Calculus 1 Quiz: Non Motion Rates Of Change
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Non Motion Rates Of ChangeQuestion 1 of 20

The amount of a certain drug in a patient's bloodstream t hours after an injection is given by A(t) = 10t / ($t^2$ + 1). At what rate is the amount of the drug in the bloodstream changing 2 hours after the injection?

Increasing at 0.8 units/hour
Decreasing at 0.8 units/hour
Increasing at 1.2 units/hour
Decreasing at 1.2 units/hour
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Calculus 1 Quiz

Calculus 1 Quiz: Non Motion Rates Of Change

Practice Non Motion Rates Of Change 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 Non Motion Rates Of Change, 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 amount of a certain drug in a patient's bloodstream t hours after an injection is given by A(t) = 10t / ($t^2$ + 1). At what rate is the amount of the drug in the bloodstream changing 2 hours after the injection?

  1. Increasing at 0.8 units/hour
  2. Decreasing at 0.8 units/hour
  3. Increasing at 1.2 units/hour
  4. Decreasing at 1.2 units/hour (correct answer)
Explanation: The rate of change is the derivative, A'(t). We use the quotient rule: A'(t) = [10($t^2$ + 1) - 10t(2t)] / (t^2 + 1)^2. Simplifying the numerator gives 10t^2 + 10 - 20t^2 = 10 - 10t^2. So, A'(t) = (10 - $10t^2$) / (t^2 + 1)^2. Now we evaluate this derivative at t = 2: A'(2) = (10 - 10($2^2$)) / (2^2 + 1)^2 = (10 - 40) / ($5^2$) = -30 / 25 = -6/5 = -1.2. The negative sign indicates the amount is decreasing at a rate of 1.2 units/hour.

Question 2

An oil slick on a lake takes the shape of a right circular cylinder. The volume of the slick is constant at 1000 m31000 \text{ m}^3. The oil is spreading, causing the radius rr to increase. Let hh be the thickness of the slick. What is the rate of change of the thickness with respect to the radius, dh/drdh/dr, at the instant when the radius is 50 m50 \text{ m}?

  1. 2/(125π) m/m2/(125\pi) \text{ m/m}
  2. −4/(5π) m/m-4/(5\pi) \text{ m/m}
  3. −2/125 m/m-2/125 \text{ m/m}
  4. −2/(125π) m/m-2/(125\pi) \text{ m/m} (correct answer)
Explanation: The volume of a cylinder is given by V=πr2hV = \pi r^2 h. We are given that the volume is constant, V=1000V=1000. So, 1000=πr2h1000 = \pi r^2 h. We need to find dh/drdh/dr. We can express hh as a function of rr and then differentiate. From the volume equation: h=1000πr2=1000πr−2h = \frac{1000}{\pi r^2} = \frac{1000}{\pi} r^{-2}. Now, differentiate hh with respect to rr using the power rule: dhdr=ddr(1000πr−2)=1000π(−2r−3)=−2000πr3\frac{dh}{dr} = \frac{d}{dr} \left(\frac{1000}{\pi} r^{-2}\right) = \frac{1000}{\pi} (-2r^{-3}) = -\frac{2000}{\pi r^3}. Finally, evaluate this derivative at the instant when r=50 mr=50 \text{ m}: dhdr=−2000π(50)3=−2000π(125000)=−2π(125)=−2125π\frac{dh}{dr} = -\frac{2000}{\pi (50)^3} = -\frac{2000}{\pi (125000)} = -\frac{2}{\pi(125)} = -\frac{2}{125\pi}. Distractor (B) results from a differentiation error where the exponent is not reduced by 1 (r−2→−2r−2r^{-2} \to -2r^{-2}). Distractor (C) comes from ignoring π\pi. Distractor (A) is a sign error.

Question 3

The concentration of a drug in a patient's bloodstream, tt hours after injection, is modeled by the function C(t)=100tt2+4C(t) = \frac{100t}{t^2 + 4} for t≥0t \ge 0. At what time is the drug concentration decreasing most rapidly?

