Calculus 1 Quiz: Intro To Related Rates
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Intro To Related RatesQuestion 1 of 12

Water is pumped into an inverted conical tank at a rate of 2 cubic meters per minute. The tank has a height of 6 meters and the radius at the top is 3 meters. How fast is the water level rising when the water is 4 meters deep?

132π\frac{1}{32\pi} m/min
14π\frac{1}{4\pi} m/min
12π\frac{1}{2\pi} m/min
2π\frac{2}{\pi} m/min
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Calculus 1 Quiz

Calculus 1 Quiz: Intro To Related Rates

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

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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

Water is pumped into an inverted conical tank at a rate of 2 cubic meters per minute. The tank has a height of 6 meters and the radius at the top is 3 meters. How fast is the water level rising when the water is 4 meters deep?

  1. 132π\frac{1}{32\pi} m/min
  2. 14π\frac{1}{4\pi} m/min
  3. 12π\frac{1}{2\pi} m/min (correct answer)
  4. 2π\frac{2}{\pi} m/min
Explanation: The volume of a cone is V=13πr2hV = \frac{1}{3}\pi r^2 h. The ratio of the radius to the height of the water in the cone is constant and equal to the ratio for the tank itself. So, rh=36=12\frac{r}{h} = \frac{3}{6} = \frac{1}{2}, which means r=h2r = \frac{h}{2}. Substituting this into the volume formula gives V=13π(h2)2h=13π(h24)h=π12h3V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{h^2}{4}\right) h = \frac{\pi}{12}h^3. Differentiating with respect to time tt, we get dVdt=π12(3h2)dhdt=π4h2dhdt\frac{dV}{dt} = \frac{\pi}{12}(3h^2) \frac{dh}{dt} = \frac{\pi}{4}h^2 \frac{dh}{dt}. We are given dVdt=2\frac{dV}{dt} = 2 m³/min and we want to find dhdt\frac{dh}{dt} when h=4h=4 m. Plugging in the values: 2=π4(42)dhdt⇒2=π4(16)dhdt⇒2=4πdhdt2 = \frac{\pi}{4}(4^2) \frac{dh}{dt} \Rightarrow 2 = \frac{\pi}{4}(16) \frac{dh}{dt} \Rightarrow 2 = 4\pi \frac{dh}{dt}. Solving for dhdt\frac{dh}{dt} gives dhdt=24π=12π\frac{dh}{dt} = \frac{2}{4\pi} = \frac{1}{2\pi} m/min.

Question 2

Gravel is falling from a conveyor belt onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the pile is always three times its height. How fast is the height of the pile increasing when the pile is 4 feet high?

  1. 572π\frac{5}{72\pi} ft/min
  2. 518π\frac{5}{18\pi} ft/min (correct answer)
  3. 18π5\frac{18\pi}{5} ft/min
  4. 458π\frac{45}{8\pi} ft/min
Explanation: Let VV, rr, and hh be the volume, radius, and height of the conical pile. We are given dVdt=10\frac{dV}{dt} = 10. The relationship between diameter DD and height is D=3hD = 3h. Since the radius is half the diameter, r=D2=3h2r = \frac{D}{2} = \frac{3h}{2}. The volume of a cone is V=13πr2hV = \frac{1}{3}\pi r^2 h. Substituting for rr: V=13π(3h2)2h=13π(9h24)h=3π4h3V = \frac{1}{3}\pi \left(\frac{3h}{2}\right)^2 h = \frac{1}{3}\pi \left(\frac{9h^2}{4}\right)h = \frac{3\pi}{4}h^3. Now, differentiate with respect to time tt: dVdt=3π4(3h2)dhdt=9π4h2dhdt\frac{dV}{dt} = \frac{3\pi}{4} (3h^2) \frac{dh}{dt} = \frac{9\pi}{4}h^2 \frac{dh}{dt}. We want to find dhdt\frac{dh}{dt} when h=4h=4. Substitute the known values: 10=9π4(42)dhdt=9π4(16)dhdt=36πdhdt10 = \frac{9\pi}{4}(4^2) \frac{dh}{dt} = \frac{9\pi}{4}(16) \frac{dh}{dt} = 36\pi \frac{dh}{dt}. Solving for dhdt\frac{dh}{dt} gives dhdt=1036π=518π\frac{dh}{dt} = \frac{10}{36\pi} = \frac{5}{18\pi} ft/min.

Question 3

The length ll of a rectangle is increasing at a rate of 8 cm/s and its width ww is decreasing at a rate of 3 cm/s. When l=20l=20 cm and w=15w=15 cm, at what rate is the length of the diagonal changing?

