Solving Related Rates Problems - AP Calculus BC

A 10-ft ladder leans on a wall; the top slides down at 1 ft/s. When the top is 8 ft high, how fast is the bottom moving outward?
Answer: $\dfrac{4}{3}$ ft/s Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The ladder's position satisfies x² + y² = 10², where x is the bottom's distance and y is the top's height. Differentiating implicitly with respect to time gives 2x dx/dt + 2y dy/dt = 0. When y=8, x=6, and dy/dt=-1, plugging in yields dx/dt = -(8/6)*(-1) = 4/3 ft/s. A tempting distractor is 3/4, perhaps from inverting the ratio or misapplying the sign. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.
A cube's edge length increases at 0.5 cm/s. When the edge is 4 cm, how fast is the cube's surface area increasing?
Answer: $24$ cm $^2$ /s Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The surface area A of the cube is A = 6a², where a is the edge length. Differentiating implicitly with respect to time gives dA/dt = 12a da/dt. When a=4 and da/dt=0.5, substituting yields dA/dt = 12*4*0.5 = 24 cm²/s. A tempting distractor is 12, perhaps from using A=6a and differentiating to 6 da/dt incorrectly. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.
Sand forms a cone whose radius equals half its height; volume increases at $8 \, \text{ft}^3/\text{min}$ . Find $\frac{dh}{dt}$ when $h=6$ ft.
Answer: $\frac{8}{9\pi}\text{ ft/min}$ This problem involves solving related rates by differentiating a geometric formula with respect to time. The volume of the cone is $V = \frac{\pi h^3}{12}$ , since $r = h/2$ ; implicit differentiation gives $\frac{dV}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$ . When $h = 6$ ft and $\frac{dV}{dt} = 8 \, \text{ft}^3/\text{min}$ , substituting yields $8 = \left( \pi \cdot 36 / 4 \right) \frac{dh}{dt}$ , so $\frac{dh}{dt} = \frac{8}{9\pi} \, \text{ft/min}$ . Thus, the height increases at $\frac{8}{9\pi}$ feet per minute. A tempting distractor like $8/(3\pi)$ occurs if one uses $r = h$ instead of $r = h/2$ in the volume formula, but correctly substituting $r = h/2$ gives the factor of $1/12$ . To solve related rates problems, identify the relating equation, differentiate with respect to time, substitute known values, and solve for the desired rate.
A streetlight is 16 ft high. A 5-ft person walks away at 3 ft/s; when 12 ft from pole, how fast is the shadow lengthening?
Answer: $\frac{15}{11}$ ft/s This problem involves solving related rates by differentiating a geometric relationship with respect to time. For the shadow, similar triangles give $s = \frac{5}{11} x$ , where s is shadow length and x is distance from pole. Implicit differentiation yields $\frac{ds}{dt} = \frac{5}{11} \frac{dx}{dt}$ . When x=12 ft and $\frac{dx}{dt} = 3$ ft/s, $\frac{ds}{dt} = \frac{5}{11} \times 3 = \frac{15}{11}$ ft/s. A tempting distractor like $\frac{11}{15}$ ft/s might result from inverting the ratio of heights or misapplying proportions, but the correct similarity uses $\frac{5}{16}$ for the fraction leading to $\frac{5}{16-5} = \frac{5}{11}$ . To solve related rates problems generally, identify the equation relating the variables, differentiate with respect to time, substitute known values and rates, and solve for the desired rate.
A circle's area increases at $12\pi$ cm $^2$ /s. When the radius is 5 cm, how fast is the radius increasing?
Answer: $\dfrac{6}{5}$ cm/s This problem requires finding how fast a circle's radius increases given the rate of area increase, a related rates problem working backwards from area to radius. The area of a circle is A = πr², so differentiating with respect to time gives dA/dt = 2πr(dr/dt). When r = 5 cm and dA/dt = 12π cm²/s, we substitute: 12π = 2π(5)(dr/dt), which simplifies to 12π = 10π(dr/dt), giving dr/dt = 12π/(10π) = 6/5 cm/s. A common mistake is forgetting the factor of 2 when differentiating r² or canceling π incorrectly. The key insight for related rates is that when given a rate of change for one quantity, you can find the rate for a related quantity by differentiating the constraint equation.
A point moves on $y=\sqrt{x}$ with $\frac{dx}{dt}=8$ at $x=9$ . What is $\frac{dy}{dt}$ at that instant?
Answer: $\frac{4}{3}$ This problem involves solving related rates by differentiating a geometric relationship with respect to time. The point moves on y = √x, so y = $x^{1/2}$ . Implicit differentiation gives dy/dt = (1/2) $x^{-1/2}$ (dx/dt). At x = 9 and dx/dt = 8, dy/dt = (1/2) (1/3) * 8 = 4/3. A tempting distractor like 8/3 might result from using dy/dx = 1/√x without halving or miscalculating the derivative, but the chain rule requires the 1/2 factor. To solve related rates problems generally, identify the equation relating the variables, differentiate with respect to time, substitute known values and rates, and solve for the desired rate.