  1. t=2t=2
  2. t=22t=2\sqrt{2}
  3. t=23t=2\sqrt{3} (correct answer)
  4. t=12t=12
Explanation: The rate of change of the drug concentration is given by the first derivative, C′(t)C'(t). To find when the concentration is decreasing most rapidly, we need to find the minimum value of C′(t)C'(t). This will occur at a critical point of C′(t)C'(t), which we can find by setting the second derivative, C′′(t)C''(t), to zero. First, find C′(t)C'(t) using the quotient rule: C′(t)=(100)(t2+4)−(100t)(2t)(t2+4)2=100t2+400−200t2(t2+4)2=400−100t2(t2+4)2C'(t) = \frac{(100)(t^2+4) - (100t)(2t)}{(t^2+4)^2} = \frac{100t^2 + 400 - 200t^2}{(t^2+4)^2} = \frac{400 - 100t^2}{(t^2+4)^2}. Concentration is decreasing when C′(t)<0C'(t) < 0, which means 400−100t2<0400 - 100t^2 < 0, so t2>4t^2 > 4, or t>2t > 2. Now, find C′′(t)C''(t) to locate the minimum of C′(t)C'(t). Let's differentiate C′(t)C'(t): C′′(t)=(−200t)(t2+4)2−(400−100t2)(2)(t2+4)(2t)(t2+4)4C''(t) = \frac{(-200t)(t^2+4)^2 - (400 - 100t^2)(2)(t^2+4)(2t)}{(t^2+4)^4} C′′(t)=(−200t)(t2+4)−4t(400−100t2)(t2+4)3C''(t) = \frac{(-200t)(t^2+4) - 4t(400 - 100t^2)}{(t^2+4)^3} C′′(t)=−200t3−800t−1600t+400t3(t2+4)3=200t3−2400t(t2+4)3=200t(t2−12)(t2+4)3C''(t) = \frac{-200t^3 - 800t - 1600t + 400t^3}{(t^2+4)^3} = \frac{200t^3 - 2400t}{(t^2+4)^3} = \frac{200t(t^2-12)}{(t^2+4)^3}. Set C′′(t)=0C''(t) = 0 to find critical points of C′(t)C'(t). For t>0t>0, this occurs when t2−12=0t^2 - 12 = 0, so t=12=23t = \sqrt{12} = 2\sqrt{3}. To confirm this is a minimum for C′(t)C'(t), we check the sign of C′′(t)C''(t). For t<23t < 2\sqrt{3}, C′′(t)<0C''(t) < 0, and for t>23t > 2\sqrt{3}, C′′(t)>0C''(t) > 0. Thus, C′(t)C'(t) has a minimum at t=23t = 2\sqrt{3}. This is the time of most rapid decrease. Note that 23≈3.46>22\sqrt{3} \approx 3.46 > 2, which is in the domain where the function is decreasing.

Question 4

A cup of coffee with temperature T(t)T(t) at time tt is in a room with a constant ambient temperature of 20∘C20^\circ\text{C}. According to Newton's Law of Cooling, its temperature changes at a rate proportional to the difference between its temperature and the ambient temperature. At t=0t=0, the coffee is 90∘C90^\circ\text{C} and is cooling at a rate of 7∘C7^\circ\text{C} per minute. What is the coffee's rate of cooling when its temperature is 55∘C55^\circ\text{C}?

  1. 3.5∘C/min3.5^\circ\text{C}/\text{min} (correct answer)
  2. 5.5∘C/min5.5^\circ\text{C}/\text{min}
  3. 3.0∘C/min3.0^\circ\text{C}/\text{min}
  4. 4.5∘C/min4.5^\circ\text{C}/\text{min}
Explanation: Newton's Law of Cooling is given by the differential equation dTdt=k(T−Ta)\frac{dT}{dt} = k(T - T_a), where TaT_a is the ambient temperature and kk is the proportionality constant. We are given Ta=20∘CT_a = 20^\circ\text{C}. So, dTdt=k(T−20)\frac{dT}{dt} = k(T - 20). We are also given initial conditions: at t=0t=0, T=90∘CT=90^\circ\text{C} and the coffee is cooling at 7∘C7^\circ\text{C} per minute, which means dTdt=−7\frac{dT}{dt} = -7. We use these initial conditions to find the constant kk: −7=k(90−20)-7 = k(90 - 20) −7=k(70)-7 = k(70) k=−7/70=−0.1k = -7/70 = -0.1. Now we have the specific model for this situation: dTdt=−0.1(T−20)\frac{dT}{dt} = -0.1(T - 20). We need to find the rate of change of temperature, dTdt\frac{dT}{dt}, when T=55∘CT = 55^\circ\text{C}. We plug this value into our model: dTdt=−0.1(55−20)=−0.1(35)=−3.5\frac{dT}{dt} = -0.1(55 - 20) = -0.1(35) = -3.5. The rate of change is −3.5∘C/min-3.5^\circ\text{C}/\text{min}. The question asks for the rate of cooling, which is the magnitude of this value, 3.5∘C/min3.5^\circ\text{C}/\text{min}. Distractor (B) comes from incorrectly calculating −0.1(55)=−5.5-0.1(55) = -5.5, ignoring the ambient temperature in the final step. Distractor (C) is a simple calculation error.