  1. 10 cm/s
  2. 23/523/5 cm/s (correct answer)
  3. 41/541/5 cm/s
  4. 60 cm/s
Explanation: Let DD be the length of the diagonal. By the Pythagorean theorem, D2=l2+w2D^2 = l^2 + w^2. We are given dldt=8\frac{dl}{dt} = 8 and dwdt=−3\frac{dw}{dt} = -3 (since the width is decreasing). We want to find dDdt\frac{dD}{dt} when l=20l=20 and w=15w=15. At this instant, the diagonal is D=202+152=400+225=625=25D = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 cm. Differentiating the relationship with respect to time tt: 2DdDdt=2ldldt+2wdwdt2D\frac{dD}{dt} = 2l\frac{dl}{dt} + 2w\frac{dw}{dt}. Divide by 2: DdDdt=ldldt+wdwdtD\frac{dD}{dt} = l\frac{dl}{dt} + w\frac{dw}{dt}. Substitute the known values: 25dDdt=20(8)+15(−3)25\frac{dD}{dt} = 20(8) + 15(-3). This gives 25dDdt=160−45=11525\frac{dD}{dt} = 160 - 45 = 115. Solving for dDdt\frac{dD}{dt} yields dDdt=11525=235\frac{dD}{dt} = \frac{115}{25} = \frac{23}{5} cm/s. Distractor C is the result of using dwdt=+3\frac{dw}{dt}=+3, a common sign error. Distractors A and D are the rates of change of the perimeter and area, respectively.

Question 4

The volume of a cube is increasing at a constant rate of 150 cm³/s. What is the rate of change of the cube's surface area at the instant when the edge length is 5 cm?

  1. 2 cm²/s
  2. 60 cm²/s
  3. 75 cm²/s
  4. 120 cm²/s (correct answer)
Explanation: Let ss be the edge length of the cube. The volume is V=s3V = s^3 and the surface area is A=6s2A = 6s^2. We are given dVdt=150\frac{dV}{dt} = 150. We want to find dAdt\frac{dA}{dt} when s=5s=5. First, use the volume information to find dsdt\frac{ds}{dt}. Differentiate VV with respect to time tt: dVdt=3s2dsdt\frac{dV}{dt} = 3s^2 \frac{ds}{dt}. Substitute the known values: 150=3(52)dsdt=3(25)dsdt=75dsdt150 = 3(5^2)\frac{ds}{dt} = 3(25)\frac{ds}{dt} = 75\frac{ds}{dt}. Solving for dsdt\frac{ds}{dt} gives dsdt=15075=2\frac{ds}{dt} = \frac{150}{75} = 2 cm/s. Next, differentiate AA with respect to time tt: dAdt=12sdsdt\frac{dA}{dt} = 12s \frac{ds}{dt}. Substitute the values for the specific instant: dAdt=12(5)(2)=120\frac{dA}{dt} = 12(5)(2) = 120 cm²/s. Distractor B results from forgetting the chain rule factor dsdt\frac{ds}{dt} (or assuming it is 1). Distractor C results from confusing the formulas for volume and surface area.

Question 5

A rocket is launched vertically and is tracked by a radar station located on the ground 5 miles from the launch pad. What is the rate of change of the rocket's angle of elevation θ\theta from the radar station when the rocket is 10 miles high and traveling at 600 miles per second?

  1. 12 rad/s
  2. 120 rad/s
  3. 60 rad/s
  4. 24 rad/s (correct answer)
Explanation: This is a classic related rates problem where you need to find how one quantity changes with respect to time when given information about another quantity's rate of change. When you see a problem involving angles and moving objects, set up the geometric relationship first, then differentiate with respect to time. Let's establish the setup: the rocket moves vertically upward at height hh, the radar station is 5 miles away horizontally, and the angle of elevation is θ\theta. From basic trigonometry: tan⁡θ=h5\tan \theta = \frac{h}{5} To find dθdt\frac{d\theta}{dt}, differentiate both sides with respect to time: sec⁡2θ⋅dθdt=15⋅dhdt\sec^2 \theta \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot \frac{dh}{dt} When the rocket is 10 miles high, we can find sec⁡2θ\sec^2 \theta. Since tan⁡θ=105=2\tan \theta = \frac{10}{5} = 2, we have sec⁡2θ=1+tan⁡2θ=1+4=5\sec^2 \theta = 1 + \tan^2 \theta = 1 + 4 = 5. Substituting the known values (dhdt=600\frac{dh}{dt} = 600 miles/s): 5⋅dθdt=15⋅600=1205 \cdot \frac{d\theta}{dt} = \frac{1}{5} \cdot 600 = 120 Therefore: dθdt=1205=24 rad/s\frac{d\theta}{dt} = \frac{120}{5} = 24 \text{ rad/s} The answer is D) 24 rad/s. A) 12 rad/s results from incorrectly using sec⁡2θ=10\sec^2 \theta = 10 instead of 5. B) 120 rad/s comes from forgetting to divide by sec⁡2θ\sec^2 \theta. C) 60 rad/s occurs from using the wrong geometric relationship or calculation error. Strategy tip: In related rates problems, always identify the geometric relationship first, differentiate with respect to time, then substitute known values in the correct order.

