ mph. Thus, the separation increases at 13 miles per hour. A tempting distractor like $\frac{169}{26}$ mph arises from forgetting to divide by s or mishandling the formula, but the correct rate includes the division by the current distance. To solve related rates problems, identify the relating equation, differentiate with respect to time, substitute known values, and solve for the desired rate.

This problem involves solving related rates by differentiating a geometric relationship with respect to time. For the circle, area A = π r². Implicit differentiation gives dA/dt = 2 π r (dr/dt). When r=10 cm and dr/dt=-0.4 cm/s, dA/dt = 2 π *10 * (-0.4) = -8 π cm²/s. A tempting distractor like -4 π cm²/s might result from omitting the factor of 2 or using circumference instead, but the area derivative correctly involves 2 π r. To solve related rates problems generally, identify the equation relating the variables, differentiate with respect to time, substitute known values and rates, and solve for the desired rate.

This problem involves solving related rates by differentiating a geometric formula with respect to time. The volume of a sphere is given by V = (4/3)πr³, so implicit differentiation yields dV/dt = 4πr² dr/dt. Substituting r = 10 cm and dr/dt = 0.4 cm/s gives dV/dt = 4π(10)²(0.4) = 160π cm³/s. Thus, the volume increases at 160π cubic centimeters per second at that instant. A tempting distractor like 40π arises if one mistakenly uses the surface area formula instead of the correct volume derivative, but the volume rate requires the 4πr² factor. To solve related rates problems, identify the relating equation, differentiate with respect to time, substitute known values, and solve for the desired rate.

This problem involves solving related rates by differentiating a geometric formula with respect to time. The volume of the cone is $V = \frac{\pi h^3}{12}$, since $r = h/2$; implicit differentiation gives $\frac{dV}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$. When $h = 6$ ft and $\frac{dV}{dt} = 8 , \text{ft}^3/\text{min}$, substituting yields $8 = \left( \pi \cdot 36 / 4 \right) \frac{dh}{dt}$, so $\frac{dh}{dt} = \frac{8}{9\pi} , \text{ft/min}$. Thus, the height increases at $\frac{8}{9\pi}$ feet per minute. A tempting distractor like $8/(3\pi)$ occurs if one uses $r = h$ instead of $r = h/2$ in the volume formula, but correctly substituting $r = h/2$ gives the factor of $1/12$. To solve related rates problems, identify the relating equation, differentiate with respect to time, substitute known values, and solve for the desired rate.

This problem involves solving related rates by differentiating a geometric relationship with respect to time. For the shadow, similar triangles give $ s = \frac{5}{11} x $, where s is shadow length and x is distance from pole. Implicit differentiation yields $ \frac{ds}{dt} = \frac{5}{11} \frac{dx}{dt} $. When x=12 ft and $ \frac{dx}{dt} = 3 $ ft/s, $ \frac{ds}{dt} = \frac{5}{11} \times 3 = \frac{15}{11} $ ft/s. A tempting distractor like $ \frac{11}{15} $ ft/s might result from inverting the ratio of heights or misapplying proportions, but the correct similarity uses $ \frac{5}{16} $ for the fraction leading to $ \frac{5}{16-5} = \frac{5}{11} $. To solve related rates problems generally, identify the equation relating the variables, differentiate with respect to time, substitute known values and rates, and solve for the desired rate.

This problem involves solving related rates by differentiating a geometric relationship with respect to time. The point moves on y = √x, so y = $x^{1/2}$. Implicit differentiation gives dy/dt = (1/2) $x^{-1/2}$ (dx/dt). At x = 9 and dx/dt = 8, dy/dt = (1/2) (1/3) * 8 = 4/3. A tempting distractor like 8/3 might result from using dy/dx = 1/√x without halving or miscalculating the derivative, but the chain rule requires the 1/2 factor. To solve related rates problems generally, identify the equation relating the variables, differentiate with respect to time, substitute known values and rates, and solve for the desired rate.