Question 5

The population of a species of fish in a lake is modeled by the logistic function P(t)=100001+99e−0.2tP(t) = \frac{10000}{1 + 99e^{-0.2t}}, where tt is the time in years. The rate of population growth is initially increasing and then later decreasing. At what time is the population growing fastest?

  1. t=5ln⁡(100)t=5\ln(100)
  2. t=5000t=5000
  3. t=5ln⁡(99)t=5\ln(99) (correct answer)
  4. t=0t=0
Explanation: The population is 'growing fastest' when its rate of growth, P′(t)P'(t), is at a maximum. This occurs at the point of inflection of the logistic curve P(t)P(t). For a logistic function of the form P(t)=L1+Ae−ktP(t) = \frac{L}{1 + Ae^{-kt}}, the maximum growth rate occurs when the population PP is at half the carrying capacity, L/2L/2. In this model, the carrying capacity is L=10000L=10000. The maximum growth rate will occur when the population is P(t)=L/2=10000/2=5000P(t) = L/2 = 10000/2 = 5000. We need to find the time tt when P(t)=5000P(t) = 5000. We set up the equation and solve for tt: 5000=100001+99e−0.2t5000 = \frac{10000}{1 + 99e^{-0.2t}} 1=21+99e−0.2t1 = \frac{2}{1 + 99e^{-0.2t}} 1+99e−0.2t=21 + 99e^{-0.2t} = 2 99e−0.2t=199e^{-0.2t} = 1 e−0.2t=199e^{-0.2t} = \frac{1}{99} Take the natural logarithm of both sides: −0.2t=ln⁡(199)=−ln⁡(99)-0.2t = \ln\left(\frac{1}{99}\right) = -\ln(99) 0.2t=ln⁡(99)0.2t = \ln(99) t=ln⁡(99)0.2=5ln⁡(99)t = \frac{\ln(99)}{0.2} = 5\ln(99). Distractor (B) confuses the population level (50005000) with the time. Distractor (D) is the initial time, where growth is starting, not at its maximum. Distractor (A) results from approximating 99 as 100.

Question 6

A cubic crystal is dissolving in a solution such that the rate of change of its volume is directly proportional to its surface area. Let ss be the side length of the cube. Which of the following equations correctly models this relationship, where kk is a positive constant?

  1. ds/dt=kds/dt = k
  2. ds/dt=−ks2ds/dt = -ks^2
  3. ds/dt=−2kds/dt = -2k (correct answer)
  4. dV/dt=−ks3dV/dt = -ks^3
Explanation: Let VV be the volume and AA be the surface area of the cube. The statement 'the rate of change of its volume is directly proportional to its surface area' can be written as dV/dt=cAdV/dt = cA for some constant cc. Since the crystal is dissolving, its volume is decreasing, so dV/dtdV/dt must be negative. As surface area AA is always positive, the constant cc must be negative. Let's define c=−kc = -k, where kk is a positive constant. So, the governing equation is dV/dt=−kAdV/dt = -kA. For a cube with side length ss, we have V=s3V = s^3 and A=6s2A = 6s^2. Substituting the expression for AA into our equation gives: dV/dt=−k(6s2)dV/dt = -k(6s^2). We can also express dV/dtdV/dt in terms of ss and its rate of change, ds/dtds/dt, using the chain rule: dV/dt=ddt(s3)=3s2dsdtdV/dt = \frac{d}{dt}(s^3) = 3s^2 \frac{ds}{dt}. Now we can equate the two expressions for dV/dtdV/dt: 3s2dsdt=−k(6s2)3s^2 \frac{ds}{dt} = -k(6s^2). Assuming the crystal exists (s>0s > 0), we can divide both sides by 3s23s^2: dsdt=−6ks23s2=−2k\frac{ds}{dt} = -\frac{6ks^2}{3s^2} = -2k. This means the side length decreases at a constant rate. Distractor (A) has the wrong sign (implying growth). Distractor (B) would mean dV/dt=−3ks4dV/dt = -3ks^4, which is not proportional to surface area. Distractor (D) implies the rate is proportional to the volume (exponential decay), not the surface area.

Question 7

The length ll of a rectangle is increasing at a rate of 3 cm/s3 \text{ cm/s} and its width ww is decreasing at a rate of 2 cm/s2 \text{ cm/s}. At the moment when l=10 cml=10 \text{ cm} and w=5 cmw=5 \text{ cm}, what is the rate of change of the rate of change of the area?