Question 6

A 10-foot ladder is leaning against a vertical wall. The base of the ladder is pulled away from the wall at a constant rate of 2 ft/s. At what rate is the angle between the ladder and the ground changing when the base of the ladder is 6 feet from the wall?

  1. The angle is decreasing at a rate of 1/41/4 rad/s. (correct answer)
  2. The angle is increasing at a rate of 1/41/4 rad/s.
  3. The angle is decreasing at a rate of 1/31/3 rad/s.
  4. The angle is decreasing at a rate of 1/51/5 rad/s.
Explanation: Let xx be the distance from the base of the ladder to the wall, and let θ\theta be the angle between the ladder and the ground. The relationship between xx and θ\theta is given by cos⁡(θ)=x/10\cos(\theta) = x/10. Differentiating both sides with respect to time tt, we get −sin⁡(θ)dθdt=110dxdt-\sin(\theta) \frac{d\theta}{dt} = \frac{1}{10} \frac{dx}{dt}. We are given dxdt=2\frac{dx}{dt} = 2 ft/s. We need to find dθdt\frac{d\theta}{dt} when x=6x=6. At this instant, the ladder forms a right triangle with sides 6 and y=102−62=64=8y = \sqrt{10^2 - 6^2} = \sqrt{64} = 8. Thus, sin⁡(θ)=oppositehypotenuse=y10=810=45\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{10} = \frac{8}{10} = \frac{4}{5}. Substituting the known values into the differentiated equation: −(45)dθdt=110(2)-\left(\frac{4}{5}\right) \frac{d\theta}{dt} = \frac{1}{10}(2). This simplifies to −45dθdt=15-\frac{4}{5} \frac{d\theta}{dt} = \frac{1}{5}. Solving for dθdt\frac{d\theta}{dt} gives dθdt=−14\frac{d\theta}{dt} = -\frac{1}{4} rad/s. The negative sign indicates that the angle is decreasing.

Question 7

A stone is dropped into a calm pond, causing circular ripples. The radius of the outer ripple is increasing at a rate given by the equation drdt=2r\frac{dr}{dt} = \frac{2}{r} cm/s, where rr is the radius in cm. At what rate is the area of the disturbed water increasing when the radius is 10 cm?

  1. 2π/52\pi/5 cm²/s
  2. 4π4\pi cm²/s (correct answer)
  3. 20π20\pi cm²/s
  4. 40π40\pi cm²/s
Explanation: The area of a circle is given by A=πr2A = \pi r^2. To find the rate at which the area is increasing, we differentiate this equation with respect to time tt: dAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \frac{dr}{dt}. We are given that drdt=2r\frac{dr}{dt} = \frac{2}{r}. Substituting this expression into our equation for dAdt\frac{dA}{dt} gives dAdt=2πr(2r)=4π\frac{dA}{dt} = 2\pi r \left(\frac{2}{r}\right) = 4\pi. The rate of change of the area is a constant 4π4\pi cm²/s, independent of the radius. The information that the radius is 10 cm is not needed to find the rate, which is a key feature of this problem.

Question 8

A right cylindrical tank with a radius of 5 meters is being filled with water at a rate of 4π4\pi cubic meters per minute. However, the tank is also expanding in such a way that its radius is increasing at a rate of 0.1 meters per minute. How fast is the water level, hh, rising when the height of the water is 2 meters?

