This problem tests derivative interpretation for a changing geometric measurement. The derivative r'(20) represents the instantaneous rate at which the balloon's radius is changing at t = 20 seconds. Since r(t) is in centimeters and t is in seconds, r'(20) has units of centimeters per second and tells us how fast the radius is expanding at that moment. Choice D incorrectly adds "per second" to what would be the total change in radius, confusing accumulated change with instantaneous rate. For any measurement changing over time, the derivative gives the instantaneous rate with appropriate rate units.

This question requires interpreting the derivative of position with respect to time. The derivative s'(12) represents the instantaneous rate of change of position at t = 12 seconds, which is the definition of instantaneous velocity. Since s(t) is in meters and t is in seconds, s'(12) has units of meters per second and gives the car's velocity at that exact moment. Choice C incorrectly describes the total distance s(12), not the rate s'(12). When position is given as a function of time, always remember that the first derivative gives velocity.

This question asks you to interpret the derivative in a geometric context. The function r(t) = 2 + 0.3t² gives the balloon's radius in centimeters at time t seconds, so r'(t) represents the instantaneous rate of change of radius with respect to time. Therefore, r'(5) tells us how fast the radius is changing at t = 5 seconds, measured in centimeters per second. Choice A incorrectly identifies this as the radius value r(5), while choice E confuses instantaneous rate with average rate over an interval. Remember that derivatives capture instantaneous behavior - they tell us the rate of change at a single moment, not over a time period.

This question requires interpreting the derivative of a position function. Since s(t) represents position in meters at time t seconds, the derivative s'(t) represents the instantaneous rate of change of position, which is velocity. Therefore, s'(2) gives the car's instantaneous velocity at t = 2 seconds, measured in meters per second. Choice B incorrectly identifies this as position s(2), while choice C confuses the first derivative (velocity) with the second derivative (acceleration). Remember that for motion problems, the derivative chain is: position → velocity → acceleration, with each derivative representing the rate of change of the previous quantity.

This problem requires interpreting the derivative of a population function. Since P(t) = 5000 + 200t - 4t² represents population size in people at time t years, the derivative P'(t) represents the instantaneous rate of change of population with respect to time. At t = 6, P'(6) tells us how fast the population is changing at that moment, measured in people per year. Choice A incorrectly gives the actual population P(6), while choice C mistakenly describes a total change rather than an instantaneous rate. When interpreting derivatives in real-world contexts, focus on the word "rate" - derivatives always represent how fast something is changing at a specific instant.

This question requires interpreting the derivative of a height function. Since h(t) = 150t - 4.9t² represents the rocket's height in meters at time t seconds, the derivative h'(t) represents the instantaneous rate of change of height, which is vertical velocity. Therefore, h'(8) gives the rocket's instantaneous vertical velocity at t = 8 seconds, measured in meters per second. Choice A incorrectly identifies this as the height h(8), while choice C would require the second derivative h''(t) for acceleration. In physics problems, remember the derivative relationships: position → velocity → acceleration, where each arrow represents taking a derivative.

This question tests interpreting derivatives in a resource usage context. The function W(t) = 2.4 + 0.06t² represents water use in million gallons per day at time t days, so W'(t) represents the instantaneous rate of change of water use with respect to time. Therefore, W'(10) tells us how fast the water use rate is changing on day 10, measured in million gallons per day per day. Choice A incorrectly identifies this as the water use W(10), while choice C mistakenly interprets this as a total rather than a rate of change. When the original function already represents a rate (gallons per day), its derivative represents the rate of change of that rate, leading to compound units.

This problem asks you to interpret the derivative in a temperature context. The function T(t) = 70 + 25e^(-0.2t) gives temperature in °C at time t minutes, so T'(t) represents the instantaneous rate of change of temperature with respect to time. At t = 10, T'(10) tells us how fast the temperature is changing at that exact moment, measured in °C per minute. Choice A incorrectly gives the actual temperature T(10), while choice C mistakenly describes a total change rather than an instantaneous rate. When working with derivatives, always distinguish between the value of a function (what is) and its rate of change (how fast it's changing).

This question tests interpreting derivatives in a motion context. The function $d(t) = 400(1 - e^{-0.05t})$ represents the runner's distance in meters at time $t$ seconds, so $d'(t)$ represents the instantaneous rate of change of distance, which is speed (or velocity magnitude). Therefore, $d'(20)$ gives the runner's instantaneous speed at $t = 20$ seconds, measured in meters per second. Choice A incorrectly identifies this as the position $d(20)$, while choice D confuses instantaneous speed with average speed over an interval. To verify units in derivative problems, remember that the derivative's units are always the original function's units divided by the input variable's units.

This question tests derivative interpretation in fuel consumption. f(t) is fuel in gallons at t hours, so f'(t) is the instantaneous consumption rate. At t=0.5 hours, f'(0.5) shows usage speed. Units are gallons per hour. Choice A confuses amount with rate. Divide volume by time for rate units.