  1. −5 cm2/s2-5 \text{ cm}^2/\text{s}^2
  2. 1 cm2/s21 \text{ cm}^2/\text{s}^2
  3. 0 cm2/s20 \text{ cm}^2/\text{s}^2
  4. −12 cm2/s2-12 \text{ cm}^2/\text{s}^2 (correct answer)
Explanation: The area of the rectangle is A=lwA = lw. We are given the rates of change of the sides: dl/dt=3dl/dt = 3 and dw/dt=−2dw/dt = -2. These rates are constant. The rate of change of the area is the first derivative, dA/dtdA/dt. Using the product rule: dA/dt=l(dw/dt)+w(dl/dt)dA/dt = l(dw/dt) + w(dl/dt). dA/dt=l(−2)+w(3)=−2l+3wdA/dt = l(-2) + w(3) = -2l + 3w. The question asks for the 'rate of change of the rate of change of the area', which is the second derivative, d2A/dt2d^2A/dt^2. We differentiate the expression for dA/dtdA/dt with respect to tt: d2A/dt2=ddt(−2l+3w)=−2(dl/dt)+3(dw/dt)d^2A/dt^2 = \frac{d}{dt}(-2l + 3w) = -2(dl/dt) + 3(dw/dt). Now we substitute the given constant rates for dl/dtdl/dt and dw/dtdw/dt: d2A/dt2=−2(3)+3(−2)=−6−6=−12d^2A/dt^2 = -2(3) + 3(-2) = -6 - 6 = -12. The rate of change of the rate of change of the area is −12 cm2/s2-12 \text{ cm}^2/\text{s}^2. The specific values l=10l=10 and w=5w=5 are not needed for this calculation, but are needed to calculate the first derivative at that instant. Distractor (A) is the value of dA/dtdA/dt at that specific moment: dA/dt=−2(10)+3(5)=−20+15=−5dA/dt = -2(10) + 3(5) = -20 + 15 = -5. This is a common error of calculating the first derivative instead of the second.

Question 8

Boyle's Law states that for a fixed amount of an ideal gas at constant temperature, pressure PP and volume VV are related by PV=CPV = C for some constant CC. A piston is compressing a gas in a cylinder. At a certain instant, the volume is 60 in360 \text{ in}^3, the pressure is 30 lb/in230 \text{ lb/in}^2, and the pressure is increasing at a rate of 2 lb/in22 \text{ lb/in}^2 per second. At what rate is the volume changing at this instant?

  1. 4 in3/s4 \text{ in}^3/\text{s}
  2. −1 in3/s-1 \text{ in}^3/\text{s}
  3. −4 in3/s-4 \text{ in}^3/\text{s} (correct answer)
  4. −1800 in3/s-1800 \text{ in}^3/\text{s}
Explanation: We are given the relationship PV=CPV = C. To find the relationship between the rates of change, we differentiate both sides with respect to time tt. Using the product rule on the left side, we get: PdVdt+VdPdt=0P\frac{dV}{dt} + V\frac{dP}{dt} = 0. We are given the following values at a specific instant: V=60 in3V = 60 \text{ in}^3 P=30 lb/in2P = 30 \text{ lb/in}^2 dPdt=2 (lb/in2)/s\frac{dP}{dt} = 2 \text{ (lb/in}^2)/\text{s} (increasing, so positive). We want to find dVdt\frac{dV}{dt}. We substitute the given values into our differentiated equation: (30)dVdt+(60)(2)=0(30)\frac{dV}{dt} + (60)(2) = 0 30dVdt+120=030\frac{dV}{dt} + 120 = 0 30dVdt=−12030\frac{dV}{dt} = -120 dVdt=−12030=−4 in3/s\frac{dV}{dt} = -\frac{120}{30} = -4 \text{ in}^3/\text{s}. The negative sign indicates that the volume is decreasing, which makes sense as the pressure is increasing. Distractor (A) is a sign error. Distractor (B) comes from mixing up the roles of PP and VV in the calculation, like −(P/V)(dP/dt)-(P/V)(dP/dt). Distractor (D) is the value of the constant C=PV=(30)(60)=1800C = PV = (30)(60) = 1800, with a negative sign.

Question 9

Sand is being poured onto a conical pile whose radius rr is always twice its height hh. The volume of the pile is increasing at a rate of 10π ft3/s10\pi \text{ ft}^3/\text{s}. What is the rate of change of the circumference of the base when the height is 5 ft5 \text{ ft}?

