AP Calculus BC Quiz: Introduction To Related Rates

What this quiz covers

This quiz focuses on Introduction To Related Rates, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$ (correct answer)
2. $\dfrac{dr}{dt}=2\pi r\dfrac{dA}{dt}$
3. $\dfrac{dA}{dt}=\pi r\dfrac{dr}{dt}$
4. $\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}$
5. $\dfrac{dA}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2$

Explanation: This question introduces related rates by setting up the differentiation to solve for $dr/dt$ from $dA/dt$ for a circle. Since $r$ depends on $t$, implicit differentiation of $A = \pi r^2$ gives $\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$. This equation can be solved for $\dfrac{dr}{dt} = \dfrac{\dfrac{dA}{dt}}{2\pi r}$. The setup requires the chain rule to relate the rates properly. Choice B incorrectly solves by multiplying instead of dividing, reversing the relationship. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to $t$ using the chain rule, and solve for the unknown rate.

Question 2

1. $\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}$ (correct answer)
2. $\dfrac{dV}{dt}=s^2\dfrac{ds}{dt}$
3. $\dfrac{dV}{dt}=3\dfrac{ds}{dt}$
4. $\dfrac{dV}{dt}=3s^2$
5. $\dfrac{dV}{dt}=\left(\dfrac{ds}{dt}\right)^3$

Explanation: This question introduces related rates by differentiating the volume formula for a cube with respect to time. Since s depends on t, implicit differentiation with the chain rule is essential. Differentiating $V = s^3$ yields $3s^2 \dfrac{ds}{dt}$. This relates volume and side length rates. Choice D omits $\dfrac{ds}{dt}$, treating s as constant. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to $t$ using the chain rule, and solve for the unknown rate.

Question 3

1. $x\dfrac{dy}{dt}+y\dfrac{dx}{dt}=1$ (correct answer)
2. $\dfrac{dx}{dt}\dfrac{dy}{dt}=1$
3. $x+y=1$
4. $\dfrac{dx}{dt}+\dfrac{dy}{dt}=1$
5. $xy\left(\dfrac{dx}{dt}+\dfrac{dy}{dt}\right)=1$

Explanation: This question introduces related rates by differentiating a product equation involving time. With x and y as functions of t in $xy = t$, implicit differentiation is required. The product rule gives $x \dfrac{dy}{dt} + y \dfrac{dx}{dt} = 1$, since $\dfrac{d(t)}{dt} = 1$. This relates the rates with the time derivative. Choice C presents the undifferentiated equation without rates. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 4

1. $\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$ (correct answer)
2. $\frac{dV}{dt}=2\pi r\frac{dr}{dt}\,h$
3. $\frac{dV}{dt}=\pi r^2h\frac{dh}{dt}$
4. $\frac{dV}{dt}=\pi r\frac{dh}{dt}$
5. $\frac{dV}{dt}=\pi r^2h$

Explanation: This problem introduces the concept of related rates, determining a cylinder's volume change with constant radius and varying height. Implicit differentiation with respect to time is required because V = πr²h, with h changing over t and r fixed, so dr/dt = 0. Differentiating yields dV/dt = πr² dh/dt, focusing on the height's rate. This simplifies due to the constant radius. A tempting distractor like choice C fails by including an extra h factor, treating volume as the rate. Always differentiate the given equation implicitly with respect to time and apply the chain rule to each variable that depends on t.

Question 5

1. $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ (correct answer)
2. $\dfrac{dV}{dt}=4\pi r^2$
3. $\dfrac{dV}{dt}=4\pi\dfrac{dr}{dt}$
4. $\dfrac{dV}{dt}=\dfrac{4}{3}\pi\left(3r^2\right)$
5. $\dfrac{dV}{dt}=\dfrac{4}{3}\pi r^3\dfrac{dr}{dt}$

Explanation: This problem requires setting up a related rates equation for a changing sphere. The volume formula V = (4/3)πr³ must be differentiated implicitly with respect to time since r depends on t. Using the chain rule, dV/dt = (4/3)π · 3r² · dr/dt = 4πr² · dr/dt, which matches choice A. Choice E incorrectly keeps the entire original formula and multiplies by dr/dt, failing to apply the chain rule properly. The key strategy is to differentiate each variable that changes with time, applying the chain rule to composite functions.

Question 6

1. $\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}$ (correct answer)
2. $\dfrac{dV}{dt}=3s\dfrac{ds}{dt}$
3. $\dfrac{dV}{dt}=s^3\dfrac{ds}{dt}$
4. $\dfrac{dV}{dt}=3s^2$
5. $\dfrac{dV}{dt}=\left(\dfrac{ds}{dt}\right)^3$

Explanation: This is a basic related rates problem for a cube's changing volume. Starting with V = s³, we differentiate with respect to time using the chain rule: dV/dt = 3s²(ds/dt), which matches choice A. The power rule gives us 3s² as the derivative of s³ with respect to s, and the chain rule requires multiplying by ds/dt. Choice C incorrectly keeps s³ in the formula, failing to apply the power rule before the chain rule. The strategy for any related rates problem is to differentiate the given relationship, treating all time-dependent variables as functions of t.