A circle's radius decreases at 3 cm/s. When $r=10$ cm, how fast is the area changing?
Answer: $-60\pi$ cm $^2$ /s Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The area A of the circle is A = π r². Differentiating implicitly with respect to time gives dA/dt = 2π r dr/dt. When r=10 and dr/dt=-3, substituting yields dA/dt = 2π*10*(-3) = -60π cm²/s. A tempting distractor is -30π, perhaps from forgetting the factor of 2 in the derivative. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.
A 10-ft ladder leans against a wall; the base slides away at $3$ ft/s. When the base is 6 ft out, how fast is the top sliding down?
Answer: $-\dfrac{9}{4}$ ft/s This problem requires solving related rates to find how fast the top of a ladder slides down a wall. The ladder forms a right triangle with the wall and ground, so we use the Pythagorean theorem: x² + y² = 100 (where x is the base distance and y is the height). Differentiating implicitly with respect to time gives 2x(dx/dt) + 2y(dy/dt) = 0, which simplifies to x(dx/dt) + y(dy/dt) = 0. When x = 6, we find y = 8 using the Pythagorean theorem, and substituting dx/dt = 3, we get 6(3) + 8(dy/dt) = 0, yielding dy/dt = -18/8 = -9/4 ft/s. A common error is forgetting the negative sign, which would give 9/4 ft/s, but the top must move down (negative direction) as the base moves out. The key to related rates problems is: identify the constraint equation, differentiate implicitly with respect to time, then substitute known values.
A 13-ft ladder slides so its base moves away at $2$ ft/s; when the base is $5$ ft out, how fast is the top sliding down?
Answer: $-\dfrac{5}{6}\text{ ft/s}$ This problem requires solving related rates to find how fast the ladder's top slides down the wall. The ladder forms a right triangle with the wall and ground, so we use the Pythagorean theorem: x² + y² = 13², where x is the base distance and y is the height. Differentiating implicitly with respect to time gives 2x(dx/dt) + 2y(dy/dt) = 0, which simplifies to x(dx/dt) + y(dy/dt) = 0. When x = 5 ft, we find y = √(169-25) = 12 ft, and with dx/dt = 2 ft/s, we get 5(2) + 12(dy/dt) = 0, yielding dy/dt = -10/12 = -5/6 ft/s. A common error is forgetting the negative sign, which indicates downward motion. The key to related rates problems is to differentiate the constraint equation implicitly with respect to time, then substitute known values.
Two cyclists start at an intersection; one rides east at 12 mph, the other north at 5 mph. How fast is their separation increasing after 2 hours?
Answer: $13\text{ mph}$ This problem involves solving related rates by differentiating a geometric formula with respect to time. The separation is $s = \sqrt{x^2 + y^2}$ , so implicit differentiation gives $\frac{ds}{dt} = \frac{x \frac{dx}{dt} + y \frac{dy}{dt}}{s}$ . After 2 hours, x = 24 mi, y = 10 mi, $\frac{dx}{dt}$ = 12 mph, $\frac{dy}{dt}$ = 5 mph, s = 26 mi, substituting yields $\frac{ds}{dt} = \frac{24 \times 12 + 10 \times 5}{26} = \frac{338}{26} = 13$ mph. Thus, the separation increases at 13 miles per hour. A tempting distractor like $\frac{169}{26}$ mph arises from forgetting to divide by s or mishandling the formula, but the correct rate includes the division by the current distance. To solve related rates problems, identify the relating equation, differentiate with respect to time, substitute known values, and solve for the desired rate.
A circle's radius decreases at $0.4$ cm/s; when $r=10$ cm, how fast is the area changing?
Answer: $-8\pi$ cm $^2$ /s This problem involves solving related rates by differentiating a geometric relationship with respect to time. For the circle, area A = π r². Implicit differentiation gives dA/dt = 2 π r (dr/dt). When r=10 cm and dr/dt=-0.4 cm/s, dA/dt = 2 π 10 (-0.4) = -8 π cm²/s. A tempting distractor like -4 π cm²/s might result from omitting the factor of 2 or using circumference instead, but the area derivative correctly involves 2 π r. To solve related rates problems generally, identify the equation relating the variables, differentiate with respect to time, substitute known values and rates, and solve for the desired rate.
A cylindrical tank has radius $3$ m; water height rises at $0.5$ m/min. What is the water volume's rate of change?
Answer: $4.5\pi\text{ m}^3/\text{min}$ This problem asks for the rate of volume change in a cylindrical tank using related rates. The volume formula for a cylinder is V = πr²h, where r is the radius and h is the height. Since the radius is constant at 3 m, differentiating with respect to time gives dV/dt = πr²(dh/dt). With r = 3 m and dh/dt = 0.5 m/min, we get dV/dt = π(3)²(0.5) = π(9)(0.5) = 4.5π m³/min. A common mistake is to differentiate r as if it were changing, which would complicate the problem unnecessarily. When solving related rates problems, always identify which quantities are constant and which are changing before differentiating.
