Introduction to Related Rates - AP Calculus AB

A savings account balance satisfies $B=500+20t$ dollars; what is the rate of change of $B$ with respect to time?
Answer: $20\text{ dollars/month}$ This problem involves interpreting the rate of change in a linear function. Given $B = 500 + 20t$ , taking the derivative with respect to time: $\frac{dB}{dt} = 20$ dollars/month. This represents the constant rate at which the balance increases each month. The constant term 500 disappears upon differentiation, leaving only the coefficient of t. Students might mistakenly include the constant term or confuse the units.
A circular bracelet's radius decreases at $0.05$ cm/s when $r=20$ cm; how fast is its area changing?
Answer: $-2\pi\text{ cm}^2/\text{s}$ This problem demonstrates area rate change for a decreasing circular radius. Given A = πr² and dr/dt = -0.05 cm/s (negative for decreasing) when r = 20 cm, taking the derivative: dA/dt = 2πr(dr/dt) = 2π(20)(-0.05) = -2π cm²/s. The negative sign correctly indicates the area is decreasing as the bracelet contracts. Students might forget the negative sign or miscalculate the coefficient.
A rectangle has area $A=xy$ with $x=4$ increasing at $3$ and $y=10$ decreasing at $1$ ; what is $\dfrac{dA}{dt}$ ?
Answer: $26$ This problem applies the product rule to find the rate of change of area. Given $A = xy$ with $x = 4$ , $y = 10$ , $\frac{dx}{dt} = 3$ , and $\frac{dy}{dt} = -1$ , taking the derivative: $\frac{dA}{dt} = x \frac{dy}{dt} + y \frac{dx}{dt} = 4(-1) + 10(3) = -4 + 30 = 26$ . The positive result indicates area is increasing despite y decreasing, because the increase from x growing outweighs the decrease from y shrinking.
A cube's edge length decreases at $0.25$ m/hr when the edge is $8$ m; how fast is the volume changing?
Answer: $-48\text{ m}^3/\text{hr}$ This problem applies volume rate change to a decreasing cube. Given V = s³ and ds/dt = -0.25 m/hr (negative for decreasing) when s = 8 m, taking the derivative: dV/dt = 3s²(ds/dt) = 3(8)²(-0.25) = 3(64)(-0.25) = -48 m³/hr. The negative sign correctly indicates volume is decreasing as the edge length decreases. Students might forget the negative sign or miscalculate the power rule application.
A spherical balloon's volume increases at $12\pi\text{ cm}^3/\text{min}$ when $r=3$ cm; how fast is $r$ changing?
Answer: $\dfrac{1}{3}\text{ cm/min}$ This problem requires applying related rates to interpret how changes in volume translate to changes in radius for a sphere. For a sphere, $V = \frac{4}{3} \pi r^3$ . Taking the derivative with respect to time: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$ . Given $\frac{dV}{dt} = 12 \pi$ cm³/min and r = 3 cm, we substitute: $12 \pi = 4 \pi (3)^2 \frac{dr}{dt} = 4 \pi (9) \frac{dr}{dt} = 36 \pi \frac{dr}{dt}$ . Solving for $\frac{dr}{dt}$ : $\frac{dr}{dt} = \frac{12 \pi}{36 \pi} = \frac{1}{3}$ cm/min. A common error would be forgetting to square the radius or confusing the volume formula with surface area, leading to incorrect calculations.
A sphere's radius decreases at $0.1$ cm/min when $r=5$ cm; what is $\dfrac{dV}{dt}$ for $V=\frac{4}{3}\pi r^3$ ?
Answer: $-10\pi \text{ cm}^3/\text{min}$ This problem demonstrates related rates with decreasing radius for spherical volume. Given $V = \frac{4}{3} \pi r^3$ and $\frac{dr}{dt} = -0.1 \, \text{cm/min}$ (negative for decreasing) when $r = 5$ cm, taking the derivative: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} = 4 \pi (5)^2 (-0.1) = 4 \pi (25) (-0.1) = -10 \pi \, \text{cm}^3/\text{min}$ . The negative sign correctly indicates volume is decreasing as radius decreases. Students might forget the negative sign or miscalculate the derivative.
A square sheet has side $s$ and diagonal $d$ with $d=s\sqrt{2}$ ; if $\dfrac{ds}{dt}=3$ cm/min, what is $\dfrac{dd}{dt}$ ?
Answer: $3\sqrt{2}\text{ cm/min}$ This problem applies the relationship between square side length and diagonal. Given $d = s\sqrt{2}$ , taking the derivative: $\dfrac{dd}{dt} = \sqrt{2} \dfrac{ds}{dt}$ . With $\dfrac{ds}{dt} = 3$ cm/min, we have $\dfrac{dd}{dt} = \sqrt{2}(3) = 3\sqrt{2}$ cm/min. This demonstrates how geometric relationships create proportional rate relationships. Students might forget the $\sqrt{2}$ factor or confuse the direction of the relationship.
