Introduction to Related Rates
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AP Calculus BC › Introduction to Related Rates
A cube’s side length $s$ changes over time. If surface area $S=6s^2$, which equation correctly relates $\frac{dS}{dt}$ and $\frac{ds}{dt}$?
$\frac{dS}{dt}=12\frac{ds}{dt}$
$\frac{dS}{dt}=6\cdot 2s$
$\frac{dS}{dt}=12s\frac{ds}{dt}$
$\frac{dS}{dt}=6s^2\frac{ds}{dt}$
$\frac{dS}{dt}=\frac{12s}{\frac{ds}{dt}}$
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.
A right triangle has legs $x(t)$ and $y(t)$ with hypotenuse $5$. If $x^2+y^2=25$, which differentiated equation is correct?
$2x+2y=0$
$\frac{dx}{dt}+\frac{dy}{dt}=0$
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$
$2x\frac{dy}{dt}+2y\frac{dx}{dt}=0$
$x\frac{dy}{dt}+y\frac{dx}{dt}=25$
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.
A camera is 20 m from a launch pad; rocket height $y(t)$ and viewing angle $\theta(t)$ satisfy $\tan\theta=\frac{y}{20}$. Which rate setup is correct?
$\tan\theta,\frac{d\theta}{dt}=\frac{y}{20}$
$\sec^2\theta,\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$
$\sec^2\theta=\frac{1}{20}\frac{dy}{dt}$
$\sec\theta,\frac{d\theta}{dt}=\frac{dy}{dt}$
$\frac{d\theta}{dt}=\frac{1}{20}\frac{dy}{dt}$
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.
A spherical balloon’s radius $r$ increases at $\frac{dr}{dt}$. Which equation correctly relates $\frac{dV}{dt}$ to $\frac{dr}{dt}$ for $V=\frac{4}{3}\pi r^3$?
$\frac{dV}{dt}=4\pi r^2$
$\frac{dV}{dt}=4\pi r\frac{dr}{dt}$
$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$
$\frac{dV}{dt}=\frac{4}{3}\pi r^3\frac{dr}{dt}$
$\frac{dV}{dt}=\frac{4}{3}\pi\cdot 3r^2$
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.
A circle’s radius $r(t)$ changes and its area is $A=\pi r^2$; which setup finds $\dfrac{dr}{dt}$ from $\dfrac{dA}{dt}$?
$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$
$\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}$
$\dfrac{dA}{dt}=\pi r\dfrac{dr}{dt}$
$\dfrac{dA}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2$
$\dfrac{dr}{dt}=2\pi r\dfrac{dA}{dt}$
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.
An expanding cube has volume $V(t)=s(t)^3$; which equation correctly relates $\dfrac{dV}{dt}$ and $\dfrac{ds}{dt}$?
$\dfrac{dV}{dt}=3s^2$
$\dfrac{dV}{dt}=s^2\dfrac{ds}{dt}$
$\dfrac{dV}{dt}=\left(\dfrac{ds}{dt}\right)^3$
$\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}$
$\dfrac{dV}{dt}=3\dfrac{ds}{dt}$
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.
A square has side $s(t)$ and diagonal $d(t)$ with $d = s \sqrt{2}$; which rate relationship is correct?
$\dfrac{dd}{dt} = 2s \dfrac{ds}{dt}$
$\dfrac{dd}{dt} = \dfrac{1}{\sqrt{2}} , \dfrac{ds}{dt}$
$\dfrac{dd}{dt} = \sqrt{2} , s$
$\dfrac{dd}{dt} = \left(\dfrac{ds}{dt}\right) \sqrt{2} s$
$\dfrac{dd}{dt} = \sqrt{2} , \dfrac{ds}{dt}$
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.
A particle satisfies $x(t)y(t)=t$; which implicit differentiation correctly relates $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$?
$\dfrac{dx}{dt}\dfrac{dy}{dt}=1$
$x\dfrac{dy}{dt}+y\dfrac{dx}{dt}=1$
$xy\left(\dfrac{dx}{dt}+\dfrac{dy}{dt}\right)=1$
$x+y=1$
$\dfrac{dx}{dt}+\dfrac{dy}{dt}=1$
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.
A cone has $V(t)=\frac{1}{3}\pi r(t)^2 h(t)$; which related-rates derivative is correct?
$\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2rh\right)$
$\dfrac{dV}{dt}=\frac{1}{3}\pi\left(r^2 h\right)'=\frac{1}{3}\pi\left(2r+r^2\right)$
$\dfrac{dV}{dt}=\frac{1}{3}\pi r^2 h \dfrac{dr}{dt} \dfrac{dh}{dt}$
$\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}+r^2\dfrac{dh}{dt}\right)$
$\dfrac{dV}{dt}=\frac{1}{3}\pi\left(2r\dfrac{dr}{dt}h+r^2\dfrac{dh}{dt}\right)$
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.
A circle’s area is $A(t)=\pi r(t)^2$; which equation correctly relates $\dfrac{dA}{dt}$ and $\dfrac{dr}{dt}$?
$\dfrac{dA}{dt}=2\pi\dfrac{dr}{dt}$
$\dfrac{dA}{dt}=\pi r^2\dfrac{dr}{dt}$
$\dfrac{dA}{dt}=\pi\left(\dfrac{dr}{dt}\right)^2$
$\dfrac{dA}{dt}=2\pi r$
$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$
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.