  1. 0.08 m/min (correct answer)
  2. 0.1 m/min
  3. 0.12 m/min
  4. 0.2 m/min
Explanation: The volume of a cylinder is given by V=πr2hV = \pi r^2 h. Since both the radius rr and height hh are changing with time, we must use the product rule when differentiating: dVdt=π(2rdrdth+r2dhdt)\frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right). We are given dVdt=4π\frac{dV}{dt} = 4\pi m³/min, drdt=0.1\frac{dr}{dt} = 0.1 m/min. We want to find dhdt\frac{dh}{dt} at the instant when r=5r=5 m and h=2h=2 m. Substitute the known values into the equation: 4π=π(2(5)(0.1)(2)+(52)dhdt)4\pi = \pi \left( 2(5)(0.1)(2) + (5^2) \frac{dh}{dt} \right). Dividing by π\pi gives 4=2(5)(0.1)(2)+25dhdt4 = 2(5)(0.1)(2) + 25 \frac{dh}{dt}. Simplify: 4=2+25dhdt4 = 2 + 25 \frac{dh}{dt}. This leads to 2=25dhdt2 = 25 \frac{dh}{dt}, so dhdt=225=0.08\frac{dh}{dt} = \frac{2}{25} = 0.08 m/min.

Question 9

A particle moves along the curve y=x2+9y = \sqrt{x^2+9}. The x-coordinate of the particle is increasing at a rate of 8 units per second. What is the rate of change of the particle's y-coordinate when the particle is at the point (4,5)(4, 5)?

  1. 5 units/s
  2. 10 units/s
  3. 8 units/s
  4. 6.4 units/s (correct answer)
Explanation: This is a related rates problem, where you need to find how fast one quantity changes when you know how another is changing. When you see "rate of change" language with connected variables, think about using implicit differentiation with respect to time. Start with the given curve equation y=x2+9y = \sqrt{x^2+9} and differentiate both sides with respect to time tt. Using the chain rule: dydt=12x2+9⋅2x⋅dxdt=xx2+9⋅dxdt\frac{dy}{dt} = \frac{1}{2\sqrt{x^2+9}} \cdot 2x \cdot \frac{dx}{dt} = \frac{x}{\sqrt{x^2+9}} \cdot \frac{dx}{dt} You're told that dxdt=8\frac{dx}{dt} = 8 units/s and need dydt\frac{dy}{dt} when the particle is at (4,5)(4,5). Substituting these values: dydt=442+9⋅8=425⋅8=45⋅8=6.4\frac{dy}{dt} = \frac{4}{\sqrt{4^2+9}} \cdot 8 = \frac{4}{\sqrt{25}} \cdot 8 = \frac{4}{5} \cdot 8 = 6.4 units/s. Answer choice A (5 units/s) likely comes from incorrectly using just the y-coordinate value. Answer choice B (10 units/s) might result from miscalculating 45⋅8\frac{4}{5} \cdot 8 as 54⋅8\frac{5}{4} \cdot 8. Answer choice C (8 units/s) is the trap of assuming dydt=dxdt\frac{dy}{dt} = \frac{dx}{dt}, ignoring that the curve's slope affects how changes in x translate to changes in y. For related rates problems, always: (1) write down the constraint equation, (2) differentiate implicitly with respect to time, (3) substitute known values including the specific point, and (4) solve for the unknown rate. The key insight is that rates of change along a curve depend on both the given rate and the curve's geometry at that point.

Question 10

Two cars start from the same point. Car A travels east at 60 mph and Car B travels northeast (45° east of north) at 50 mph. At what rate is the distance between the cars increasing after 2 hours?

  1. 106110\sqrt{61} mph
  2. 6100−30002\sqrt{6100 - 3000\sqrt{2}} mph (correct answer)
  3. 6100−15002\sqrt{6100 - 1500\sqrt{2}} mph
  4. 6100+30002\sqrt{6100 + 3000\sqrt{2}} mph
Explanation: Let a(t)a(t) be the distance of Car A from the start and b(t)b(t) be the distance of Car B. We have dadt=60\frac{da}{dt} = 60 and dbdt=50\frac{db}{dt} = 50. The angle θ\theta between their paths is 45∘45^\circ. Let c(t)c(t) be the distance between them. By the Law of Cosines, c2=a2+b2−2abcos⁡(45∘)=a2+b2−2abc^2 = a^2 + b^2 - 2ab\cos(45^\circ) = a^2 + b^2 - \sqrt{2}ab. Differentiate with respect to tt: 2cdcdt=2adadt+2bdbdt−2(dadtb+adbdt)2c\frac{dc}{dt} = 2a\frac{da}{dt} + 2b\frac{db}{dt} - \sqrt{2}\left(\frac{da}{dt}b + a\frac{db}{dt}\right). After t=2t=2 hours, a=60(2)=120a = 60(2) = 120 miles and b=50(2)=100b = 50(2) = 100 miles. At this time, c2=1202+1002−2(120)(100)=14400+10000−120002=24400−120002c^2 = 120^2 + 100^2 - \sqrt{2}(120)(100) = 14400 + 10000 - 12000\sqrt{2} = 24400 - 12000\sqrt{2}. So c=24400−120002c = \sqrt{24400 - 12000\sqrt{2}}. Now substitute into the derivative equation: 2cdcdt=2(120)(60)+2(100)(50)−2(60⋅100+120⋅50)=14400+10000−2(6000+6000)=24400−1200022c\frac{dc}{dt} = 2(120)(60) + 2(100)(50) - \sqrt{2}(60 \cdot 100 + 120 \cdot 50) = 14400 + 10000 - \sqrt{2}(6000 + 6000) = 24400 - 12000\sqrt{2}. So, 2cdcdt=c22c\frac{dc}{dt} = c^2, which implies dcdt=c2=1224400−120002=24400−1200024=6100−30002\frac{dc}{dt} = \frac{c}{2} = \frac{1}{2}\sqrt{24400 - 12000\sqrt{2}} = \sqrt{\frac{24400 - 12000\sqrt{2}}{4}} = \sqrt{6100 - 3000\sqrt{2}} mph.

