Introduction to Related Rates - AP Calculus BC

A spherical balloon's radius $r$ increases over time; with $V=\frac{4}{3}\pi r^3$ , which related-rates setup is correct?
Answer: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ This problem requires setting up a related rates equation for a changing sphere. The volume formula V = (4/3)πr³ contains the variable r that changes with time, so we must differentiate both sides with respect to t. Applying the chain rule to the right side: dV/dt = (4/3)π · 3r² · dr/dt = 4πr² · dr/dt. Choice D incorrectly omits dr/dt, treating the derivative as if we're finding dV/dr instead of dV/dt. The key strategy in related rates is to differentiate the entire equation with respect to time, ensuring every changing variable gets multiplied by its rate of change.
A camera is 30 m from a rocket launch pad; with height $y$ and angle $\theta$ , $\tan\theta=\frac{y}{30}$ ; which setup relates rates?
Answer: $\sec^2\theta\,\dfrac{d\theta}{dt}=\dfrac{1}{30}\dfrac{dy}{dt}$ This camera-rocket problem involves related rates with a trigonometric function. Starting with tan θ = y/30, we differentiate both sides with respect to time. The left side uses the chain rule: d/dt(tan θ = sec²θ · (dθ/dt). The right side gives (1/30)(dy/dt) since 30 is constant. Therefore: sec²θ(dθ/dt) = (1/30)(dy/dt). Choice C incorrectly omits dθ/dt, treating the problem as if finding an instantaneous relationship rather than a rate relationship. When differentiating trigonometric functions in related rates, always include the derivative of the angle with respect to time.
Water fills a spherical tank with radius $r(t)$ and volume $V(t)=\frac{4}{3}\pi r(t)^3$ . Which implicit differentiation step is correct?
Answer: $\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ This problem introduces the concept of related rates, connecting volume and radius changes in a filling spherical tank. Implicit differentiation with respect to time is required as V is a function of r(t), so we treat r as time-dependent. Differentiating V = (4/3)πr³ yields dV/dt = (4/3)π * 3r² dr/dt, simplifying to 4πr² dr/dt. This relates the rates directly. A tempting distractor like choice A fails by missing the r factor in the derivative, using 3r instead of 3r². Always differentiate the given equation implicitly with respect to time and apply the chain rule to each variable that depends on t.
A car is $x(t)$ miles east and $y(t)$ miles north of a station, with distance $s(t)=\sqrt{x^2+y^2}$ ; which derivative is correct?
Answer: $\dfrac{ds}{dt}=\dfrac{x\dfrac{dx}{dt}+y\dfrac{dy}{dt}}{\sqrt{x^2+y^2}}$ 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.
A cube's surface area is $S(t)=6s(t)^2$ ; which related-rates equation correctly relates $\dfrac{dS}{dt}$ and $\dfrac{ds}{dt}$ ?
Answer: $\dfrac{dS}{dt}=12s\dfrac{ds}{dt}$ This question introduces related rates by differentiating the surface area formula for a cube with respect to time. With s as a function of t, implicit differentiation is required. Differentiating $S = 6s^2$ gives $12s \dfrac{ds}{dt}$ using the chain rule. This connects surface area and side rates. Choice C omits $\dfrac{ds}{dt}$ , ignoring the chain rule. 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 spherical balloon's radius $r$ increases with time; its volume is $V=\frac{4}{3}\pi r^3$ . Which related-rates setup is correct?
Answer: $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ 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.
A rectangle has perimeter 40, with length $L(t)$ and width $W(t)$ . If $2L+2W=40$ , which rate equation is correct?
Answer: $2\frac{dL}{dt}+2\frac{dW}{dt}=0$ This problem involves setting up a related rates equation for a rectangle with constant perimeter 40, where L and W change. From $2L + 2W = 40$ , differentiate with respect to t to obtain $2 \frac{dL}{dt} + 2 \frac{dW}{dt} = 0$ . Since the perimeter is constant, its derivative is zero, and chain rule applies to L and W as functions of t. This implicit differentiation captures how the rates balance to keep the perimeter fixed. A tempting distractor like choice D is the undifferentiated form, not relating the rates. 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 cube has side length $s(t)$ and volume $V=s^3$ . Which related-rates equation correctly connects $\dfrac{dV}{dt}$ and $\dfrac{ds}{dt}$ ?
Answer: $\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}$ 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.
For a cylinder, $V(t)=\pi r(t)^2h$ where $h$ is constant; what is the correct related-rates equation?
Answer: $\dfrac{dV}{dt}=2\pi r h\dfrac{dr}{dt}$ This question introduces related rates by differentiating the volume formula for a cylinder with constant height. With r as a function of t and h fixed, implicit differentiation focuses on the varying radius. Differentiating V = πr²h gives π(2r dr/dt)h = 2πr h dr/dt using the chain rule. This captures the volume change due to radius variation. Choice B omits the factor of 2 and h, incorrectly simplifying 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.
Two cars start at the same intersection, one moving east $x(t)$ and one north $y(t)$ , so separation $s$ satisfies $s^2=x^2+y^2$ . Which differentiated relation is correct?
Answer: $2s\dfrac{ds}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$ This problem models the separation distance between two moving cars using the Pythagorean theorem for related rates. The equation s² = x² + y² must be differentiated with respect to time, yielding 2s(ds/dt) = 2x(dx/dt) + 2y(dy/dt) by applying the chain rule to each term. Choice B incorrectly assumes a linear relationship between the rates, ignoring that we're differentiating squared terms. The crucial concept is that when differentiating equations involving squares, the chain rule introduces the variable itself as a coefficient of its rate of change.
