AP Calculus BC Quiz: Introduction To Related Rates
20 questions · exam conditions
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Introduction To Related RatesQuestion 1 of 20

A circle's radius r(t)r(t) changes and its area is A=πr2A=\pi r^2; which setup finds drdt\dfrac{dr}{dt} from dAdt\dfrac{dA}{dt}?

dAdt=2πrdrdt\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}
drdt=2πrdAdt\dfrac{dr}{dt}=2\pi r\dfrac{dA}{dt}
dAdt=πrdrdt\dfrac{dA}{dt}=\pi r\dfrac{dr}{dt}
dAdt=2πdrdt\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}
dAdt=π(drdt)2\dfrac{dA}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2
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AP Calculus BC Quiz

AP Calculus BC Quiz: Introduction To Related Rates

Practice Introduction To Related Rates in AP Calculus BC with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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.

All questions

Question 1

A circle's radius r(t)r(t) changes and its area is A=πr2A=\pi r^2; which setup finds drdt\dfrac{dr}{dt} from dAdt\dfrac{dA}{dt}?

  1. dAdt=2πrdrdt\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt} (correct answer)
  2. drdt=2πrdAdt\dfrac{dr}{dt}=2\pi r\dfrac{dA}{dt}
  3. dAdt=πrdrdt\dfrac{dA}{dt}=\pi r\dfrac{dr}{dt}
  4. dAdt=2πdrdt\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}
  5. dAdt=π(drdt)2\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/dtdr/dt from dA/dtdA/dt for a circle. Since rr depends on tt, implicit differentiation of A=πr2A = \pi r^2 gives dAdt=2πrdrdt\dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}. This equation can be solved for drdt=dAdt2πr\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 tt using the chain rule, and solve for the unknown rate.

Question 2

An expanding cube has volume V(t)=s(t)3V(t)=s(t)^3; which equation correctly relates dVdt\dfrac{dV}{dt} and dsdt\dfrac{ds}{dt}?

  1. dVdt=3s2dsdt\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt} (correct answer)
  2. dVdt=s2dsdt\dfrac{dV}{dt}=s^2\dfrac{ds}{dt}
  3. dVdt=3dsdt\dfrac{dV}{dt}=3\dfrac{ds}{dt}
  4. dVdt=3s2\dfrac{dV}{dt}=3s^2
  5. dVdt=(dsdt)3\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=s3V = s^3 yields 3s2dsdt3s^2 \dfrac{ds}{dt}. This relates volume and side length rates. Choice D omits dsdt\dfrac{ds}{dt}, treating s as constant. A transferable related-rates strategy is to identify the governing equation, differentiate both sides with respect to tt using the chain rule, and solve for the unknown rate.

Question 3

A particle satisfies x(t)y(t)=tx(t)y(t)=t; which implicit differentiation correctly relates dxdt\dfrac{dx}{dt} and dydt\dfrac{dy}{dt}?

  1. xdydt+ydxdt=1x\dfrac{dy}{dt}+y\dfrac{dx}{dt}=1 (correct answer)
  2. dxdtdydt=1\dfrac{dx}{dt}\dfrac{dy}{dt}=1
  3. x+y=1x+y=1
  4. dxdt+dydt=1\dfrac{dx}{dt}+\dfrac{dy}{dt}=1
  5. xy(dxdt+dydt)=1xy\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=txy = t, implicit differentiation is required. The product rule gives xdydt+ydxdt=1x \dfrac{dy}{dt} + y \dfrac{dx}{dt} = 1, since d(t)dt=1\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

A cylinder has constant radius rr and changing height hh; volume V=πr2hV=\pi r^2h. Which setup correctly gives dVdt\frac{dV}{dt} in terms of dhdt\frac{dh}{dt}?

  1. dVdt=πr2dhdt\frac{dV}{dt}=\pi r^2\frac{dh}{dt} (correct answer)
  2. dVdt=2πrdrdt h\frac{dV}{dt}=2\pi r\frac{dr}{dt}\,h
  3. dVdt=πr2hdhdt\frac{dV}{dt}=\pi r^2h\frac{dh}{dt}
  4. dVdt=πrdhdt\frac{dV}{dt}=\pi r\frac{dh}{dt}
  5. dVdt=πr2h\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

A spherical balloon's radius rr increases with time; its volume is V=43πr3V=\frac{4}{3}\pi r^3. Which related-rates setup is correct?​