  1. 0.2 ft/s0.2 \text{ ft/s}
  2. 0.4π ft/s0.4\pi \text{ ft/s} (correct answer)
  3. 2π ft/s2\pi \text{ ft/s}
  4. 0.1 ft/s0.1 \text{ ft/s}
Explanation: Let VV be the volume, rr the radius, hh the height, and CC the circumference. We are given r=2hr=2h, dV/dt=10πdV/dt = 10\pi, and we want to find dC/dtdC/dt when h=5h=5. The volume of a cone is V=13πr2hV = \frac{1}{3}\pi r^2 h. The circumference is C=2πrC = 2\pi r. First, express VV in terms of a single variable, hh: V=13π(2h)2h=43πh3V = \frac{1}{3}\pi (2h)^2 h = \frac{4}{3}\pi h^3. Differentiate with respect to time tt: dV/dt=4πh2(dh/dt)dV/dt = 4\pi h^2 (dh/dt). Use the given information to find dh/dtdh/dt when h=5h=5: 10π=4π(5)2(dh/dt)  ⟹  10π=100π(dh/dt)  ⟹  dh/dt=0.1 ft/s10\pi = 4\pi (5)^2 (dh/dt) \implies 10\pi = 100\pi (dh/dt) \implies dh/dt = 0.1 \text{ ft/s}. Next, express the circumference CC in terms of hh: C=2πr=2π(2h)=4πhC = 2\pi r = 2\pi(2h) = 4\pi h. Differentiate with respect to time tt: dC/dt=4π(dh/dt)dC/dt = 4\pi (dh/dt). Substitute the value of dh/dtdh/dt: dC/dt=4π(0.1)=0.4π ft/sdC/dt = 4\pi(0.1) = 0.4\pi \text{ ft/s}. Distractor (A) is the value of dr/dt=2(dh/dt)=0.2dr/dt = 2(dh/dt) = 0.2. Distractor (D) is the value of dh/dtdh/dt. Distractor (C) results from a differentiation error, d(h3)/dt=h2(dh/dt)d(h^3)/dt = h^2(dh/dt), which leads to dh/dt=0.5dh/dt=0.5 and dC/dt=2πdC/dt=2\pi.

Question 10

A light source is on the ground, and a vertical wall is 20 meters away. A person who is 2 meters tall walks from the wall towards the light source at a speed of 1.5 m/s. Let hh be the height of the person's shadow on the wall and let xx be the person's distance from the wall. Find the rate of change of the shadow's height with respect to the person's distance from the wall, dhdx\frac{dh}{dx}, at the instant the person is 8 meters from the wall.

  1. 518\frac{5}{18} m per m (correct answer)
  2. 512\frac{5}{12} m per m
  3. 103\frac{10}{3} m per m
  4. −518-\frac{5}{18} m per m
Explanation: Let the light source be at the origin and the wall be at horizontal distance 20 m. The person, with height 2 m, is at distance xx from the wall, so their distance from the light source is 20−x20-x. The shadow has height hh on the wall. By similar triangles, the ratio of height to base is the same for the shadow triangle and the person triangle: h20=220−x\frac{h}{20} = \frac{2}{20-x} Solving for hh as a function of xx: h(x)=4020−xh(x) = \frac{40}{20-x} Differentiating with respect to xx using the chain rule: dhdx=ddx[40(20−x)−1]=40(−1)(20−x)−2(−1)=40(20−x)2\frac{dh}{dx} = \frac{d}{dx}[40(20-x)^{-1}] = 40(-1)(20-x)^{-2}(-1) = \frac{40}{(20-x)^2} At x=8x = 8 meters: dhdx=40(20−8)2=40144=518\frac{dh}{dx} = \frac{40}{(20-8)^2} = \frac{40}{144} = \frac{5}{18} Note: The person's speed would be needed to find dhdt\frac{dh}{dt}, but this question asks for dhdx\frac{dh}{dx}.

Question 11

A spherical snowball is melting such that its surface area decreases at a constant rate of 2 cm2/min2 \text{ cm}^2/\text{min}. What is the rate of change of the volume of the snowball at the instant its radius is 5 cm5 \text{ cm}?

