Calculus 1 Quiz: Solving Related Rates
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Solving Related RatesQuestion 1 of 8

A spherical balloon is inflated inside a cube so that it is always tangent to the cube's six faces. If the volume of the balloon is increasing at a constant rate of 10π10\pi cm³/s, what is the rate of increase of the volume of the cube at the instant the balloon's radius is 5 cm?

1515 cm³/s
3030 cm³/s
6060 cm³/s
7575 cm³/s
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Calculus 1 Quiz

Calculus 1 Quiz: Solving Related Rates

Practice Solving Related Rates in Calculus 1 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 Solving Related Rates, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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.

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Question 1

A spherical balloon is inflated inside a cube so that it is always tangent to the cube's six faces. If the volume of the balloon is increasing at a constant rate of 10π10\pi cm³/s, what is the rate of increase of the volume of the cube at the instant the balloon's radius is 5 cm?

  1. 1515 cm³/s
  2. 3030 cm³/s
  3. 6060 cm³/s (correct answer)
  4. 7575 cm³/s
Explanation: Let rr be the radius of the sphere and ss be the side length of the cube. Since the sphere is tangent to the faces of the cube, s=2rs=2r. The volume of the sphere is Vs=43πr3V_s = \frac{4}{3}\pi r^3 and the volume of the cube is Vc=s3=(2r)3=8r3V_c = s^3 = (2r)^3 = 8r^3. We are given dVsdt=10π\frac{dV_s}{dt} = 10\pi. We want to find dVcdt\frac{dV_c}{dt}. First, differentiate both volume formulas with respect to tt: dVsdt=4πr2drdt\frac{dV_s}{dt} = 4\pi r^2 \frac{dr}{dt} and dVcdt=24r2drdt\frac{dV_c}{dt} = 24r^2 \frac{dr}{dt}. From the first equation, we can find an expression for drdt=14πr2dVsdt\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV_s}{dt}. Substitute this into the second equation: dVcdt=24r2(14πr2dVsdt)=244πdVsdt=6πdVsdt\frac{dV_c}{dt} = 24r^2 \left( \frac{1}{4\pi r^2}\frac{dV_s}{dt} \right) = \frac{24}{4\pi}\frac{dV_s}{dt} = \frac{6}{\pi}\frac{dV_s}{dt}. Plugging in the given rate dVsdt=10π\frac{dV_s}{dt} = 10\pi, we get dVcdt=6π(10π)=60\frac{dV_c}{dt} = \frac{6}{\pi}(10\pi) = 60 cm³/s. Note that the rate is constant and does not depend on the radius rr.

Question 2

A pebble dropped in a pond creates a circular ripple. The area of the ripple increases at a constant rate of 6π6\pi cm²/s. What is the rate of increase of the circumference of the ripple when its area is 9π9\pi cm²?

  1. 11 cm/s
  2. 2π2\pi cm/s (correct answer)
  3. 3π3\pi cm/s
  4. 18π18\pi cm/s
Explanation: Let AA be the area, CC be the circumference, and rr be the radius of the ripple. The formulas are A=πr2A = \pi r^2 and C=2πrC = 2\pi r. We are given dAdt=6π\frac{dA}{dt} = 6\pi. We want to find dCdt\frac{dC}{dt}. First, let's find the radius when the area A=9πA=9\pi. From 9π=πr29\pi = \pi r^2, we get r2=9r^2=9, so r=3r=3 cm. Next, we differentiate the area and circumference formulas with respect to time tt: dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt} and dCdt=2πdrdt\frac{dC}{dt} = 2\pi \frac{dr}{dt}. Using the first differentiated equation with the given values: 6π=2π(3)drdt6\pi = 2\pi (3) \frac{dr}{dt}, which gives 6π=6πdrdt6\pi = 6\pi \frac{dr}{dt}, so drdt=1\frac{dr}{dt} = 1 cm/s. Now substitute this into the second differentiated equation: dCdt=2π(1)=2π\frac{dC}{dt} = 2\pi (1) = 2\pi cm/s.

Question 3

A spherical snowball is melting such that its surface area decreases at a constant rate of 8π8\pi cm²/min. What is the rate at which the radius is decreasing at the moment when the snowball's volume is 288π288\pi cm³?

