Non-Motion Rates of Change - Calculus 1

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?
   Answer: Decreasing at 1.2 units/hour 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.
2. A spherical snowball is melting such that its surface area decreases at a constant rate of $2 \text{ cm}^2/\text{min}$. What is the rate of change of the volume of the snowball at the instant its radius is $5 \text{ cm}$?
   Answer: $-5 \text{ cm}^3/\text{min}$ Let $V$ be the volume, $A$ be the surface area, and $r$ be the radius of the sphere. The formulas are $V = \frac{4}{3}\pi r^3$ and $A = 4\pi r^2$. We are given $dA/dt = -2 \text{ cm}^2/\text{min}$ and we want to find $dV/dt$ when $r=5 \text{ cm}$. First, we relate the rates by differentiating both formulas with respect to time $t$: $dV/dt = 4\pi r^2 (dr/dt)$ $dA/dt = 8\pi r (dr/dt)$ Using the given information about the surface area, we can find $dr/dt$ when $r=5$: $-2 = 8\pi (5) (dr/dt)$ $-2 = 40\pi (dr/dt)$ $dr/dt = -2/(40\pi) = -1/(20\pi) \text{ cm/min}$. Now, substitute this value into the equation for $dV/dt$: $dV/dt = 4\pi (5)^2 (-1/(20\pi)) = 4\pi(25) (-1/(20\pi)) = 100\pi (-1/(20\pi)) = -5 \text{ cm}^3/\text{min}$. Alternatively, one can establish a direct relationship between $V$ and $A$. $r = \sqrt{A/(4\pi)}$. Substituting into $V$ gives $V = \frac{1}{6\sqrt{\pi}}A^{3/2}$. A more direct method is to note that $dV/dt = (dV/dr)(dr/dt)$ and $dA/dt = (dA/dr)(dr/dt)$. So $dr/dt = (dA/dt)/(dA/dr)$. This gives $dV/dt = (dV/dr) \frac{dA/dt}{dA/dr} = (4\pi r^2) \frac{dA/dt}{8\pi r} = \frac{r}{2} \frac{dA}{dt}$. Plugging in the values: $dV/dt = \frac{5}{2}(-2) = -5$.
3. 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?
   Answer: $2250 * ln(1.5) per year 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.
4. A spherical balloon is being inflated such that its volume is increasing at a rate of 100 cm^3/s. At what rate is the surface area of the balloon increasing when its radius is 5 cm? (The volume of a sphere is V = (4/3)πr^3 and the surface area is A = 4πr^2.)
   Answer: 40 $cm^2$/s We are given dV/dt = 100. We want to find dA/dt when r = 5. First, relate dV/dt to dr/dt. V = (4/3)πr^3, so dV/dt = 4πr^2(dr/dt). At r=5, 100 = 4π(5^2)(dr/dt) = 100π(dr/dt), so dr/dt = 1/π cm/s. Next, relate dA/dt to dr/dt. A = 4πr^2, so dA/dt = 8πr(dr/dt). Now substitute the values for r and dr/dt: dA/dt = 8π(5)(1/π) = 40 $cm^2$/s.
5. The radius of a circular oil slick is expanding at a rate of 2 meters/minute. How fast is the area of the slick growing when the circumference is 20π meters?
   Answer: 40π m²/min The area of a circle is A = πr^2 and the circumference is C = 2πr. We are given dr/dt = 2. We need to find dA/dt. First, find the radius r when the circumference is 20π. 20π = 2πr, so r = 10 meters. The rate of change of the area is dA/dt = 2πr(dr/dt). Substitute the known values: dA/dt = 2π(10)(2) = 40π m²/min.
6. The population of a species of fish in a lake is modeled by the logistic function $P(t) = \frac{10000}{1 + 99e^{-0.2t}}$, where $t$ 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?