Question 7

1. $\frac{dS}{dt}=12s\frac{ds}{dt}$ (correct answer)
2. $\frac{dS}{dt}=6\cdot 2s$
3. $\frac{dS}{dt}=6s^2\frac{ds}{dt}$
4. $\frac{dS}{dt}=\frac{12s}{\frac{ds}{dt}}$
5. $\frac{dS}{dt}=12\frac{ds}{dt}$

Explanation: This problem involves setting up a related rates equation for the surface area of a cube with changing side length s. Given $S = 6s^2$, differentiate both sides with respect to time t. Applying the chain rule to $s^2$ gives $\frac{dS}{dt} = 6 \cdot 2s \frac{ds}{dt} = 12s \frac{ds}{dt}$. This implicit differentiation accounts for s varying with t, linking the rates properly. A tempting distractor like choice B omits $\frac{ds}{dt}$, as if computing a static value rather than a rate. A transferable strategy in related rates is to differentiate the given equation with respect to time, applying the chain rule to all variables that depend on t.

Question 8

1. $\sec^2\theta\,\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$ (correct answer)
2. $\sec\theta\,\frac{d\theta}{dt}=\frac{dy}{dt}$
3. $\tan\theta\,\frac{d\theta}{dt}=\frac{y}{20}$
4. $\sec^2\theta=\frac{1}{20}\frac{dy}{dt}$
5. $\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$

Explanation: This problem involves setting up a related rates equation for the viewing angle $\theta$ of a rocket with height $y$, given $\tan \theta = \frac{y}{20}$. Differentiate both sides with respect to t: $\sec^2 \theta \frac{d\theta}{dt} = \frac{1}{20} \frac{dy}{dt}$. The chain rule is used on the left for $\tan \theta$, as $\theta$ depends on t, while on the right it's straightforward for y. This implicit method relates the angular rate to the height rate. A tempting distractor like choice E omits the $\sec^2 \theta$ factor, ignoring the derivative of $\tan \theta$. A transferable strategy in related rates is to differentiate the governing equation with respect to time, applying the chain rule to all variables that depend on t.

Question 9

1. $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$ (correct answer)
2. $2x+2y=0$
3. $x\frac{dy}{dt}+y\frac{dx}{dt}=25$
4. $2x\frac{dy}{dt}+2y\frac{dx}{dt}=0$
5. $\frac{dx}{dt}+\frac{dy}{dt}=0$

Explanation: This problem involves setting up a related rates equation for a right triangle with legs x and y and fixed hypotenuse 5. From $x^2 + y^2 = 25$, differentiate with respect to t to get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$. The chain rule is applied to each squared term since x and y change over time. This implicit approach relates the rates while maintaining the constant hypotenuse. A tempting distractor like choice B is the undifferentiated equation, failing to incorporate time derivatives. A transferable strategy in related rates is to differentiate the governing equation with respect to time, applying the chain rule to all variables that depend on t.

Question 10

1. $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ (correct answer)
2. $\frac{dV}{dt}=4\pi r^2$
3. $\frac{dV}{dt}=\frac{4}{3}\pi\cdot 3r^2$
4. $\frac{dV}{dt}=\frac{4}{3}\pi r^3\frac{dr}{dt}$
5. $\frac{dV}{dt}=4\pi r\frac{dr}{dt}$

Explanation: This problem introduces the concept of related rates, where we connect the rate of change of a sphere's volume to its radius using differentiation. Implicit differentiation with respect to time is required because the volume V is a function of r, which itself changes over time, so we treat r as a function of t. By differentiating both sides of V = (4/3)πr³ with respect to t, we apply the chain rule to get dV/dt = (4/3)π * 3r² dr/dt, which simplifies to 4πr² dr/dt. This step accounts for how the rate of volume change depends on both the current radius and its rate of change. A tempting distractor like choice B fails because it omits the dr/dt term, treating the rate as constant without accounting for the changing radius. Always differentiate the given equation implicitly with respect to time and apply the chain rule to each variable that depends on t.