  1. 1/91/9 cm/min
  2. 1/31/3 cm/min
  3. 24π24\pi cm/min
  4. 1/61/6 cm/min (correct answer)
Explanation: This is a related rates problem involving a melting sphere. When you see rates of change for different properties of the same object, you need to connect them using differentiation and the geometric formulas. Start by identifying what you know and need. The surface area decreases at dAdt=−8π\frac{dA}{dt} = -8\pi cm²/min (negative because it's decreasing), and you need drdt\frac{dr}{dt} when the volume is 288π288\pi cm³. First, find the radius when V=288πV = 288\pi. Since V=43πr3V = \frac{4}{3}\pi r^3: 288π=43πr3288\pi = \frac{4}{3}\pi r^3 288=43r3288 = \frac{4}{3}r^3 r3=216r^3 = 216 r=6r = 6 cm Next, relate the surface area to the radius. For a sphere, A=4πr2A = 4\pi r^2. Differentiate both sides with respect to time: dAdt=8πrdrdt\frac{dA}{dt} = 8\pi r \frac{dr}{dt} Substitute the known values: −8π=8π(6)drdt-8\pi = 8\pi(6)\frac{dr}{dt} −8π=48πdrdt-8\pi = 48\pi\frac{dr}{dt} drdt=−8π48π=−16\frac{dr}{dt} = -\frac{8\pi}{48\pi} = -\frac{1}{6} cm/min The radius decreases at 16\frac{1}{6} cm/min, making D correct. Choice A (19\frac{1}{9}) likely comes from miscalculating r3=216r^3 = 216 as r=9r = 9. Choice B (13\frac{1}{3}) results from using the wrong surface area formula or arithmetic errors. Choice C (24π24\pi) probably comes from confusing which derivative formula to use or forgetting to solve for drdt\frac{dr}{dt}. For related rates problems, always: identify all rates, find intermediate values using given conditions, write the relationship equation, then differentiate with respect to time.

Question 4

The volume VV of a spherical cap of height hh from a sphere of radius RR is given by the formula V=πh23(3R−h)V = \frac{\pi h^2}{3}(3R - h).

A hemispherical bowl of radius 10 cm is being filled with water at a constant rate of 3π3\pi cm³/s. Using the formula provided, find the rate at which the water level is rising when the depth of the water is 5 cm.

  1. 1/51/5 cm/s
  2. 1/501/50 cm/s
  3. 3/753/75 cm/s
  4. 1/251/25 cm/s (correct answer)
Explanation: This is a related rates problem, where you need to find how fast one quantity changes based on how fast another quantity changes. When you see "rate of change" language with geometric shapes being filled, think about connecting the rates using the chain rule. Given the volume formula V=πh23(3R−h)V = \frac{\pi h^2}{3}(3R - h) for a spherical cap, you need to find dhdt\frac{dh}{dt} when h=5h = 5 cm, given that dVdt=3π\frac{dV}{dt} = 3\pi cm³/s and R=10R = 10 cm. First, differentiate the volume formula with respect to time: dVdt=π3ddt[h2(3R−h)]=π3[2h(3R−h)+h2(−1)]=π3[6Rh−3h2]\frac{dV}{dt} = \frac{\pi}{3}\frac{d}{dt}[h^2(3R - h)] = \frac{\pi}{3}[2h(3R - h) + h^2(-1)] = \frac{\pi}{3}[6Rh - 3h^2] Substituting the known values (R=10R = 10, h=5h = 5, dVdt=3π\frac{dV}{dt} = 3\pi): 3π=π3[6(10)(5)−3(5)2]=π3[300−75]=225π3=75π3\pi = \frac{\pi}{3}[6(10)(5) - 3(5)^2] = \frac{\pi}{3}[300 - 75] = \frac{225\pi}{3} = 75\pi This gives us: 3π=75π⋅dhdt3\pi = 75\pi \cdot \frac{dh}{dt}, so dhdt=3π75π=125\frac{dh}{dt} = \frac{3\pi}{75\pi} = \frac{1}{25} cm/s. Choice A (15\frac{1}{5}) likely comes from incorrect differentiation or arithmetic errors. Choice B (150\frac{1}{50}) might result from doubling the denominator incorrectly. Choice C (375\frac{3}{75}) equals 125\frac{1}{25} but wasn't simplified, or represents a calculation error. Study tip: In related rates problems, always differentiate the entire equation with respect to time, substitute known values only after differentiating, and double-check your product rule applications when dealing with composite expressions.