  1. −1/(20π) cm3/min-1/(20\pi) \text{ cm}^3/\text{min}
  2. −5 cm3/min-5 \text{ cm}^3/\text{min} (correct answer)
  3. −10 cm3/min-10 \text{ cm}^3/\text{min}
  4. 5 cm3/min5 \text{ cm}^3/\text{min}
Explanation: Let VV be the volume, AA be the surface area, and rr be the radius of the sphere. The formulas are V=43πr3V = \frac{4}{3}\pi r^3 and A=4πr2A = 4\pi r^2. We are given dA/dt=−2 cm2/mindA/dt = -2 \text{ cm}^2/\text{min} and we want to find dV/dtdV/dt when r=5 cmr=5 \text{ cm}. First, we relate the rates by differentiating both formulas with respect to time tt: dV/dt=4πr2(dr/dt)dV/dt = 4\pi r^2 (dr/dt) dA/dt=8πr(dr/dt)dA/dt = 8\pi r (dr/dt) Using the given information about the surface area, we can find dr/dtdr/dt when r=5r=5: −2=8π(5)(dr/dt)-2 = 8\pi (5) (dr/dt) −2=40π(dr/dt)-2 = 40\pi (dr/dt) dr/dt=−2/(40π)=−1/(20π) cm/mindr/dt = -2/(40\pi) = -1/(20\pi) \text{ cm/min}. Now, substitute this value into the equation for dV/dtdV/dt: dV/dt=4π(5)2(−1/(20π))=4π(25)(−1/(20π))=100π(−1/(20π))=−5 cm3/mindV/dt = 4\pi (5)^2 (-1/(20\pi)) = 4\pi(25) (-1/(20\pi)) = 100\pi (-1/(20\pi)) = -5 \text{ cm}^3/\text{min}. Alternatively, one can establish a direct relationship between VV and AA. r=A/(4π)r = \sqrt{A/(4\pi)}. Substituting into VV gives V=16πA3/2V = \frac{1}{6\sqrt{\pi}}A^{3/2}. A more direct method is to note that dV/dt=(dV/dr)(dr/dt)dV/dt = (dV/dr)(dr/dt) and dA/dt=(dA/dr)(dr/dt)dA/dt = (dA/dr)(dr/dt). So dr/dt=(dA/dt)/(dA/dr)dr/dt = (dA/dt)/(dA/dr). This gives dV/dt=(dV/dr)dA/dtdA/dr=(4πr2)dA/dt8πr=r2dAdtdV/dt = (dV/dr) \frac{dA/dt}{dA/dr} = (4\pi r^2) \frac{dA/dt}{8\pi r} = \frac{r}{2} \frac{dA}{dt}. Plugging in the values: dV/dt=52(−2)=−5dV/dt = \frac{5}{2}(-2) = -5.

Question 12

The cost, C(q)C(q), in dollars to produce qq units of a product is given by C(q)=0.01q3−1.5q2+100q+500C(q) = 0.01q^3 - 1.5q^2 + 100q + 500. For which of the following production levels qq is the total cost increasing while the marginal cost is decreasing?

  1. q=40q=40 (correct answer)
  2. q=50q=50
  3. q=60q=60
  4. q=75q=75
Explanation: Total cost is increasing when the first derivative, C′(q)C'(q), is positive. Marginal cost is decreasing when its derivative, C′′(q)C''(q), is negative. First, find the marginal cost function, C′(q)C'(q): C′(q)=ddq(0.01q3−1.5q2+100q+500)=0.03q2−3q+100C'(q) = \frac{d}{dq}(0.01q^3 - 1.5q^2 + 100q + 500) = 0.03q^2 - 3q + 100. To check if cost is increasing, we check if C′(q)>0C'(q) > 0. The discriminant of this quadratic is (−3)2−4(0.03)(100)=9−12=−3(-3)^2 - 4(0.03)(100) = 9 - 12 = -3. Since the leading coefficient (0.03) is positive and the discriminant is negative, the quadratic C′(q)C'(q) is always positive. Thus, the total cost is always increasing for q≥0q \ge 0. Next, find the derivative of the marginal cost, C′′(q)C''(q): C′′(q)=ddq(0.03q2−3q+100)=0.06q−3C''(q) = \frac{d}{dq}(0.03q^2 - 3q + 100) = 0.06q - 3. The marginal cost is decreasing when C′′(q)<0C''(q) < 0. 0.06q−3<00.06q - 3 < 0 0.06q<30.06q < 3 q<3/0.06q < 3/0.06 q<50q < 50. We need to find a value of qq that satisfies both conditions: q≥0q \ge 0 (cost is always increasing) and q<50q < 50. Of the given options, only q=40q=40 is in the interval [0,50)[0, 50). At q=50q=50, C′′(q)=0C''(q)=0, so marginal cost is at a minimum, not decreasing. For q=60q=60 and q=75q=75, C′′(q)>0C''(q)>0, so marginal cost is increasing.

Question 13

A 13-foot ladder is leaning against a vertical wall. The base of the ladder is being pulled away from the wall at a constant rate of 2 feet per second. At what rate is the angle θ between the ladder and the ground changing when the base of the ladder is 5 feet from the wall?

  1. Decreasing at 1/6 radians per second (correct answer)
  2. Increasing at 1/6 radians per second
  3. Decreasing at 5/12 radians per second
  4. Increasing at 1/12 radians per second
Explanation: Let x be the distance from the base of the ladder to the wall and y be the height of the ladder on the wall. We are given dx/dt = 2. We want to find dθ/dt when x = 5. We can relate x and θ using cos(θ) = x/13. Differentiating with respect to time t, we get -sin(θ)(dθ/dt) = (1/13)(dx/dt). When x = 5, the ladder forms a right triangle with hypotenuse 13 and base 5. The height y is sqrt($13^2$ - $5^2$) = sqrt(169 - 25) = sqrt(144) = 12. At this moment, sin(θ) = y/13 = 12/13. Now substitute the known values into the differentiated equation: -(12/13)(dθ/dt) = (1/13)(2). Solving for dθ/dt, we get dθ/dt = - (2/13) * (13/12) = -2/12 = -1/6. The negative sign indicates the angle is decreasing.