A medication amount satisfies $A=80e^{-0.1t}$ mg; at $t=0$ , what is the instantaneous rate of change of $A$ ?
Answer: $-8\text{ mg/hr}$ This problem requires finding the instantaneous rate of change using exponential differentiation. Given $A = 80e^{-0.1t}$ , taking the derivative: $dA/dt = 80(-0.1)e^{-0.1t} = -8e^{-0.1t}$ . At $t = 0$ , $dA/dt = -8e^0 = -8(1) = -8$ mg/hr. The negative sign indicates the medication amount is decreasing over time, which is expected for drug elimination. Students might forget the chain rule or the negative coefficient.
A circular logo has diameter increasing at $0.8$ cm/s when $d=10$ cm; how fast is the area changing then?
Answer: $4\pi\text{ cm}^2/\text{s}$ This problem applies area rate change using diameter for a circle. Given A = π(d/2)² = πd²/4 and dd/dt = 0.8 cm/s when d = 10 cm, taking the derivative: dA/dt = π(d/2)(dd/dt) = π(10/2)(0.8) = π(5)(0.8) = 4π cm²/s. This demonstrates how diameter-based formulations affect the derivative calculation compared to radius-based approaches.
A company's revenue is $R=50p$ and price $p$ decreases at $0.4$ dollars/day; what is $\dfrac{dR}{dt}$ ?
Answer: $-20\text{ dollars/day}$ This problem demonstrates interpreting rates in a linear business context. Given $R = 50p$ , taking the derivative: $\dfrac{dR}{dt} = 50 \dfrac{dp}{dt}$ . Since price decreases at 0.4 dollars/day, $\dfrac{dp}{dt} = -0.4$ . Therefore, $\dfrac{dR}{dt} = 50(-0.4) = -20$ dollars/day. The negative sign indicates revenue is decreasing as price decreases. A common error would be forgetting the negative sign or confusing the direction of change.
A cube has volume increasing at $150 \, \text{cm}^3/\text{min}$ when side length is $5 \, \text{cm}$ ; what is the rate of change of its surface area then?
Answer: $120 \, \text{cm}^2 / \text{min}$ This problem connects volume and surface area rates for a cube. Given $V = s^3$ and $SA = 6s^2$ , with $\frac{dV}{dt} = 150 \, \text{cm}^3/\text{min}$ when $s = 5 \, \text{cm}$ . First, find $ds/dt$ : $150 = 3s^2 (ds/dt) = 3(25)(ds/dt) = 75(ds/dt)$ , so $ds/dt = 2 \, \text{cm/min}$ . Then $dSA/dt = 12s(ds/dt) = 12(5)(2) = 120 \, \text{cm}^2/\text{min}$ . This shows how volume changes translate to surface area changes through the common rate of side length change.
A rectangle has constant area $144$ ; when length is $9$ decreasing at $2$ ft/min, what is the rate of change of the width?
Answer: $\dfrac{32}{9}\text{ ft/min}$ This problem uses constant area constraint for a rectangle. Given $A = lw = 144$ (constant), when $l = 9$ and $\frac{dl}{dt} = -2$ (negative for decreasing), we find $w = \frac{144}{9} = 16$ . Taking the derivative: $l \frac{dw}{dt} + w \frac{dl}{dt} = 0$ . Substituting: $9 \frac{dw}{dt} + 16(-2) = 0$ , so $9 \frac{dw}{dt} = 32$ . Therefore, $\frac{dw}{dt} = \frac{32}{9} \text{ ft/min}$ . The positive result indicates width increases as length decreases to maintain constant area.
For a rectangle with constant perimeter $40$ , when length $12$ increases at $1$ ft/min, how fast is the width changing?
Answer: $-1\text{ ft/min}$ This problem involves related rates with constant perimeter constraint. For a rectangle with perimeter P = 2l + 2w = 40, we have l + w = 20. Taking the derivative: dl/dt + dw/dt = 0. Given l = 12 and dl/dt = 1 ft/min, we substitute: 1 + dw/dt = 0. Solving: dw/dt = -1 ft/min. The negative sign correctly indicates that as length increases, width must decrease to maintain constant perimeter. A common error would be forgetting the constraint or the negative relationship.
A rectangle has length $x$ and width $y$ with $y=\dfrac{100}{x}$ ; when $x=20$ and $\dfrac{dx}{dt}=1$ , what is $\dfrac{dy}{dt}$ ?
