Rate of Change at a Point

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AP Calculus AB › Rate of Change at a Point

Questions 1 - 10
1

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

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

The value $f(a)$.

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

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

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

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.

2

For temperature $T(t)$, which expression represents the instantaneous rate of change at $t=a$?

$\dfrac{T(a+1)-T(a)}{1}$

$\dfrac{T(b)-T(a)}{b-a}$

$\lim_{h\to 0}\dfrac{T(a+h)-T(a)}{h}$

$T(a+h)-T(a)$ for small $h$

$\dfrac{T(a)-T(0)}{a}$

Explanation

The instantaneous rate of change at t=a captures how rapidly the temperature is changing at that precise moment. Choice C correctly uses the limit definition, examining the rate as the time interval h approaches zero. Choice A approximates the instantaneous rate using a one-unit interval, but this gives the average rate over [a, a+1], not the exact instantaneous rate. Choice B explicitly calculates an average rate over the interval [a,b]. Choice D gives a temperature difference without dividing by time, so it's not a rate at all. Choice E calculates the average rate from time 0 to time a. The fundamental principle: instantaneous rates require limits to isolate the rate at a single point.

3

If $x(t)$ is a particle’s position, which expression gives instantaneous velocity at $t=b$?

$\dfrac{x(b)}{b}$

$\lim_{h\to 0}\dfrac{x(b+h)-x(b)}{h}$

$x(b+h)-x(b)$

$\dfrac{x(b)-x(a)}{b-a}$

$\dfrac{x(b+1)-x(b-1)}{2}$

Explanation

Instantaneous velocity at $t=b$ represents the particle's speed and direction at that exact moment. Choice B correctly uses the limit definition, where we examine the velocity as the time interval $h$ approaches zero. Choice A calculates the average velocity over the interval $[a, b]$, which tells us the overall motion between two times but not the specific velocity at $t=b$. Choice C gives a position difference without dividing by time, so it's not a velocity. Choice D approximates using a centered difference over a 2-unit interval, giving an average velocity. Choice E would only represent average velocity if the particle started at position 0 at time 0. Remember: instantaneous velocity requires a limit to isolate the rate at a single instant.

4

Water volume is $V(t)$. Which expression is the instantaneous inflow rate at $t=10$ minutes?​

$\displaystyle V(10+h)-V(10)$

$\displaystyle \frac{V(15)-V(5)}{10}$

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

$\displaystyle \frac{V(12)-V(10)}{2}$

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

Explanation

The instantaneous inflow rate at a specific time is the derivative of volume with respect to time, found by taking the limit of average rates over vanishingly small intervals. Choice C correctly expresses this as the limit as h approaches 0 of [V(10+h)-V(10)]/h, which defines V'(10). Choice A calculates the average rate over the first 10 minutes, Choice B gives the average rate from t=10 to t=12, and Choice E gives the average rate from t=5 to t=15. Choice D is missing the crucial division by h, giving only a change in volume rather than a rate. The distinguishing feature of instantaneous rates is the limit process—without it, you only have average rates over finite intervals.

5

A tank contains $V(t)$ liters after $t$ minutes. Which gives the instantaneous filling rate at $t=10$?​​

$\dfrac{V(12)-V(8)}{12-8}$

$\dfrac{V(10)-V(0)}{10}$

$\dfrac{V(15)-V(10)}{15-10}$

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

$V(11)-V(10)$

Explanation

The instantaneous filling rate represents how fast the volume is changing at exactly $t=10$ minutes, which requires the derivative at that point. Choice B correctly expresses this as $\lim_{h\to 0}\frac{V(10+h)-V(10)}{h}$, the limit definition of the derivative. Choices A, D, and E calculate average rates over various time intervals, telling us the average filling rate but not the instantaneous rate. Choice C computes a volume difference without dividing by time, giving units of liters rather than liters per minute. Students often confuse average rates (which use two distinct points) with instantaneous rates (which use a limit). The key distinction: instantaneous rates capture the rate at a single moment through limits, while average rates measure change over an interval.

6

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

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

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

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

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

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

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.

7

For a function $f$, which statement best characterizes the instantaneous rate of change at $x=1$?

The total change $f(2)-f(0)$

The average rate of change on $[0,1]$

The slope of the secant line from $x=1$ to $x=2$

The slope of the tangent line to $y=f(x)$ at $x=1$

The value $f(1)$

Explanation

The instantaneous rate of change at x=1 describes how rapidly f(x) is changing at that exact point. Choice C correctly identifies this as the slope of the tangent line at x=1, which geometrically captures the instantaneous rate. Choice A describes the average rate of change over the interval [1,2], telling us the overall trend between those points but not the specific rate at x=1. Choice B gives the function value itself, not a rate of change. Choice D calculates the average rate from x=0 to x=1. Choice E gives the total change over a 2-unit interval, not a rate. The fundamental distinction: instantaneous rates use tangent lines (limits), while average rates use secant lines (difference quotients).

8

If $V(t)$ is water volume, which expression is the instantaneous filling rate at $t=10$ minutes?

$\dfrac{V(10)-V(0)}{10}$

$\dfrac{V(12)-V(8)}{12-8}$

$V(10)-V(9)$

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

$\dfrac{V(20)-V(0)}{20}$

Explanation

The key distinction is that average rates average change over time periods, such as [V(b) - V(a)] / (b - a), while instantaneous rates capture the rate at an exact instant via the derivative limit. For V(t) at t=10, the instantaneous filling rate is the limit as h approaches 0 of [V(10+h) - V(10)] / h, matching option D. Options A, C, and E are average rates, and B is a volume difference without normalization by time. A frequent mix-up is thinking a tiny interval like B approximates it well enough, but it lacks the division and limit for true instantaneity. Always compare by looking for the limit form to identify instantaneous rates across contexts.

9

For position $s(t)$ of a runner, which expression represents the instantaneous velocity at $t=3$?

$\dfrac{s(5)-s(1)}{5-1}$

$s(3)-s(2)$

$\dfrac{s(3)-s(0)}{3-0}$

$\dfrac{s(4)-s(3)}{4-3}$

$\lim_{h\to 0}\dfrac{s(3+h)-s(3)}{h}$

Explanation

The instantaneous velocity at a specific time is the rate of change of position at that exact moment, not over an interval. Choice C correctly uses the limit definition: as the time interval h approaches 0, we get the instantaneous rate at t=3. Choices A, B, and E calculate average velocities over intervals (like [1,5], [0,3], or [3,4]), which give the average speed during those time periods. Choice D simply finds a position difference without dividing by time, so it doesn't represent velocity at all. Remember: instantaneous rates always involve limits, while average rates use the difference quotient over a finite interval.

10

For profit $R(t)$, which quantity is the instantaneous rate of change of profit at time $t=c$?

$R(c+1)-R(c)$

$\lim_{h\to 0}\dfrac{R(c+h)-R(c)}{h}$

$\dfrac{R(c)-R(0)}{c}$

$\dfrac{R(c+2)-R(c-2)}{4}$

The total profit earned up to time $c$

Explanation

The instantaneous rate of change of profit at t=c represents how quickly profit is changing at that exact moment. Choice D correctly uses the limit definition, examining how the profit rate behaves as the time interval h shrinks to zero around time c. Choice A gives the profit change over one time unit, which is an average rate over [c, c+1], not the instantaneous rate. Choice B calculates the average profit rate from the beginning. Choice C finds an average rate over a 4-unit interval centered at c. Choice E describes total profit, not a rate of change. Remember: instantaneous rates capture the rate at a single point using limits, while finite differences give average rates.

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