Question 14

The total cost to operate a manufacturing plant is C(x) = 0.5x^2 + 15x + 5000, where x is the number of units produced. The production level at t hours from the start of the day is x(t) = 20t. At what rate is the cost changing with respect to time 3 hours after the start of the day?

  1. $60 per hour
  2. $1500 per hour (correct answer)
  3. $1700 per hour
  4. $8300 per hour
Explanation: We want to find dC/dt at t=3. We use the chain rule: dC/dt = (dC/dx) * (dx/dt). First, find the derivatives: dC/dx = x + 15 and dx/dt = 20. Therefore, dC/dt = (x + 15) * 20. At t=3 hours, the production level is x(3) = 20 * 3 = 60 units. Now substitute x=60 into the expression for dC/dt: dC/dt = (60 + 15) * 20 = 75 * 20 = 1500. The cost is changing at a rate of $1500 per hour.

Question 15

The population P of a city is increasing at a rate of 2000 people per year, while its radius r is expanding at 0.1 kilometers per year. The population density is given by D = P / ($πr^2$). Find the rate at which the population density is changing when the radius is 10 km and the population is 400,000 people.

  1. Decreasing by about 2.5 people/km²/year
  2. Increasing by about 6.4 people/km²/year
  3. Decreasing by about 19.1 people/km²/year (correct answer)
  4. Increasing by about 25.5 people/km²/year
Explanation: We need to find dD/dt. Using the quotient rule on D = P/A where A=πr^2: dD/dt = ( (dP/dt)A - P(dA/dt) ) / A^2. We also need dA/dt = 2πr(dr/dt). We are given dP/dt = 2000, dr/dt = 0.1, r = 10, and P = 400,000. First, calculate A and dA/dt at this instant. A = π(10)^2 = 100π. dA/dt = 2π(10)(0.1) = 2π. Now substitute into the quotient rule: dD/dt = ( (2000)(100π) - (400,000)(2π) ) / (100π)^2. dD/dt = ( 200,000π - 800,000π ) / ($10000π^2$) = -600,000π / ($10000π^2$) = -60/π. Using π ≈ 3.14, dD/dt ≈ -60 / 3.14 ≈ -19.1. The density is decreasing.

Question 16

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm²?

  1. Decreasing at 1.6 cm/min (correct answer)
  2. Increasing at 1.6 cm/min
  3. Decreasing at 0.8 cm/min
  4. Increasing at 0.8 cm/min
Explanation: The area of a triangle is A = (1/2)bh. We are given dh/dt = 1, dA/dt = 2. We want to find db/dt when h = 10 and A = 100. First, find the base b at this instant: 100 = (1/2)b(10), so 100 = 5b, which means b = 20. Now differentiate the area formula with respect to time t using the product rule: dA/dt = (1/2)[(db/dt)h + b(dh/dt)]. Substitute the known values: 2 = (1/2)[(db/dt)(10) + (20)(1)]. 4 = 10(db/dt) + 20. -16 = 10(db/dt). db/dt = -1.6. The base is decreasing at a rate of 1.6 cm/min.

Question 17

The value V(t) of a stock is increasing, but the rate of increase is slowing down at time t_0. Which of the following statements about the derivatives of V(t) at t_0 must be true?

  1. V'($t_0$) > 0 and V''($t_0$) > 0
  2. V'($t_0$) > 0 and V''($t_0$) < 0 (correct answer)
  3. V'($t_0$) < 0 and V''($t_0$) > 0
  4. V'($t_0$) < 0 and V''($t_0$) < 0
Explanation: The rate of change of the value is given by the first derivative, V'(t). The phrase 'value is increasing' means that V'($t_0$) > 0. The rate of change of the rate of increase is given by the second derivative, V''(t). The phrase 'the rate of increase is slowing down' means that the slope of the tangent line is decreasing, which corresponds to the function being concave down. Therefore, V''($t_0$) < 0.

Question 18

A company's profit from selling x units of a product is P(x) = 2x^2 - 40x + 1000. The sales are increasing over time according to x(t) = 5t, where t is time in weeks. At what rate is the profit changing with respect to time when the company is selling 50 units?