  1. dVdt=4πr2drdt\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt} (correct answer)
  2. dVdt=4πr2\dfrac{dV}{dt}=4\pi r^2
  3. dVdt=4πdrdt\dfrac{dV}{dt}=4\pi\dfrac{dr}{dt}
  4. dVdt=43π(3r2)\dfrac{dV}{dt}=\dfrac{4}{3}\pi\left(3r^2\right)
  5. dVdt=43πr3drdt\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

A cube has side length s(t)s(t) and volume V=s3V=s^3. Which related-rates equation correctly connects dVdt\dfrac{dV}{dt} and dsdt\dfrac{ds}{dt}?​

  1. dVdt=3s2dsdt\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt} (correct answer)
  2. dVdt=3sdsdt\dfrac{dV}{dt}=3s\dfrac{ds}{dt}
  3. dVdt=s3dsdt\dfrac{dV}{dt}=s^3\dfrac{ds}{dt}
  4. dVdt=3s2\dfrac{dV}{dt}=3s^2
  5. dVdt=(dsdt)3\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

A cube's side length ss changes over time. If surface area S=6s2S=6s^2, which equation correctly relates dSdt\frac{dS}{dt} and dsdt\frac{ds}{dt}?

  1. dSdt=12sdsdt\frac{dS}{dt}=12s\frac{ds}{dt} (correct answer)
  2. dSdt=6⋅2s\frac{dS}{dt}=6\cdot 2s
  3. dSdt=6s2dsdt\frac{dS}{dt}=6s^2\frac{ds}{dt}
  4. dSdt=12sdsdt\frac{dS}{dt}=\frac{12s}{\frac{ds}{dt}}
  5. dSdt=12dsdt\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=6s2S = 6s^2, differentiate both sides with respect to time t. Applying the chain rule to s2s^2 gives dSdt=6⋅2sdsdt=12sdsdt\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 dsdt\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

A camera is 20 m from a launch pad; rocket height y(t)y(t) and viewing angle θ(t)\theta(t) satisfy tan⁡θ=y20\tan\theta=\frac{y}{20}. Which rate setup is correct?

  1. sec⁡2θ dθdt=120dydt\sec^2\theta\,\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt} (correct answer)
  2. sec⁡θ dθdt=dydt\sec\theta\,\frac{d\theta}{dt}=\frac{dy}{dt}
  3. tan⁡θ dθdt=y20\tan\theta\,\frac{d\theta}{dt}=\frac{y}{20}
  4. sec⁡2θ=120dydt\sec^2\theta=\frac{1}{20}\frac{dy}{dt}
  5. dθdt=120dydt\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}
Explanation: This problem involves setting up a related rates equation for the viewing angle θ\theta of a rocket with height yy, given tan⁡θ=y20\tan \theta = \frac{y}{20}. Differentiate both sides with respect to t: sec⁡2θdθdt=120dydt\sec^2 \theta \frac{d\theta}{dt} = \frac{1}{20} \frac{dy}{dt}. The chain rule is used on the left for tan⁡θ\tan \theta, as θ\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⁡2θ\sec^2 \theta factor, ignoring the derivative of tan⁡θ\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

A right triangle has legs x(t)x(t) and y(t)y(t) with hypotenuse 55. If x2+y2=25x^2+y^2=25, which differentiated equation is correct?

  1. 2xdxdt+2ydydt=02x\frac{dx}{dt}+2y\frac{dy}{dt}=0 (correct answer)
  2. 2x+2y=02x+2y=0
  3. xdydt+ydxdt=25x\frac{dy}{dt}+y\frac{dx}{dt}=25
  4. 2xdydt+2ydxdt=02x\frac{dy}{dt}+2y\frac{dx}{dt}=0
  5. dxdt+dydt=0\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 x2+y2=25x^2 + y^2 = 25, differentiate with respect to t to get 2xdxdt+2ydydt=02x \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

A spherical balloon's radius rr increases at drdt\frac{dr}{dt}. Which equation correctly relates dVdt\frac{dV}{dt} to drdt\frac{dr}{dt} for V=43πr3V=\frac{4}{3}\pi r^3?