Question 5

A fixed amount of gas in a cylinder is being compressed. The relationship between pressure PP (in kPa), volume VV (in L), and temperature TT (in K) is given by the law PV=kTPV=kT for some constant kk. At an instant when the volume is 50 L, the pressure is 400 kPa, and the temperature is 400 K, the volume is decreasing at 1 L/s and the temperature is increasing at 2 K/s. What is the rate of change of the pressure at this instant?

  1. It is increasing at 2 kPa/s.
  2. It is decreasing at 6 kPa/s.
  3. It is increasing at 10 kPa/s. (correct answer)
  4. It is increasing at 136 kPa/s.
Explanation: This is a related rates problem involving three variables connected by the ideal gas law. When you see multiple changing quantities governed by an equation, you need to differentiate implicitly with respect to time to find how their rates of change relate. Starting with PV=kTPV = kT, differentiate both sides with respect to time: ddt(PV)=kdTdt\frac{d}{dt}(PV) = k\frac{dT}{dt}. Using the product rule on the left side: PdVdt+VdPdt=kdTdtP\frac{dV}{dt} + V\frac{dP}{dt} = k\frac{dT}{dt}. First, find the constant kk using the given conditions: k=PVT=400×50400=50k = \frac{PV}{T} = \frac{400 \times 50}{400} = 50. Now substitute all known values into the differentiated equation. Given: V=50V = 50 L, P=400P = 400 kPa, dVdt=−1\frac{dV}{dt} = -1 L/s (negative because volume is decreasing), and dTdt=2\frac{dT}{dt} = 2 K/s. 400(−1)+50dPdt=50(2)400(-1) + 50\frac{dP}{dt} = 50(2) −400+50dPdt=100-400 + 50\frac{dP}{dt} = 100 50dPdt=50050\frac{dP}{dt} = 500 dPdt=10\frac{dP}{dt} = 10 kPa/s The pressure is increasing at 10 kPa/s, which is answer C. Answer A (2 kPa/s) likely results from calculation errors or forgetting the product rule. Answer B (-6 kPa/s) suggests sign confusion with the decreasing volume. Answer D (136 kPa/s) probably comes from incorrectly using the original values instead of properly applying the differentiated equation. For related rates problems, always differentiate the constraint equation first, then substitute known values—never substitute before differentiating.

Question 6

A water trough is 10 m long and has a cross-section in the shape of an isosceles trapezoid. The trapezoid is 2 m wide at the bottom, 4 m wide at the top, and has a height of 4 m. If the trough is being filled with water at a rate of 5 m³/min, how fast is the water level rising when the water is 2 m deep?

  1. 1/61/6 m/min (correct answer)
  2. 1/51/5 m/min
  3. 1/41/4 m/min
  4. 5/125/12 m/min
Explanation: Let VV be the volume of water, hh be the water level, and ww be the width of the water surface. The volume is V=Area×Length=12(2+w)h×10V = \text{Area} \times \text{Length} = \frac{1}{2}(2+w)h \times 10. We need to express ww in terms of hh. The total change in width is 4−2=24-2=2 m over a height of 44 m. The trapezoid has slanted sides. By similar triangles formed by the slanted sides, the extra width on each side is xx. So x/h=(1)/4x/h = (1)/4, which means x=h/4x = h/4. The total width of the water surface is w=2+2x=2+2(h/4)=2+h/2w = 2 + 2x = 2 + 2(h/4) = 2 + h/2. Substitute this into the volume formula: V=5(2+(2+h/2))h=5(4+h/2)h=20h+52h2V = 5(2 + (2+h/2))h = 5(4+h/2)h = 20h + \frac{5}{2}h^2. Now, differentiate with respect to tt: dVdt=20dhdt+5hdhdt=(20+5h)dhdt\frac{dV}{dt} = 20\frac{dh}{dt} + 5h\frac{dh}{dt} = (20+5h)\frac{dh}{dt}. We are given dVdt=5\frac{dV}{dt} = 5 m³/min and we want to find dhdt\frac{dh}{dt} when h=2h=2 m. So, 5=(20+5(2))dhdt=30dhdt5 = (20 + 5(2))\frac{dh}{dt} = 30\frac{dh}{dt}. Thus, dhdt=530=16\frac{dh}{dt} = \frac{5}{30} = \frac{1}{6} m/min.

Question 7

A camera is positioned on the ground 2000 feet from the launchpad of a vertically ascending rocket. The rocket's velocity is a constant 500 ft/s. What is the rate of change of the camera's angle of elevation when the rocket is 4000 feet high?

