Rate of Change at a Point

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AP Calculus AB › Rate of Change at a Point

Questions 1 - 10
1

Let $A(t)$ be area of an oil spill; which represents the instantaneous rate of change at $t=7$?

$\dfrac{A(7)-A(0)}{7}$

The average value of $A$ on $[0,7]$

$\lim_{h\to 0}\dfrac{A(7+h)-A(7)}{h}$

$\dfrac{A(9)-A(7)}{2}$

$A(7)-A(6)$

Explanation

The instantaneous rate of change at t=7 tells us exactly how fast the oil spill area is expanding at that moment. Choice D correctly uses the limit definition, examining the rate as the time interval h approaches zero. Choice A calculates the average rate over the 2-hour interval [7,9], which gives the average expansion rate during that period, not the instantaneous rate at t=7. Choice B approximates using a one-hour interval but doesn't divide by time. Choice C gives the average rate from the beginning. Choice E refers to the average area value, not a rate of change. The key insight: instantaneous rates require limits to capture the exact rate at a single point in time, distinguishing them from average rates over intervals.

2

A runner’s distance is $d(t)$. Which represents the runner’s instantaneous speed at $t=6$ seconds?​

$\displaystyle \frac{d(8)-d(4)}{4}$

$\displaystyle d(6)-d(0)$

$\displaystyle \frac{d(6)-d(5)}{1}$

$\displaystyle d(6)$

$\displaystyle \lim_{h\to 0}\frac{d(6+h)-d(6)}{h}$

Explanation

Instantaneous speed is the magnitude of instantaneous velocity, which is the derivative of position (distance) with respect to time. Choice D correctly represents this as the limit as h approaches 0 of [d(6+h)-d(6)]/h, giving d'(6). Choice A computes the average speed between t=5 and t=6, Choice B gives the average speed from t=4 to t=8, and both are fixed-interval calculations. Choice C is just the position at t=6, not a rate, while Choice E is the total distance traveled in 6 seconds, not a rate of change. The critical distinction is that instantaneous rates require a limit of difference quotients, while average rates use specific time intervals without limits.

3

A population is modeled by $P(t)$. Which expression represents the instantaneous growth rate at $t=6$?

$\displaystyle \lim_{h\to 0}\dfrac{P(6+h)-P(6)}{h}$

$\dfrac{P(6)-P(5)}{6-5}$

$\dfrac{P(8)-P(4)}{8-4}$

$P(6)-P(0)$

$\dfrac{P(7)-P(6)}{7-6}$

Explanation

The instantaneous growth rate requires finding how fast the population is changing at exactly t=6, which means we need the derivative. Options A and E calculate average growth rates over the intervals [5,6] and [6,7] respectively, while option C gives the average rate from t=4 to t=8. Option B provides the total population change from t=0 to t=6, which isn't a rate. Only option D uses the limit definition, examining what happens as we consider smaller and smaller time intervals around t=6. This limiting process is essential for capturing the instantaneous rate. A helpful strategy: if you see "instantaneous," look for the limit notation—average rates never use limits.

4

A tank’s volume is $V(t)$ liters. Which quantity gives the instantaneous filling rate at $t=10$ minutes?​​​

$\displaystyle V(10)-V(9)$

$\displaystyle \frac{V(12)-V(8)}{4}$

$\displaystyle \frac{V(10)-V(0)}{10}$

The total amount of water added from $t=0$ to $t=10$, $V(10)-V(0)$

$\displaystyle \lim_{h\to 0}\frac{V(10+h)-V(10)}{h}$

Explanation

To find the instantaneous filling rate at a specific time, we need the derivative at that point, not an average over any interval. Choice A gives the average rate over 10 minutes, choices C and D give rates over fixed intervals (1 minute and 4 minutes respectively), and choice E describes the total change, not a rate. Only choice B uses the limit definition of the derivative, which captures the instantaneous rate by considering what happens as the time interval approaches zero. Many students mistakenly think that V(10)-V(9) gives the instantaneous rate because it uses a small interval, but this is still an average rate over that 1-minute period. The key insight is that instantaneous rates require limits, while average rates use fixed intervals.

5

A car’s distance from home is $d(t)$. Which expression is the instantaneous speed at $t=6$ hours?

$\displaystyle d(6)-d(0)$

$\displaystyle \frac{d(8)-d(4)}{4}$

$\displaystyle \lim_{h\to 0}\frac{d(6+h)-d(6)}{h}$

$\displaystyle \frac{d(6)-d(5)}{1}$

$\displaystyle \frac{d(10)-d(6)}{4}$

Explanation

Instantaneous speed is the magnitude of instantaneous velocity, which requires the derivative of the distance function. Choice D uses the limit definition of the derivative, capturing the instantaneous rate at exactly t=6. Choices A, B, and E all calculate average speeds over various intervals (1 hour, 4 hours, and 4 hours respectively). Choice C gives the total distance traveled in 6 hours, not a rate. Many students think that d(6)-d(5) represents instantaneous speed because it uses a small interval, but this is still the average speed over that hour. The key principle: instantaneous rates require limits as the time interval approaches zero, while any calculation over a fixed interval gives an average rate.