Question 11

Air is being pumped into a spherical balloon so that its volume increases at a rate of 100 cm³/s. How fast is the surface area of the balloon increasing when its radius is 5 cm?

  1. 1/π1/\pi cm²/s
  2. 20 cm²/s
  3. 40 cm²/s (correct answer)
  4. 40π40\pi cm²/s
Explanation: Let VV be the volume and AA be the surface area of the sphere with radius rr. The formulas are V=43πr3V = \frac{4}{3}\pi r^3 and A=4πr2A = 4\pi r^2. We are given dVdt=100\frac{dV}{dt} = 100 and want to find dAdt\frac{dA}{dt} when r=5r=5. First, we find the rate of change of the radius, drdt\frac{dr}{dt}. Differentiate the volume formula: dVdt=4πr2drdt\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}. Substitute the given values: 100=4π(52)drdt=100πdrdt100 = 4\pi (5^2) \frac{dr}{dt} = 100\pi \frac{dr}{dt}. So, drdt=1π\frac{dr}{dt} = \frac{1}{\pi} cm/s. Next, differentiate the surface area formula: dAdt=8πrdrdt\frac{dA}{dt} = 8\pi r \frac{dr}{dt}. Now substitute r=5r=5 and drdt=1π\frac{dr}{dt} = \frac{1}{\pi}: dAdt=8π(5)(1π)=40\frac{dA}{dt} = 8\pi (5) \left(\frac{1}{\pi}\right) = 40 cm²/s. Alternatively, note that dAdt=8πrdrdt\frac{dA}{dt} = 8\pi r \frac{dr}{dt} and drdt=14πr2dVdt\frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}. Substituting gives dAdt=8πr(14πr2dVdt)=2rdVdt\frac{dA}{dt} = 8\pi r \left(\frac{1}{4\pi r^2}\frac{dV}{dt}\right) = \frac{2}{r}\frac{dV}{dt}. At r=5r=5, dAdt=25(100)=40\frac{dA}{dt} = \frac{2}{5}(100) = 40 cm²/s.

Question 12

A person who is 6 feet tall is walking away from a 15-foot tall lamppost at a constant rate of 3 ft/s. What is the rate at which the length of the person's shadow is increasing?

  1. 2 ft/s (correct answer)
  2. 3 ft/s
  3. 4.5 ft/s
  4. 5 ft/s
Explanation: Let xx be the distance of the person from the lamppost and ss be the length of the shadow. Let H=15H=15 be the height of the lamppost and h=6h=6 be the height of the person. By similar triangles, we have the proportion sh=x+sH\frac{s}{h} = \frac{x+s}{H}. Substituting the heights, we get s6=x+s15\frac{s}{6} = \frac{x+s}{15}. Cross-multiplying gives 15s=6(x+s)=6x+6s15s = 6(x+s) = 6x + 6s. Solving for ss, we get 9s=6x9s = 6x, or s=23xs = \frac{2}{3}x. Differentiating with respect to time tt, we get dsdt=23dxdt\frac{ds}{dt} = \frac{2}{3}\frac{dx}{dt}. Given that the person is walking at 3 ft/s, dxdt=3\frac{dx}{dt} = 3. Therefore, dsdt=23(3)=2\frac{ds}{dt} = \frac{2}{3}(3) = 2 ft/s. Distractor D represents the rate at which the tip of the shadow is moving, which is ddt(x+s)=dxdt+dsdt=3+2=5\frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt} = 3 + 2 = 5 ft/s, a common point of confusion.