  1. $4100 per week
  2. $160 per week
  3. $1000 per week
  4. $800 per week (correct answer)
Explanation: This is a chain rule problem involving related rates. When you see profit as a function of units sold, and units sold as a function of time, you need to find how profit changes with respect to time using dPdt=dPdx⋅dxdt\frac{dP}{dt} = \frac{dP}{dx} \cdot \frac{dx}{dt}. First, find dPdx\frac{dP}{dx} by differentiating the profit function: P(x)=2x2−40x+1000P(x) = 2x^2 - 40x + 1000, so dPdx=4x−40\frac{dP}{dx} = 4x - 40. Next, find dxdt\frac{dx}{dt} from the sales function: x(t)=5tx(t) = 5t, so dxdt=5\frac{dx}{dt} = 5 units per week. When the company is selling 50 units, substitute x=50x = 50: dPdx=4(50)−40=200−40=160\frac{dP}{dx} = 4(50) - 40 = 200 - 40 = 160 dollars per unit. Therefore: dPdt=160⋅5=800\frac{dP}{dt} = 160 \cdot 5 = 800 dollars per week. Answer D ($800 per week) is correct. Answer A ($4100 per week) likely comes from incorrectly calculating $P(50)=2(50)2−40(50)+1000=3000P(50) = 2(50)^2 - 40(50) + 1000 = 3000 andsomehowusingthisvalueinsteadofthederivative.AnswerB( and somehow using this value instead of the derivative. Answer B (160 per week) is the common error of finding dPdx\frac{dP}{dx} but forgetting to multiply by dxdt\frac{dx}{dt}. Answer C ($1000 per week) might result from using only the constant term or making an arithmetic error. Remember: for related rates problems, identify all the given rates and use the chain rule to connect them. Always check that your final answer has the correct units for what the question asks.

Question 19

The value of a particular investment is modeled by the function V(t) = 1000(1.5)^t, where V is the value in dollars and t is the time in years. At what rate is the investment's value growing at the instant its value is $2250?

  1. $2250 * ln(1.5) per year (correct answer)
  2. $1500 * ln(1.5) per year
  3. $1000 * ln(1.5) per year
  4. $2250 per year
Explanation: The rate of growth is V'(t). First, find the derivative of V(t). Recall that the derivative of a^x is a^x * ln(a). So, V'(t) = 1000 * (1.5)^t * ln(1.5). Notice that V(t) = 1000(1.5)^t, so we can write V'(t) = V(t) * ln(1.5). The question asks for the rate when the value V(t) is $2250. We can substitute this directly into our simplified derivative expression: V'(t) = 2250 * ln(1.5). There is no need to solve for t explicitly.

Question 20

For a group of 100 people, the number of people N(t) who have heard a rumor t days after it started is modeled by N(t) = 100(1 - e^(-0.2t)). How fast is the number of people who have not heard the rumor changing 5 days after the rumor started?

  1. Increasing by approximately 2.7 people/day
  2. Increasing by approximately 7.4 people/day
  3. Decreasing by approximately 2.7 people/day
  4. Decreasing by approximately 7.4 people/day (correct answer)
Explanation: When you encounter a problem asking about the rate of change of people who have not heard something, you need to set up a function for that quantity and find its derivative. Since N(t)=100(1−e−0.2t)N(t) = 100(1 - e^{-0.2t}) represents people who have heard the rumor, the number who have not heard it is 100−N(t)=100−100(1−e−0.2t)=100e−0.2t100 - N(t) = 100 - 100(1 - e^{-0.2t}) = 100e^{-0.2t}. To find how fast this is changing, take the derivative: ddt[100e−0.2t]=100⋅(−0.2)⋅e−0.2t=−20e−0.2t\frac{d}{dt}[100e^{-0.2t}] = 100 \cdot (-0.2) \cdot e^{-0.2t} = -20e^{-0.2t}. At t=5t = 5 days: −20e−0.2(5)=−20e−1=−20(0.368)≈−7.4-20e^{-0.2(5)} = -20e^{-1} = -20(0.368) \approx -7.4 people/day. The negative sign means decreasing, so the number of people who haven't heard the rumor is decreasing by approximately 7.4 people/day. Answer A incorrectly uses 2.7 people/day and claims it's increasing. Answer B gets the magnitude wrong (7.4) but incorrectly says increasing. Answer C gets the direction right (decreasing) but uses the wrong magnitude (2.7). These errors likely come from calculation mistakes or forgetting to subtract from 100 to get the "not heard" function. Answer D correctly identifies both the magnitude (7.4) and direction (decreasing). Study tip: Always read carefully whether the question asks about the original quantity or its complement. Set up the correct function first, then differentiate. The sign of the derivative tells you whether it's increasing (positive) or decreasing (negative).