Question 11

1. $\dfrac{dd}{dt} = \sqrt{2} \, \dfrac{ds}{dt}$ (correct answer)
2. $\dfrac{dd}{dt} = 2s \dfrac{ds}{dt}$
3. $\dfrac{dd}{dt} = \dfrac{1}{\sqrt{2}} \, \dfrac{ds}{dt}$
4. $\dfrac{dd}{dt} = \sqrt{2} \, s$
5. $\dfrac{dd}{dt} = \left(\dfrac{ds}{dt}\right) \sqrt{2} s$

Explanation: This question introduces related rates by differentiating the diagonal formula for a square with respect to time. With s as a function of t, implicit differentiation is required for $d = s \sqrt{2}$. Differentiating gives $\frac{dd}{dt} = \sqrt{2} \frac{ds}{dt}$ using the chain rule. This connects side and diagonal rates. Choice B incorrectly uses $2s \frac{ds}{dt}$, perhaps confusing with area differentiation. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 12

1. $\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}h+r^2\dfrac{dh}{dt}\right)$ (correct answer)
2. $\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}+r^2\dfrac{dh}{dt}\right)$
3. $\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2rh\right)$
4. $\dfrac{dV}{dt}=\frac{1}{3}\pi r^2 h \dfrac{dr}{dt} \dfrac{dh}{dt}$
5. $\dfrac{dV}{dt}=\frac{1}{3}\pi\left(r^2 h\right)'=\frac{1}{3}\pi\left(2r+r^2\right)$

Explanation: This question introduces related rates by differentiating the volume formula for a cone with respect to time. Since both r and h are functions of t, implicit differentiation with respect to t is required. Differentiating $V = \frac{1}{3} \pi r^2 h$ involves the product rule: $\frac{1}{3} \pi (2r \frac{dr}{dt} h + r^2 \frac{dh}{dt})$. This accounts for simultaneous changes in radius and height. Choice C omits the rates dr/dt and dh/dt, incorrectly treating the variables as constants. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 13

1. $\dfrac{dA}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2$
2. $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$ (correct answer)
3. $\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}$
4. $\dfrac{dA}{dt}=2\pi r$
5. $\dfrac{dA}{dt}=\pi r^2\dfrac{dr}{dt}$

Explanation: This question introduces related rates by differentiating the area formula for a circle with respect to time. Since the radius r depends on time t, implicit differentiation with respect to t is required to connect $\dfrac{dA}{dt}$ and $\dfrac{dr}{dt}$. Differentiating $A = \pi r^2$ gives $2\pi r \dfrac{dr}{dt}$ via the chain rule, as the derivative of $r^2$ is $2r \dfrac{dr}{dt}$. This captures the rate at which the area changes as the radius varies. Choice D fails by omitting r and $\dfrac{dr}{dt}$, mistakenly treating the derivative as if it were a constant circumference. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 14

1. $\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$ (correct answer)
2. $\dfrac{dS}{dt}=4\pi r\dfrac{dr}{dt}$
3. $\dfrac{dS}{dt}=8\pi\dfrac{dr}{dt}$
4. $\dfrac{dS}{dt}=4\pi r^2$
5. $\dfrac{dS}{dt}=4\pi\left(\dfrac{dr}{dt}\right)^2$

Explanation: This question introduces related rates by differentiating the surface area formula for a sphere with respect to time. Since r depends on t, implicit differentiation with the chain rule is essential. Differentiating $S = 4\pi r^2$ yields $8\pi r \dfrac{dr}{dt}$. This relates surface area and radius rates. Choice D omits dr/dt and miscounts the coefficient. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 15

1. $\dfrac{ds}{dt}=\dfrac{x\dfrac{dx}{dt}+y\dfrac{dy}{dt}}{\sqrt{x^2+y^2}}$ (correct answer)
2. $\dfrac{ds}{dt}=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}$
3. $\dfrac{ds}{dt}=\dfrac{\dfrac{dx}{dt}+\dfrac{dy}{dt}}{\sqrt{x^2+y^2}}$
4. $\dfrac{ds}{dt}=\dfrac{x+y}{\sqrt{x^2+y^2}}$
5. $\dfrac{ds}{dt}=\dfrac{x^2\dfrac{dx}{dt}+y^2\dfrac{dy}{dt}}{x^2+y^2}$

Explanation: This question introduces related rates by differentiating the distance formula for a moving object with respect to time. With x and y as functions of t, implicit differentiation of $s = \sqrt{x^2 + y^2}$ is needed. Differentiating $s^2 = x^2 + y^2$ first gives $2s \dfrac{ds}{dt} = 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}$, solving to $\dfrac{ds}{dt} = \dfrac{x \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{s}$. This relates the radial rate to component rates. Choice B confuses $\dfrac{ds}{dt}$ with the speed magnitude, which is different from the rate of distance change. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 16

1. $\dfrac{dC}{dt}=2\pi\dfrac{dr}{dt}$ (correct answer)
2. $\dfrac{dC}{dt}=2\pi r\dfrac{dr}{dt}$
3. $\dfrac{dC}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2$
4. $\dfrac{dC}{dt}=2\pi r$
5. $\dfrac{dC}{dt}=\dfrac{1}{2\pi}\dfrac{dr}{dt}$

Explanation: This question introduces related rates by differentiating the circumference formula for a circle with respect to time. Since $r$ depends on $t$, implicit differentiation with respect to $t$ is needed. Differentiating $C = 2\pi r$ gives $2\pi \dfrac{dr}{dt}$ directly, as the chain rule applies to $r$. This relates the rates of circumference and radius change. Choice D omits $\dfrac{dr}{dt}$, incorrectly assuming a static radius. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to $t$ using the chain rule, and solve for the unknown rate.