   Answer: $t=5\ln(99)$ The population is 'growing fastest' when its rate of growth, $P'(t)$, is at a maximum. This occurs at the point of inflection of the logistic curve $P(t)$. For a logistic function of the form $P(t) = \frac{L}{1 + Ae^{-kt}}$, the maximum growth rate occurs when the population $P$ is at half the carrying capacity, $L/2$. In this model, the carrying capacity is $L=10000$. The maximum growth rate will occur when the population is $P(t) = L/2 = 10000/2 = 5000$. We need to find the time $t$ when $P(t) = 5000$. We set up the equation and solve for $t$: $5000 = \frac{10000}{1 + 99e^{-0.2t}}$ $1 = \frac{2}{1 + 99e^{-0.2t}}$ $1 + 99e^{-0.2t} = 2$ $99e^{-0.2t} = 1$ $e^{-0.2t} = \frac{1}{99}$ Take the natural logarithm of both sides: $-0.2t = \ln\left(\frac{1}{99}\right) = -\ln(99)$ $0.2t = \ln(99)$ $t = \frac{\ln(99)}{0.2} = 5\ln(99)$. Distractor (B) confuses the population level ($5000$) 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.
7. A particle moves along the curve y = x^3 - 2x. If the x-coordinate is increasing at a rate of 4 units per second, what is the rate of change of the slope of the curve when the particle is at the point (2, 4)?
   Answer: 48 units/s This is a multi-step problem that tests careful reading. We need the rate of change of the slope. Let m be the slope. The slope of the curve is the first derivative: m(x) = y' = dy/dx = 3x^2 - 2. We are looking for dm/dt. We can find this using the chain rule: dm/dt = (dm/dx) * (dx/dt). First, find dm/dx, which is the derivative of the slope function m(x). This is the second derivative of the original function y: dm/dx = y'' = 6x. We are given dx/dt = 4. At the point (2, 4), x=2. Substitute these values into the chain rule expression: dm/dt = (6 * 2) * (4) = 12 * 4 = 48. The slope is changing at a rate of 48 units per second.
8. A cup of coffee with temperature $T(t)$ at time $t$ is in a room with a constant ambient temperature of $20^\circ\text{C}$. 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 $t=0$, the coffee is $90^\circ\text{C}$ and is cooling at a rate of $7^\circ\text{C}$ per minute. What is the coffee's rate of cooling when its temperature is $55^\circ\text{C}$?
   Answer: $3.5^\circ\text{C}/\text{min}$ Newton's Law of Cooling is given by the differential equation $\frac{dT}{dt} = k(T - T_a)$, where $T_a$ is the ambient temperature and $k$ is the proportionality constant. We are given $T_a = 20^\circ\text{C}$. So, $\frac{dT}{dt} = k(T - 20)$. We are also given initial conditions: at $t=0$, $T=90^\circ\text{C}$ and the coffee is cooling at $7^\circ\text{C}$ per minute, which means $\frac{dT}{dt} = -7$. We use these initial conditions to find the constant $k$: $-7 = k(90 - 20)$ $-7 = k(70)$ $k = -7/70 = -0.1$. Now we have the specific model for this situation: $\frac{dT}{dt} = -0.1(T - 20)$. We need to find the rate of change of temperature, $\frac{dT}{dt}$, when $T = 55^\circ\text{C}$. We plug this value into our model: $\frac{dT}{dt} = -0.1(55 - 20) = -0.1(35) = -3.5$. The rate of change is $-3.5^\circ\text{C}/\text{min}$. The question asks for the rate of cooling, which is the magnitude of this value, $3.5^\circ\text{C}/\text{min}$. Distractor (B) comes from incorrectly calculating $-0.1(55) = -5.5$, ignoring the ambient temperature in the final step. Distractor (C) is a simple calculation error.
9. 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?
   Answer: $1500 per hour 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.
10. The concentration of a drug in a patient's bloodstream, $t$ hours after injection, is modeled by the function $C(t) = \frac{100t}{t^2 + 4}$ for $t \ge 0$. At what time is the drug concentration decreasing most rapidly?
