Calculus 1 Quiz: Related Rates With Geometry
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Related Rates With GeometryQuestion 1 of 7

The length ll of a rectangle is increasing at a rate of 3 cm/s while its width ww is decreasing at a rate of 4 cm/s. At the moment when l=8l = 8 cm and w=6w = 6 cm, what is the rate of change of the length of the rectangle's diagonal?

−14-14 cm/s
−1-1 cm/s
0 cm/s
1 cm/s
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Calculus 1 Quiz

Calculus 1 Quiz: Related Rates With Geometry

Practice Related Rates With Geometry in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Related Rates With Geometry, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

The length ll of a rectangle is increasing at a rate of 3 cm/s while its width ww is decreasing at a rate of 4 cm/s. At the moment when l=8l = 8 cm and w=6w = 6 cm, what is the rate of change of the length of the rectangle's diagonal?

  1. −14-14 cm/s
  2. −1-1 cm/s
  3. 0 cm/s (correct answer)
  4. 1 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=3\frac{dl}{dt} = 3 cm/s and dwdt=−4\frac{dw}{dt} = -4 cm/s. We want to find dDdt\frac{dD}{dt} when l=8l=8 cm and w=6w=6 cm. First, at this instant, the diagonal's length is D=82+62=64+36=100=10D = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 cm. Differentiating the Pythagorean equation with respect to time tt gives 2DdDdt=2ldldt+2wdwdt2D \frac{dD}{dt} = 2l \frac{dl}{dt} + 2w \frac{dw}{dt}, which simplifies to DdDdt=ldldt+wdwdtD \frac{dD}{dt} = l \frac{dl}{dt} + w \frac{dw}{dt}. Substituting the known values: 10dDdt=(8)(3)+(6)(−4)10 \frac{dD}{dt} = (8)(3) + (6)(-4). 10dDdt=24−24=010 \frac{dD}{dt} = 24 - 24 = 0. Therefore, dDdt=0\frac{dD}{dt} = 0 cm/s. At this specific instant, the length of the diagonal is momentarily not changing. Distractor A, −14-14 cm/s, is the rate of change of the area of the rectangle: dAdt=ldwdt+wdldt=8(−4)+6(3)=−32+18=−14\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt} = 8(-4) + 6(3) = -32 + 18 = -14. Distractor B, −1-1 cm/s, is the sum of the rates, dldt+dwdt=3+(−4)=−1\frac{dl}{dt} + \frac{dw}{dt} = 3 + (-4) = -1. Distractor D, 1 cm/s, is a sign error on the sum of the rates.

Question 2

The radius of a right circular cylinder is increasing at a rate of 2 in/min, and the height is decreasing at a rate of 5 in/min. At the instant when the radius is 10 inches and the height is 8 inches, what is the rate of change of the volume of the cylinder?

  1. 220π220\pi in$^3$/min
  2. −120π-120\pi in$^3$/min
  3. 180π180\pi in$^3$/min
  4. −180π-180\pi in$^3$/min (correct answer)
Explanation: This is a related rates problem involving a cylinder whose dimensions are changing over time. When you see multiple quantities changing simultaneously and need to find how another quantity changes, you're dealing with related rates - use implicit differentiation with respect to time. The volume of a cylinder is V=πr2hV = \pi r^2 h. To find how volume changes with time, differentiate both sides with respect to time: dVdt=πddt(r2h)\frac{dV}{dt} = \pi \frac{d}{dt}(r^2 h). Using the product rule: dVdt=π(2rdrdt⋅h+r2dhdt)\frac{dV}{dt} = \pi(2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt}). Now substitute the given values: drdt=2\frac{dr}{dt} = 2 in/min, dhdt=−5\frac{dh}{dt} = -5 in/min (negative because height is decreasing), r=10r = 10 inches, and h=8h = 8 inches. dVdt=π(2(10)(2)(8)+(10)2(−5))=π(320−500)=−180π\frac{dV}{dt} = \pi(2(10)(2)(8) + (10)^2(-5)) = \pi(320 - 500) = -180\pi cubic inches per minute. Choice A (220π220\pi) likely comes from adding instead of subtracting: 320+500=820320 + 500 = 820, then making an error. Choice B (−120π-120\pi) might result from incorrectly using the derivative formula or arithmetic mistakes. Choice C (180π180\pi) gives the correct magnitude but wrong sign - this happens when you forget that decreasing height contributes negatively to the rate of change. The key insight for related rates problems: always pay careful attention to whether rates are positive (increasing) or negative (decreasing), and remember to use the product rule when variables multiply together in your formula.