Answer: $-\dfrac{1}{4}$ This problem uses the constraint relationship y = 100/x for a hyperbolic function. Taking the derivative: dy/dt = -100/x² × dx/dt. When x = 20 and dx/dt = 1, we have dy/dt = -100/(20)² × 1 = -100/400 = -1/4. The negative sign indicates that as x increases, y decreases to maintain the constant product relationship. Students might forget the negative sign from the chain rule.
A square window's diagonal is increasing at $3$ cm/s when the diagonal is $10$ cm; how fast is its area changing then?
Answer: $30\text{ cm}^2/\text{s}$ This problem connects diagonal and area rates for a square. For a square with diagonal $d$ , the side length is $s = \frac{d}{\sqrt{2}}$ , so area $A = s^2 = \frac{d^2}{2}$ . Taking the derivative: $\frac{dA}{dt} = d \frac{dd}{dt}$ . When $d = 10$ cm and $\frac{dd}{dt} = 3$ cm/s, we have $\frac{dA}{dt} = 10(3) = 30$ cm²/s. Students might incorrectly use the side length relationship or forget to properly convert between diagonal and area formulas.
A lab culture's total biomass is $B=3n$ grams where cell count $n$ increases at $200$ cells/hr; what is $\dfrac{dB}{dt}$ ?
Answer: $600\text{ g/hr}$ This problem applies linear related rates to a biological context. Given $B = 3n$ , taking the derivative: $\dfrac{dB}{dt} = 3 \dfrac{dn}{dt}$ . Since $\dfrac{dn}{dt} = 200$ cells/hr, we have $\dfrac{dB}{dt} = 3(200) = 600$ g/hr. This is a straightforward application of the constant multiple rule for derivatives. Students might overthink this problem or forget to apply the coefficient 3 correctly.
A circular crop field keeps area $400\pi$ m $^2$ constant; when radius is $20$ m increasing at $0.1$ m/day, what is $\dfrac{d(\text{area})}{dt}$ ?
Answer: $0\text{ m}^2/\text{day}$ This problem involves a constant area constraint where the area is unchanging. Given A = πr² = 400π (constant), taking the derivative: dA/dt = 2πr(dr/dt) = 0. Since area is constant, its rate of change must be zero regardless of how radius might be changing. This demonstrates that constraints can override individual parameter changes. Students might mistakenly calculate a non-zero rate based on the radius change.
A circular cell colony has area $A=\pi r^2$ ; if $\dfrac{dr}{dt}=0.2$ mm/hr when $r=10$ mm, what is $\dfrac{dA}{dt}$ ?
Answer: $4\pi\text{ mm}^2/\text{hr}$ This problem applies the basic formula for the rate of change of circular area. Given $A = \pi r^2$ and $\dfrac{dr}{dt} = 0.2$ mm/hr when $r = 10$ mm, taking the derivative: $\dfrac{dA}{dt} = 2\pi r (\dfrac{dr}{dt}) = 2\pi(10)(0.2) = 4\pi$ mm $^2$ /hr. This is a direct application of the chain rule to the area formula. Students might forget the coefficient 2 from differentiating r $^2$ or miscalculate the substitution.
A cylindrical tank with fixed radius $5$ m fills so $\dfrac{dV}{dt} = 50\pi$ m $^3$ /hr; how fast is the height increasing?
Answer: $2\text{ m/hr}$ This problem demonstrates interpreting related rates for a cylindrical tank with fixed dimensions. For a cylinder, $V = \pi r^2 h$ where r is constant at 5 m. Taking the derivative: $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ since r is constant. Given $\frac{dV}{dt} = 50\pi$ m $^3$ /hr and r = 5 m, we substitute: $50\pi = \pi(5)^2 \frac{dh}{dt} = 25\pi \frac{dh}{dt}$ . Solving: $\frac{dh}{dt} = \frac{50\pi}{25\pi} = 2$ m/hr. Students might incorrectly include dr/dt terms even though radius is fixed, or miscalculate the area $\pi r^2$ .
A rectangle has perimeter $P=2x+2y$ ; if $\dfrac{dP}{dt}=0$ , and $\dfrac{dx}{dt}=3$ , what must $\dfrac{dy}{dt}$ be?
Answer: $-3$ This problem demonstrates constraint relationships in perimeter. Given $P = 2x + 2y$ and $\dfrac{dP}{dt} = 0$ (constant perimeter), taking the derivative: $2\dfrac{dx}{dt} + 2\dfrac{dy}{dt} = 0$ . With $\dfrac{dx}{dt} = 3$ , we have $2(3) + 2\dfrac{dy}{dt} = 0$ , so $6 + 2\dfrac{dy}{dt} = 0$ . Therefore, $\dfrac{dy}{dt} = -3$ . The negative sign indicates that as one dimension increases, the other must decrease to maintain constant perimeter.

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