A circle's radius $r$ changes over time; with circumference $C=2\pi r$ , which related-rates equation is correct?
Answer: $\dfrac{dC}{dt}=2\pi\dfrac{dr}{dt}$ This circle problem involves the simplest related rates setup with a linear relationship. Given C = 2πr, we differentiate both sides with respect to time t. Since 2π is a constant, we get: dC/dt = 2π(dr/dt). Choice B incorrectly includes an extra factor of r, as if applying the product rule where none is needed. When differentiating linear relationships in related rates, constants remain constants and only the variable term gets differentiated with respect to time.
A particle satisfies $x(t)y(t)=t$ ; which implicit differentiation correctly relates $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ ?
Answer: $x\dfrac{dy}{dt}+y\dfrac{dx}{dt}=1$ 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.
Oil spreads in a circular slick: radius $r$ grows at $\frac{dr}{dt}$ . With area $A=\pi r^2$ , which relation is correct?
Answer: $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ This problem involves setting up a related rates equation for the area of a growing circular oil slick. Start with $A = \pi r^2$ and differentiate both sides with respect to time $t$ to find $\frac{dA}{dt}$ . Using the chain rule, this gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$ , since $r$ is changing over time. Implicit differentiation is required here to account for the dependency of $r$ on $t$ , ensuring the rate of area change incorporates the radius growth rate. A tempting distractor like choice B incorrectly applies the power rule without the chain rule, multiplying by $\frac{dr}{dt}$ in the wrong place. 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 ladder of length 10 ft rests on a wall: $x^2+y^2=100$ ; which equation relates $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ ?
Answer: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$ This ladder problem involves a related rates setup where both x and y change with time. Starting with the constraint equation x² + y² = 100, we differentiate both sides with respect to time t. Using the chain rule: d/dt(x²) + d/dt(y²) = d/dt(100), which gives 2x(dx/dt) + 2y(dy/dt) = 0. Choice C incorrectly differentiates as if x and y were constants, missing the essential time derivatives. In related rates problems, always apply d/dt to the entire equation and use the chain rule for any variable that depends on time.
For $y(t)=\sin(x(t))$ , which related-rates equation correctly relates $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ ?
Answer: $\dfrac{dy}{dt}=\cos(x)\dfrac{dx}{dt}$ This question introduces related rates by differentiating a composite function with respect to time. With x as a function of t in y = sin(x), implicit differentiation via the chain rule is needed. Differentiating gives dy/dt = cos(x) dx/dt. This applies the derivative of sine to the inner rate. Choice B omits dx/dt, treating x 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 cone has fixed slant similarity so $r(t)=3h(t)$ and $V=\frac13\pi r^2h$ ; which $\dfrac{dV}{dt}$ setup is correct?
Answer: $V=3\pi h^3\Rightarrow\dfrac{dV}{dt}=9\pi h^2\dfrac{dh}{dt}$ This question introduces related rates by expressing and differentiating the volume of a cone with proportional radius and height. Since r = 3h, substitute into V = (1/3)πr²h to get V = 3πh³, then differentiate implicitly. Differentiating gives dV/dt = 9πh² dh/dt using the chain rule. This relates volume to height rate. Choice C omits dh/dt, failing the chain rule. 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 right triangle has legs $x(t)$ and $y(t)$ with hypotenuse fixed at $10$ : $x^2+y^2=100$ . Which relation between $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ is correct?
Answer: $2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0$ This is a related rates problem involving a right triangle with a fixed hypotenuse. The constraint equation x² + y² = 100 must be differentiated implicitly with respect to time. Applying the chain rule to both terms gives 2x(dx/dt) + 2y(dy/dt) = 0, since the derivative of the constant 100 is zero. Choice C incorrectly treats 100 as if it were changing with time, placing it on the right side after differentiation. The fundamental principle is that when differentiating a constraint equation, constants remain constants and their derivatives are zero.
A circle's radius $r(t)$ changes and its area is $A=\pi r^2$ ; which setup finds $\dfrac{dr}{dt}$ from $\dfrac{dA}{dt}$ ?
Answer: $\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$ 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.
A point moves so its distances to the axes are $x(t)$ and $y(t)$ with $xy=12$ . Which related-rates equation is correct?
Answer: $x\frac{dy}{dt}+y\frac{dx}{dt}=0$ This problem involves setting up a related rates equation for a point moving such that the product of distances x and y to the axes is constant at 12. From xy = 12, differentiate both sides with respect to t using the product rule. This yields x dy/dt + y dx/dt = 0, connecting the rates of change. Implicit differentiation is needed because both x and y are functions of t, so each requires multiplication by its time derivative. A tempting distractor like choice A reverses the product rule terms, leading to an incorrect relation. 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 cone has constant height $h$ and changing radius $r$ ; volume $V=\frac{1}{3}\pi r^2h$ . Which expression for $\frac{dV}{dt}$ is correct?
Answer: $\frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt}$ This problem introduces the concept of related rates, finding how a cone's volume changes with varying radius but constant height. Implicit differentiation with respect to time is required since V depends on r, which changes with t, while h is fixed, so dh/dt = 0. Differentiating V = (1/3)πr²h gives dV/dt = (1/3)π 2r dr/dt h, or (2/3)π r h dr/dt. This isolates the effect of the changing radius on volume. A tempting distractor like choice A fails by incorrectly using dh/dt instead of dr/dt, ignoring the constant height. Always differentiate the given equation implicitly with respect to time and apply the chain rule to each variable that depends on t.

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