Question 3

Water is being pumped into an inverted conical tank at a rate of 10 ft3$/min.Atthesametime,waterisleakingoutofthebottomatarateof2ft^3$/min. At the same time, water is leaking out of the bottom at a rate of 2 ft^3$/min. The tank has a height of 9 ft and a top radius of 3 ft. What is the rate at which the water level is rising when the water is 6 ft deep?

  1. 1/(π)1 / (\pi) ft/min
  2. 2/(π)2 / (\pi) ft/min (correct answer)
  3. 3/(2π)3 / (2\pi) ft/min
  4. 4/(π)4 / (\pi) ft/min
Explanation: Let VV be the volume of water in the tank. The net rate of change of the volume is the rate in minus the rate out: dVdt=10−2=8\frac{dV}{dt} = 10 - 2 = 8 ft3$/min.Thevolumeofthewater,whichisintheshapeofacone,is^3$/min. The volume of the water, which is in the shape of a cone, is V = \frac{1}{3}\pi r^2 h.Forthetank,theratioofradiustoheightis. For the tank, the ratio of radius to height is R/H = 3/9 = 1/3.Thisratioholdsforthewateratanylevel,so. This ratio holds for the water at any level, so r/h = 1/3,whichmeans, which means r = h/3.Substitutethisintothevolumeformulatoexpress. Substitute this into the volume formula to express Vintermsofin terms ofhonly:only:V = \frac{1}{3}\pi (\frac{h}{3})^2 h = \frac{1}{3}\pi (h29\frac{h^2}{9}) h = \frac{\pi}{27}h^3.Now,differentiatewithrespecttotime. Now, differentiate with respect to time t:: \frac{dV}{dt} = \frac{\pi}{27}(3h23h^2) \frac{dh}{dt} = \frac{\pi}{9}h^2 \frac{dh}{dt}.Wewanttofind. We want to find \frac{dh}{dt}whenwhenh=6ft.Wehaveft. We have\frac{dV}{dt} = 8.So,. So, 8 = \frac{\pi}{9}(626^2) \frac{dh}{dt} = \frac{\pi}{9}(36) \frac{dh}{dt} = 4\pi \frac{dh}{dt}.Solvingfor. Solving for \frac{dh}{dt}givesgives\frac{dh}{dt} = \frac{8}{4\pi} = \frac{2}{\pi}$ ft/min. Distractor A, 1/π1/\pi, results from an arithmetic error or from using r=h/6r=h/6. Distractor C, 3/(23/(2\pi)), might arise from a calculation error. Distractor D, 4/π4/\pi, results from using only the inflow rate (10 ft3$/min)insteadofthenetrate,whichwouldgive^3$/min) instead of the net rate, which would give \frac{dh}{dt} = 10/(4\pi) = 5/(2\pi),orfromanothercalculationerrorleadingto, or from another calculation error leading to 4/\pi$.

Question 4

A particle moves along the curve y=xy = \sqrt{x}. As the particle passes through the point (4,2)(4, 2), its xx-coordinate is increasing at a rate of 3 units/sec. How fast is the distance from the particle to the origin changing at this instant?