  1. dVdt=4πr2drdt\frac{dV}{dt}=4\pi r^2\frac{dr}{dt} (correct answer)
  2. dVdt=4πr2\frac{dV}{dt}=4\pi r^2
  3. dVdt=43π⋅3r2\frac{dV}{dt}=\frac{4}{3}\pi\cdot 3r^2
  4. dVdt=43πr3drdt\frac{dV}{dt}=\frac{4}{3}\pi r^3\frac{dr}{dt}
  5. dVdt=4πrdrdt\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

A square has side s(t)s(t) and diagonal d(t)d(t) with d=s2d = s \sqrt{2}; which rate relationship is correct?

  1. dddt=2 dsdt\dfrac{dd}{dt} = \sqrt{2} \, \dfrac{ds}{dt} (correct answer)
  2. dddt=2sdsdt\dfrac{dd}{dt} = 2s \dfrac{ds}{dt}
  3. dddt=12 dsdt\dfrac{dd}{dt} = \dfrac{1}{\sqrt{2}} \, \dfrac{ds}{dt}
  4. dddt=2 s\dfrac{dd}{dt} = \sqrt{2} \, s
  5. dddt=(dsdt)2s\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=s2d = s \sqrt{2}. Differentiating gives dddt=2dsdt\frac{dd}{dt} = \sqrt{2} \frac{ds}{dt} using the chain rule. This connects side and diagonal rates. Choice B incorrectly uses 2sdsdt2s \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

A cone has V(t)=13πr(t)2h(t)V(t)=\frac{1}{3}\pi r(t)^2 h(t); which related-rates derivative is correct?

  1. dVdt=13π(2rdrdth+r2dhdt)\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}h+r^2\dfrac{dh}{dt}\right) (correct answer)
  2. dVdt=13π(2rdrdt+r2dhdt)\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}+r^2\dfrac{dh}{dt}\right)
  3. dVdt=13π(2rh)\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2rh\right)
  4. dVdt=13πr2hdrdtdhdt\dfrac{dV}{dt}=\frac{1}{3}\pi r^2 h \dfrac{dr}{dt} \dfrac{dh}{dt}
  5. dVdt=13π(r2h)′=13π(2r+r2)\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=13πr2hV = \frac{1}{3} \pi r^2 h involves the product rule: 13π(2rdrdth+r2dhdt)\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

A circle's area is A(t)=πr(t)2A(t)=\pi r(t)^2; which equation correctly relates dAdt\dfrac{dA}{dt} and drdt\dfrac{dr}{dt}?

  1. dAdt=π(drdt)2\dfrac{dA}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2
  2. dAdt=2πrdrdt\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt} (correct answer)
  3. dAdt=2πdrdt\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}
  4. dAdt=2πr\dfrac{dA}{dt}=2\pi r
  5. dAdt=πr2drdt\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 dAdt\dfrac{dA}{dt} and drdt\dfrac{dr}{dt}. Differentiating A=πr2A = \pi r^2 gives 2πrdrdt2\pi r \dfrac{dr}{dt} via the chain rule, as the derivative of r2r^2 is 2rdrdt2r \dfrac{dr}{dt}. This captures the rate at which the area changes as the radius varies. Choice D fails by omitting r and drdt\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

A sphere's surface area is S(t)=4πr(t)2S(t)=4\pi r(t)^2; which equation correctly relates dSdt\dfrac{dS}{dt} and drdt\dfrac{dr}{dt}?

  1. dSdt=8πrdrdt\dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt} (correct answer)
  2. dSdt=4πrdrdt\dfrac{dS}{dt}=4\pi r\dfrac{dr}{dt}
  3. dSdt=8πdrdt\dfrac{dS}{dt}=8\pi\dfrac{dr}{dt}
  4. dSdt=4πr2\dfrac{dS}{dt}=4\pi r^2
  5. dSdt=4π(drdt)2\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=4πr2S = 4\pi r^2 yields 8πrdrdt8\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

A car is x(t)x(t) miles east and y(t)y(t) miles north of a station, with distance s(t)=x2+y2s(t)=\sqrt{x^2+y^2}; which derivative is correct?