Two cars leave an intersection, one north at 30 mph and one east at 40 mph. How fast is the distance between them increasing after 2 hours?
Answer: $50$ mph Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The distance s between the cars satisfies s² = x² + y², where x and y are their positions. Differentiating implicitly with respect to time gives 2s ds/dt = 2x dx/dt + 2y dy/dt. After 2 hours, x=80, y=60, s=100, dx/dt=40, dy/dt=30, so ds/dt = (80*40 + 60*30)/100 = 50 mph. A tempting distractor is 70, perhaps from adding speeds directly without using the related rates formula. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.
A streetlight is 15 ft tall; a 6-ft person walks away at 5 ft/s. When the person is 10 ft from the pole, how fast is the shadow lengthening?
Answer: $\dfrac{10}{3}$ ft/s Solving related rates problems involves finding how rates of change are connected through a geometric or physical relationship. The shadow length s relates to the person's distance x by s = (2/3) x from similar triangles with heights 15 ft and 6 ft. Differentiating implicitly with respect to time gives ds/dt = (2/3) dx/dt. When dx/dt=5, this yields ds/dt = (2/3)*5 = 10/3 ft/s, independent of x=10. A tempting distractor is 25/3, which might come from finding the tip's speed d(x+s)/dt instead of the lengthening rate. To solve related rates problems generally, identify variables and their relationship, differentiate with respect to time, substitute known values, and solve for the desired rate.
A point moves on $x^2+y^2=25$ with $\dfrac{dx}{dt}=4$ . When $x=3$ and $y=4$ , what is $\dfrac{dy}{dt}$ ?
Answer: $-3$ This problem involves solving related rates by relating coordinates on a circle. Given $x^2 + y^2 = 25$ , differentiating implicitly with respect to time $t$ gives $2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0$ , which simplifies to $\dfrac{dy}{dt} = -\dfrac{x}{y} \dfrac{dx}{dt}$ . When $x=3$ and $y=4$ , $\dfrac{dx}{dt}=4$ , substituting yields $\dfrac{dy}{dt} = -\dfrac{3}{4} \times 4 = -3$ . This assumes y is positive, consistent with the point (3,4) on the circle. A tempting distractor like $-\dfrac{16}{3}$ fails because it might arise from inverting the ratio incorrectly or misapplying signs. In general, for related rates problems, identify the variables and their relationship, differentiate with respect to time, plug in known values, and solve for the desired rate.
A spherical balloon's radius increases at $0.4$ cm/s; what is $\frac{dV}{dt}$ when $r=10$ cm?
Answer: $160\pi\text{ cm}^3/\text{s}$ This problem involves solving related rates by differentiating a geometric formula with respect to time. The volume of a sphere is given by V = (4/3)πr³, so implicit differentiation yields dV/dt = 4πr² dr/dt. Substituting r = 10 cm and dr/dt = 0.4 cm/s gives dV/dt = 4π(10)²(0.4) = 160π cm³/s. Thus, the volume increases at 160π cubic centimeters per second at that instant. A tempting distractor like 40π arises if one mistakenly uses the surface area formula instead of the correct volume derivative, but the volume rate requires the 4πr² factor. To solve related rates problems, identify the relating equation, differentiate with respect to time, substitute known values, and solve for the desired rate.
A boat is pulled toward a dock by a rope 20 ft long, shortened at 3 ft/s; when the boat is 12 ft from the dock, how fast is it moving toward the dock?
Answer: $-5\text{ ft/s}$ This boat problem uses related rates to find how fast the boat approaches the dock. Let L be the rope length, x the horizontal distance to the dock, and h the vertical distance from water to pulley; then L² = x² + h². Since the pulley height h is constant, differentiating gives 2L(dL/dt) = 2x(dx/dt), or L(dL/dt) = x(dx/dt). When x = 12 ft and L = 20 ft, we verify h = √(400-144) = 16 ft, and with dL/dt = -3 ft/s (negative because rope shortens), we get 20(-3) = 12(dx/dt), so dx/dt = -60/12 = -5 ft/s. A common mistake is using the wrong sign for dL/dt; rope shortening means dL/dt < 0. The key insight is that the rope length, not the horizontal distance, is being directly controlled.
A circle's area increases at $12\pi$ cm $^2$ /min. When the radius is $6$ cm, how fast is the radius changing?
Answer: $1$ cm/min This problem involves solving related rates by relating the area of a circle to its radius. The area A = πr², and differentiating implicitly with respect to time t gives dA/dt = 2πr dr/dt. When r=6 cm and dA/dt=12π cm²/min, substituting yields 12π = 2π(6) dr/dt, so dr/dt = 1 cm/min. This uses the chain rule to connect the area change to the radius change accurately. A tempting distractor like 1/π cm/min fails because it might come from incorrectly dividing by 2πr instead of solving properly. In general, for related rates problems, identify the variables and their relationship, differentiate with respect to time, plug in known values, and solve for the desired rate.

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