  1. 9/29/2 units/sec
  2. 95/49\sqrt{5}/4 units/sec
  3. 27/427/4 units/sec
  4. 275/2027\sqrt{5}/20 units/sec (correct answer)
Explanation: This is a related rates problem where you need to find how fast the distance to the origin changes as the particle moves along a curve. When you see "how fast is changing," think derivatives with respect to time. Let's set up the problem systematically. The particle is at point (x,y)(x, y) where y=xy = \sqrt{x}, and the distance from the origin is d=x2+y2d = \sqrt{x^2 + y^2}. Since y=xy = \sqrt{x}, we have d=x2+x=x(x+1)d = \sqrt{x^2 + x} = \sqrt{x(x+1)}. To find dddt\frac{dd}{dt}, differentiate both sides: dddt=12x(x+1)⋅ddt[x(x+1)]=12x(x+1)⋅(2x+1)⋅dxdt\frac{dd}{dt} = \frac{1}{2\sqrt{x(x+1)}} \cdot \frac{d}{dt}[x(x+1)] = \frac{1}{2\sqrt{x(x+1)}} \cdot (2x+1) \cdot \frac{dx}{dt}. At point (4,2)(4,2), we have x=4x = 4 and dxdt=3\frac{dx}{dt} = 3. Substituting: dddt=124⋅5⋅(8+1)⋅3=9⋅3220=272⋅25=2745\frac{dd}{dt} = \frac{1}{2\sqrt{4 \cdot 5}} \cdot (8+1) \cdot 3 = \frac{9 \cdot 3}{2\sqrt{20}} = \frac{27}{2 \cdot 2\sqrt{5}} = \frac{27}{4\sqrt{5}}. Rationalizing the denominator: 2745⋅55=27520\frac{27}{4\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{27\sqrt{5}}{20}, which is choice D. Choice A (9/29/2) likely comes from forgetting to account for the curve constraint y=xy = \sqrt{x}. Choice B (95/49\sqrt{5}/4) appears to miss the factor of 55 in the denominator. Choice C (27/427/4) forgets to rationalize the denominator and incorrectly simplifies 20\sqrt{20}. Key strategy: In related rates problems involving distance, always write distance as a function of the given constraint, then differentiate the entire expression with respect to time.

Question 5

A spherical snowball is melting in such a way that its surface area decreases at a rate of 2 cm2$/min.Whatistherateofchangeoftheradius,incm/min,attheinstantthevolumeis^2$/min. What is the rate of change of the radius, in cm/min, at the instant the volume is 288\picm cm^3$?

  1. −1/(24π)-1/(24\pi) (correct answer)
  2. −1/(12π)-1/(12\pi)
  3. −1/(6π)-1/(6\pi)
  4. −1/(4π)-1/(4\pi)
Explanation: Let AA be the surface area, VV be the volume, and rr be the radius of the snowball. The formulas are A=4πr2A = 4\pi r^2 and V=43πr3V = \frac{4}{3}\pi r^3. We are given dAdt=−2\frac{dA}{dt} = -2 cm2$/min(negativebecauseit′sdecreasing).Weneedtofind^2$/min (negative because it's decreasing). We need to find \frac{dr}{dt}attheinstantat the instantV = 288\picm cm^3.First,wemustfindtheradius. First, we must find the radius ratthisinstant.Usingthevolumeformula:at this instant. Using the volume formula:288\pi = \frac{4}{3}\pi r^3.Dividingby. Dividing by \piandmultiplyingbyand multiplying by3/4givesgivesr^3 = 288 \times \frac{3}{4} = 72 \times 3 = 216.Thus,. Thus, r = \sqrt[3]{216} = 6cm.Now,weusethesurfaceareaformula′sderivativetofindcm. Now, we use the surface area formula's derivative to find\frac{dr}{dt}.Differentiating. Differentiating A = 4\pi r^2withrespecttotimewith respect to timetgivesgives\frac{dA}{dt} = 8\pi r \frac{dr}{dt}.Substitutetheknownvalues:. Substitute the known values: -2 = 8\pi (6) \frac{dr}{dt},whichis, which is -2 = 48\pi \frac{dr}{dt}.Solvingfor. Solving for \frac{dr}{dt}givesgives\frac{dr}{dt} = -\frac{2}{48\pi} = -\frac{1}{24\pi}$ cm/min. Distractor B, −1/(12-1/(12\pi)), results from an error in finding the radius, possibly using r=3r=3. Distractor C, −1/(6-1/(6\pi)), may result from forgetting the radius in the derivative, computing drdt=dAdt/(8\frac{dr}{dt} = \frac{dA}{dt} / (8\pi)). Distractor D, −1/(4-1/(4\pi)), might arise from a mistake in the surface area formula or its derivative.

Question 6

Two cars start moving from the same point. One travels south at 48 mi/h, and the other travels west at 36 mi/h. At what rate is the distance between the cars increasing two hours later?