Question 17

1. $V=\frac{1}{12}\pi h^3\Rightarrow\dfrac{dV}{dt}=\frac{1}{4}\pi h^2\dfrac{dh}{dt}$ (correct answer)
2. $V=\frac{1}{3}\pi\left(\frac{1}{2}h\right)^2\Rightarrow\dfrac{dV}{dt}=\frac{1}{3}\pi\cdot\frac{1}{4}\cdot2h\dfrac{dh}{dt}$
3. $V=\frac{1}{12}\pi h^3\Rightarrow\dfrac{dV}{dt}=\frac{1}{12}\pi\cdot3h^2$
4. $V=\frac{1}{3}\pi r^2h\Rightarrow\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}+\dfrac{dh}{dt}\right)$
5. $V=\frac{1}{12}\pi h^3\Rightarrow\dfrac{dV}{dt}=\frac{1}{12}\pi\left(\dfrac{dh}{dt}\right)^3$

Explanation: This question introduces related rates by expressing and differentiating the volume of a cone with proportional radius and height. Since r = (1/2)h and both depend on t, substitute to make V a function of h, then differentiate implicitly. Differentiating V = (1/12)πh³ gives (1/12)π * 3h² dh/dt = (1/4)π h² dh/dt. This relates volume rate to height rate. Choice C forgets dh/dt, treating h as constant in the derivative. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 18

1. $\dfrac{dd}{dt}=2\dfrac{dr}{dt}$ (correct answer)
2. $\dfrac{dd}{dt}=2r\dfrac{dr}{dt}$
3. $\dfrac{dd}{dt}=\dfrac{1}{2}\dfrac{dr}{dt}$
4. $\dfrac{dd}{dt}=2r$
5. $\dfrac{dd}{dt}=\left(\dfrac{dr}{dt}\right)^2$

Explanation: This question introduces related rates by differentiating the diameter formula for a circle with respect to time. Since r depends on t, implicit differentiation of $d = 2r$ is straightforward. Differentiating gives $\frac{dd}{dt} = 2 \frac{dr}{dt}$ using the chain rule implicitly. This directly relates the rates. Choice D omits dr/dt, ignoring the rate. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to t using the chain rule, and solve for the unknown rate.

Question 19

1. $\frac{dx}{dt}\frac{dy}{dt}=0$
2. $x\frac{dy}{dt}+y\frac{dx}{dt}=0$ (correct answer)
3. $\frac{dx}{dt}+\frac{dy}{dt}=0$
4. $xy\left(\frac{dx}{dt}+\frac{dy}{dt}\right)=0$
5. $x\frac{dx}{dt}+y\frac{dy}{dt}=50$

Explanation: This problem introduces the concept of related rates, relating side length changes in a rectangle with constant area. Implicit differentiation with respect to time is required because x and y both depend on t, constrained by xy = 50. Differentiating yields x dy/dt + y dx/dt = 0, linking the rates inversely. This reflects how one side's increase compensates for the other's decrease to maintain area. A tempting distractor like choice E fails by mistakenly differentiating as if it were a sum instead of a product. Always differentiate the given equation implicitly with respect to time and apply the chain rule to each variable that depends on t.

Question 20

1. $2x\frac{dx}{dt}+8y\frac{dy}{dt}=0$ (correct answer)
2. $2x+8y=100$
3. $\frac{dx}{dt}+4\frac{dy}{dt}=0$
4. $2x\frac{dx}{dt}+4y\frac{dy}{dt}=100$
5. $x\frac{dx}{dt}+4y\frac{dy}{dt}=0$

Explanation: This problem introduces the concept of related rates, tracking a point's movement on an elliptic curve over time. Implicit differentiation with respect to time is required as x and y are functions of t, related by x² + 4y² = 100. Differentiating gives 2x dx/dt + 8y dy/dt = 0, connecting the velocity components. This maintains the curve constraint dynamically. A tempting distractor like choice D fails by incorrectly differentiating the constant to 100 rather than 0. Always differentiate the given equation implicitly with respect to time and apply the chain rule to each variable that depends on t.