  1. dsdt=xdxdt+ydydtx2+y2\dfrac{ds}{dt}=\dfrac{x\dfrac{dx}{dt}+y\dfrac{dy}{dt}}{\sqrt{x^2+y^2}} (correct answer)
  2. dsdt=(dxdt)2+(dydt)2\dfrac{ds}{dt}=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}
  3. dsdt=dxdt+dydtx2+y2\dfrac{ds}{dt}=\dfrac{\dfrac{dx}{dt}+\dfrac{dy}{dt}}{\sqrt{x^2+y^2}}
  4. dsdt=x+yx2+y2\dfrac{ds}{dt}=\dfrac{x+y}{\sqrt{x^2+y^2}}
  5. dsdt=x2dxdt+y2dydtx2+y2\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=x2+y2s = \sqrt{x^2 + y^2} is needed. Differentiating s2=x2+y2s^2 = x^2 + y^2 first gives 2sdsdt=2xdxdt+2ydydt2s \dfrac{ds}{dt} = 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}, solving to dsdt=xdxdt+ydydts\dfrac{ds}{dt} = \dfrac{x \dfrac{dx}{dt} + y \dfrac{dy}{dt}}{s}. This relates the radial rate to component rates. Choice B confuses dsdt\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

For a circle, circumference C(t)=2πr(t)C(t)=2\pi r(t); which related-rates equation correctly connects dCdt\dfrac{dC}{dt} and drdt\dfrac{dr}{dt}?

  1. dCdt=2πdrdt\dfrac{dC}{dt}=2\pi\dfrac{dr}{dt} (correct answer)
  2. dCdt=2πrdrdt\dfrac{dC}{dt}=2\pi r\dfrac{dr}{dt}
  3. dCdt=π(drdt)2\dfrac{dC}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2
  4. dCdt=2πr\dfrac{dC}{dt}=2\pi r
  5. dCdt=12πdrdt\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 rr depends on tt, implicit differentiation with respect to tt is needed. Differentiating C=2πrC = 2\pi r gives 2πdrdt2\pi \dfrac{dr}{dt} directly, as the chain rule applies to rr. This relates the rates of circumference and radius change. Choice D omits drdt\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 tt using the chain rule, and solve for the unknown rate.

Question 17

Water fills a cone where r(t)=12h(t)r(t)=\frac{1}{2}h(t) and V=13πr2hV=\frac{1}{3}\pi r^2h; which setup for dVdt\dfrac{dV}{dt} is correct?

  1. V=112πh3⇒dVdt=14πh2dhdtV=\frac{1}{12}\pi h^3\Rightarrow\dfrac{dV}{dt}=\frac{1}{4}\pi h^2\dfrac{dh}{dt} (correct answer)
  2. V=13π(12h)2⇒dVdt=13π⋅14⋅2hdhdtV=\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=112πh3⇒dVdt=112π⋅3h2V=\frac{1}{12}\pi h^3\Rightarrow\dfrac{dV}{dt}=\frac{1}{12}\pi\cdot3h^2
  4. V=13πr2h⇒dVdt=13π(2rdrdt+dhdt)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=112πh3⇒dVdt=112π(dhdt)3V=\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

A circle's radius r(t)r(t) changes and diameter is d(t)=2r(t)d(t)=2r(t); which rate relationship is correct?

  1. dddt=2drdt\dfrac{dd}{dt}=2\dfrac{dr}{dt} (correct answer)
  2. dddt=2rdrdt\dfrac{dd}{dt}=2r\dfrac{dr}{dt}
  3. dddt=12drdt\dfrac{dd}{dt}=\dfrac{1}{2}\dfrac{dr}{dt}
  4. dddt=2r\dfrac{dd}{dt}=2r
  5. dddt=(drdt)2\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=2rd = 2r is straightforward. Differentiating gives dddt=2drdt\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

A rectangle has sides xx and yy with constant area xy=50xy=50, and both change over time. Which relation between dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} is correct?

  1. dxdtdydt=0\frac{dx}{dt}\frac{dy}{dt}=0
  2. xdydt+ydxdt=0x\frac{dy}{dt}+y\frac{dx}{dt}=0 (correct answer)
  3. dxdt+dydt=0\frac{dx}{dt}+\frac{dy}{dt}=0
  4. xy(dxdt+dydt)=0xy\left(\frac{dx}{dt}+\frac{dy}{dt}\right)=0
  5. xdxdt+ydydt=50x\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

A point moves on the curve x2+4y2=100x^2+4y^2=100, with x,yx,y functions of tt. Which differentiated equation is correct?

  1. 2xdxdt+8ydydt=02x\frac{dx}{dt}+8y\frac{dy}{dt}=0 (correct answer)
  2. 2x+8y=1002x+8y=100
  3. dxdt+4dydt=0\frac{dx}{dt}+4\frac{dy}{dt}=0
  4. 2xdxdt+4ydydt=1002x\frac{dx}{dt}+4y\frac{dy}{dt}=100
  5. xdxdt+4ydydt=0x\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.