  1. 60 mi/h (correct answer)
  2. 84 mi/h
  3. 96 mi/h
  4. 120 mi/h
Explanation: Let yy be the distance the first car has traveled south, and xx be the distance the second car has traveled west. Let DD be the distance between the cars. By the Pythagorean theorem, D2=x2+y2D^2 = x^2 + y^2. We are given dydt=48\frac{dy}{dt} = 48 mi/h and dxdt=36\frac{dx}{dt} = 36 mi/h. We want to find dDdt\frac{dD}{dt} after t=2t=2 hours. Differentiating the equation with respect to time tt, we get 2DdDdt=2xdxdt+2ydydt2D \frac{dD}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}, which simplifies to DdDdt=xdxdt+ydydtD \frac{dD}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt}. After 2 hours, the distances are y=48×2=96y = 48 \times 2 = 96 miles and x=36×2=72x = 36 \times 2 = 72 miles. The distance DD between them at this time is D=x2+y2=722+962=5184+9216=14400=120D = \sqrt{x^2 + y^2} = \sqrt{72^2 + 96^2} = \sqrt{5184 + 9216} = \sqrt{14400} = 120 miles. Now substitute all values into the differentiated equation: 120dDdt=(72)(36)+(96)(48)120 \frac{dD}{dt} = (72)(36) + (96)(48). 120dDdt=2592+4608=7200120 \frac{dD}{dt} = 2592 + 4608 = 7200. Solving for dDdt\frac{dD}{dt} gives dDdt=7200120=60\frac{dD}{dt} = \frac{7200}{120} = 60 mi/h. A common shortcut is to notice that if the velocities are constant, the rate of separation is constant and equal to (dxdt)2+(dydt)2=362+482=1296+2304=3600=60\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{36^2+48^2} = \sqrt{1296+2304} = \sqrt{3600} = 60 mi/h. This problem is designed so that this shortcut works, but the full method is required for more complex scenarios. Distractor B, 8484 mi/h, is the sum of the speeds, 48+3648 + 36. Distractor C, 9696 mi/h, is one of the car's positions at t=2t=2. Distractor D, 120120 mi/h, is the distance between the cars at t=2t=2, not the rate of change of the distance.

Question 7

The surface area of a spherical balloon is increasing at a constant rate of 12π12 \pi cm$^2$/s. What is the rate of change of the volume of the balloon when its radius is 3 cm?

  1. 9π9 \pi cm$^3$/s
  2. 18π18 \pi cm$^3$/s (correct answer)
  3. 36π36 \pi cm$^3$/s
  4. 72π72 \pi cm$^3$/s
Explanation: Let AA be the surface area, VV be the volume, and rr be the radius of the sphere. The formulas are A=4πr2A = 4\pi r^2 and V=43πr3V = \frac{4}{3}\pi r^3. We are given dAdt=12π\frac{dA}{dt} = 12\pi cm2$/sandwewanttofind^2$/s and we want to find \frac{dV}{dt}whenwhenr=3cm.Thisisatwo−stepproblem.First,wefindcm. This is a two-step problem. First, we find\frac{dr}{dt}usingthesurfaceareainformation.Differentiateusing the surface area information. DifferentiateAwithrespecttotimewith respect to timet:: \frac{dA}{dt} = 8\pi r \frac{dr}{dt}.Substitutethegivenvalues:. Substitute the given values: 12\pi = 8\pi (3) \frac{dr}{dt},whichgives, which gives 12\pi = 24\pi \frac{dr}{dt},so, so \frac{dr}{dt} = \frac{12\pi}{24\pi} = \frac{1}{2}cm/s.Next,wefindcm/s. Next, we find\frac{dV}{dt}.Differentiate. Differentiate Vwithrespecttotimewith respect to timet:: \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}.Nowsubstitute. Now substitute r=3andand\frac{dr}{dt} = \frac{1}{2}:: \frac{dV}{dt} = 4\pi (3)^2 (12\frac{1}{2}) = 4\pi(9)(12\frac{1}{2}) = 18\picm cm^3$/s. Distractor A, 9π9\pi cm3$/s,mightresultfromanalgebraicerror,perhapsdividingby2twice.DistractorC,^3$/s, might result from an algebraic error, perhaps dividing by 2 twice. Distractor C, 36\picm cm^3$/s, is the value of the volume when r=3r=3, not its rate of change. It could also result from forgetting the 12\frac{1}{2} factor for drdt\frac{dr}{dt}. Distractor D, 72π72\pi cm$^3$/s, might result from misremembering the derivative of surface area or volume.