6

Let $C(t)$ be the cost to produce $t$ items. Which quantity is the instantaneous marginal cost at $t=100$?

$\displaystyle \lim_{h\to 0}\frac{C(100+h)-C(100)}{h}$

$\displaystyle \frac{C(120)-C(80)}{40}$

The added cost for the next item, $C(101)-C(100)$

The average cost per item at $t=100$, $\dfrac{C(100)}{100}$

$\displaystyle \frac{C(100)-C(0)}{100}$

Explanation

In economics, the instantaneous marginal cost is the derivative of the cost function, representing the rate of change of cost at a specific production level. Choice C correctly uses the limit definition of the derivative at t=100. Choice A gives the average cost per item over the first 100 items, choice B is the average cost per item (not marginal cost), choice D approximates the marginal cost but is technically the average rate over [100,101], and choice E gives the average marginal cost over a 40-item interval. A common confusion is between C(101)-C(100) and the true marginal cost; while economists often use this as an approximation, the exact instantaneous marginal cost requires the limit. Remember: instantaneous rates always involve limits, distinguishing them from discrete approximations.

7

For differentiable $f$, which limit gives the instantaneous rate of change of $f$ at $x=0$?​​​

$\displaystyle \frac{f(0)-f(-2)}{2}$

$\displaystyle \lim_{h\to 0}\frac{f(h)-f(0)}{h}$

$\displaystyle f(0)-f(-1)$

$\displaystyle \frac{f(1)-f(-1)}{2}$

$\displaystyle \frac{f(2)-f(0)}{2}$

Explanation

The instantaneous rate of change at x=0 is the derivative f'(0), which is defined as the limit of difference quotients. Choice A correctly represents this as lim[h→0] (f(0+h)-f(0))/h = lim[h→0] (f(h)-f(0))/h. Choices B, D, and E all calculate average rates of change over fixed intervals centered at or near x=0. Choice C gives only a difference without dividing by the change in x, so it's not a rate at all. A common mistake is thinking that using a symmetric interval like [-1,1] gives the instantaneous rate at the center, but this still yields an average. The distinguishing feature of instantaneous rates is the limit process, which captures what happens as the interval shrinks to nothing.

8

For a curve $y=f(x)$, what does $\displaystyle \lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ represent?

The y-intercept of the graph.

Slope of the tangent line at $x=2$.

Slope of the secant line from $x=2$ to $x=2+h$ for a fixed $h$.

Average rate of change on $[0,2]$.

The value $f(2+h)-f(2)$ when $h=1$.

Explanation

The limit given is the derivative $f'(2)$, which is the slope of the tangent at $x=2$, choice B, representing instantaneous rate, unlike average which is secant slope over an interval. Choice A is average over $[0,2]$, and choice C is secant for fixed h. A common confusion is mistaking net change (choice E) or y-intercept (choice D) for the rate. The limit makes it instantaneous. For comparison, link limits to tangents for instantaneous and fixed intervals to secants in curve analyses.

9

If $f(x)$ is differentiable, which best describes the instantaneous rate of change at $x=a$?

The value $f(a)$.

The average value of $f$ on $[a,a+1]$.

The change in $f$ from $x=0$ to $x=a$.

The slope of the secant line through $(a,f(a))$ and $(a+1,f(a+1))$.

The slope of the tangent line to $y=f(x)$ at $x=a$.

Explanation

Instantaneous rate of change is the slope of the tangent at x=a, choice C, differing from average, which is the secant slope over an interval like in choice A. This tangent slope captures the rate precisely at a. Choice A is average over [a,a+1], and choice D is an average value, not rate. Students commonly confuse the function value (choice E) or total change (choice B) with the rate. The derivative embodies the instantaneous concept. For strategy, associate tangent slopes with instantaneous and secants with averages in graphical interpretations.

10

If $C(t)$ is a company’s cost, which expression gives the instantaneous rate of change of cost at $t=8$?

$\displaystyle \lim_{h\to 0}\dfrac{C(8+h)-C(8)}{h}$

$C(8+h)$ for small $h$

$\dfrac{C(8)-C(0)}{8}$

$C(8)-C(7)$

$\dfrac{C(10)-C(6)}{10-6}$

Explanation

Average rates of change divide total change by interval length, reflecting overall trends, whereas instantaneous rates use the limit to isolate the rate at one moment. For C(t) at t=8, the instantaneous rate is the limit as h approaches 0 of [C(8+h) - C(8)] / h, which is D. A and B are averages, C is a difference, and E is a future value without rate calculation. Often, E is confused as indicating change, but it doesn't provide a rate. To distinguish, always look for the difference quotient limited to h=0 as the indicator of instantaneous over average rates.

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