  1. 1/101/10 rad/s
  2. 1/201/20 rad/s (correct answer)
  3. 1/51/5 rad/s
  4. 1/41/4 rad/s
Explanation: This is a classic related rates problem where you need to find how fast one quantity changes when you know how fast another quantity changes. When you see a setup involving changing distances and angles, think about forming a triangle and using trigonometry to relate the variables. Set up the problem by drawing a right triangle where the horizontal leg is the 2000-foot distance from camera to launchpad, the vertical leg is the rocket's height hh, and the angle of elevation is θ\theta. From trigonometry: tan⁡(θ)=h2000\tan(\theta) = \frac{h}{2000}. To find dθdt\frac{d\theta}{dt}, differentiate both sides with respect to time: sec⁡2(θ)⋅dθdt=12000⋅dhdt\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{2000} \cdot \frac{dh}{dt}. Since the rocket rises at 500 ft/s, dhdt=500\frac{dh}{dt} = 500. When the rocket is 4000 feet high, you can find sec⁡2(θ)\sec^2(\theta) using the triangle. At this moment, the hypotenuse has length 20002+40002=20,000,000=20005\sqrt{2000^2 + 4000^2} = \sqrt{20,000,000} = 2000\sqrt{5}. Therefore, cos⁡(θ)=200020005=15\cos(\theta) = \frac{2000}{2000\sqrt{5}} = \frac{1}{\sqrt{5}}, so sec⁡2(θ)=5\sec^2(\theta) = 5. Substituting: 5⋅dθdt=5002000=145 \cdot \frac{d\theta}{dt} = \frac{500}{2000} = \frac{1}{4}, giving dθdt=120\frac{d\theta}{dt} = \frac{1}{20} rad/s. Answer B is correct. Answer A (110\frac{1}{10}) likely comes from forgetting to square the secant. Answer C (15\frac{1}{5}) might result from using cos⁡2(θ)\cos^2(\theta) instead of sec⁡2(θ)\sec^2(\theta). Answer D (14\frac{1}{4}) occurs if you forget to account for the sec⁡2(θ)\sec^2(\theta) factor entirely. Remember: in related rates problems, always identify your triangle, write the trigonometric relationship, then differentiate with respect to time.

Question 8

Ship A leaves a port at noon and sails north at 15 knots. Ship B leaves the same port at 1:00 PM and sails east at 20 knots. At 3:00 PM, what is the rate of change of the angle θ\theta of the line of sight from Ship A to Ship B, measured clockwise from the north direction?

  1. 12/14512/145 rad/hr (correct answer)
  2. 00 rad/hr
  3. 1/51/5 rad/hr
  4. 6/256/25 rad/hr
Explanation: Let yy be the distance of Ship A from the port and xx be the distance of Ship B. At 3:00 PM, Ship A has been sailing for 3 hours and Ship B for 2 hours. So, y=15×3=45y = 15 \times 3 = 45 nautical miles, and x=20×2=40x = 20 \times 2 = 40 nautical miles. We have dydt=15\frac{dy}{dt} = 15 knots and dxdt=20\frac{dx}{dt} = 20 knots. The angle θ\theta satisfies tan⁡(θ)=x/y\tan(\theta) = x/y. Differentiating with respect to time tt: sec⁡2(θ)dθdt=y(dx/dt)−x(dy/dt)y2\sec^2(\theta) \frac{d\theta}{dt} = \frac{y(dx/dt) - x(dy/dt)}{y^2}. We can find dθdt=y(dx/dt)−x(dy/dt)y2sec⁡2(θ)\frac{d\theta}{dt} = \frac{y(dx/dt) - x(dy/dt)}{y^2 \sec^2(\theta)}. The distance between the ships is D=x2+y2D = \sqrt{x^2+y^2}, so sec⁡(θ)=D/y\sec(\theta) = D/y. Thus, y2sec⁡2(θ)=y2(D2/y2)=D2=x2+y2y^2 \sec^2(\theta) = y^2(D^2/y^2) = D^2 = x^2+y^2. So, dθdt=y(dx/dt)−x(dy/dt)x2+y2\frac{d\theta}{dt} = \frac{y(dx/dt) - x(dy/dt)}{x^2+y^2}. Plugging in the values: dθdt=(45)(20)−(40)(15)402+452=900−6001600+2025=3003625=12145\frac{d\theta}{dt} = \frac{(45)(20) - (40)(15)}{40^2 + 45^2} = \frac{900 - 600}{1600 + 2025} = \frac{300}{3625} = \frac{12}{145} rad/hr.