    Answer: $t=2\sqrt{3}$ The rate of change of the drug concentration is given by the first derivative, $C'(t)$. To find when the concentration is decreasing most rapidly, we need to find the minimum value of $C'(t)$. This will occur at a critical point of $C'(t)$, which we can find by setting the second derivative, $C''(t)$, to zero. First, find $C'(t)$ using the quotient rule: $C'(t) = \frac{(100)(t^2+4) - (100t)(2t)}{(t^2+4)^2} = \frac{100t^2 + 400 - 200t^2}{(t^2+4)^2} = \frac{400 - 100t^2}{(t^2+4)^2}$. Concentration is decreasing when $C'(t) < 0$, which means $400 - 100t^2 < 0$, so $t^2 > 4$, or $t > 2$. Now, find $C''(t)$ to locate the minimum of $C'(t)$. Let's differentiate $C'(t)$: $C''(t) = \frac{(-200t)(t^2+4)^2 - (400 - 100t^2)(2)(t^2+4)(2t)}{(t^2+4)^4}$ $C''(t) = \frac{(-200t)(t^2+4) - 4t(400 - 100t^2)}{(t^2+4)^3}$ $C''(t) = \frac{-200t^3 - 800t - 1600t + 400t^3}{(t^2+4)^3} = \frac{200t^3 - 2400t}{(t^2+4)^3} = \frac{200t(t^2-12)}{(t^2+4)^3}$. Set $C''(t) = 0$ to find critical points of $C'(t)$. For $t>0$, this occurs when $t^2 - 12 = 0$, so $t = \sqrt{12} = 2\sqrt{3}$. To confirm this is a minimum for $C'(t)$, we check the sign of $C''(t)$. For $t < 2\sqrt{3}$, $C''(t) < 0$, and for $t > 2\sqrt{3}$, $C''(t) > 0$. Thus, $C'(t)$ has a minimum at $t = 2\sqrt{3}$. This is the time of most rapid decrease. Note that $2\sqrt{3} \approx 3.46 > 2$, which is in the domain where the function is decreasing.
11. Boyle's Law states that for a fixed amount of an ideal gas at constant temperature, pressure $P$ and volume $V$ are related by $PV = C$ for some constant $C$. A piston is compressing a gas in a cylinder. At a certain instant, the volume is $60 \text{ in}^3$, the pressure is $30 \text{ lb/in}^2$, and the pressure is increasing at a rate of $2 \text{ lb/in}^2$ per second. At what rate is the volume changing at this instant?
    Answer: $-4 \text{ in}^3/\text{s}$ We are given the relationship $PV = C$. To find the relationship between the rates of change, we differentiate both sides with respect to time $t$. Using the product rule on the left side, we get: $P\frac{dV}{dt} + V\frac{dP}{dt} = 0$. We are given the following values at a specific instant: $V = 60 \text{ in}^3$ $P = 30 \text{ lb/in}^2$ $\frac{dP}{dt} = 2 \text{ (lb/in}^2)/\text{s}$ (increasing, so positive). We want to find $\frac{dV}{dt}$. We substitute the given values into our differentiated equation: $(30)\frac{dV}{dt} + (60)(2) = 0$ $30\frac{dV}{dt} + 120 = 0$ $30\frac{dV}{dt} = -120$ $\frac{dV}{dt} = -\frac{120}{30} = -4 \text{ in}^3/\text{s}$. 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 $P$ and $V$ in the calculation, like $-(P/V)(dP/dt)$. Distractor (D) is the value of the constant $C = PV = (30)(60) = 1800$, with a negative sign.
12. 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.
    Answer: Decreasing by about 19.1 people/km²/year 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.
13. According to Boyle's Law, the pressure P and volume V of a gas at constant temperature are related by PV = k, where k is a constant. If the pressure of a gas is increasing at a rate of 10 kPa/min, what is the rate of change of the volume when the pressure is 100 kPa and the volume is 50 L? Assume k is constant.
    Answer: Decreasing at 5 L/min We are given PV = k, dP/dt = 10 kPa/min, P = 100 kPa, and V = 50 L. We want to find dV/dt. We differentiate the equation PV = k with respect to time t using the product rule: (dP/dt)V + P(dV/dt) = 0. Now, substitute the given values: (10)(50) + (100)(dV/dt) = 0. This simplifies to 500 + 100(dV/dt) = 0. Solving for dV/dt, we get 100(dV/dt) = -500, so dV/dt = -5 L/min. The negative sign indicates that the volume is decreasing.
14. The temperature T of a cooling object is given by T(t) = 20 + 70e^(-0.1t), where T is in degrees Celsius and t is in minutes. At what rate is the temperature decreasing at the instant the object's temperature is 55°C?
    Answer: 3.5 °C/min The rate of change of temperature is T'(t) = 70 * e^(-0.1t) * (-0.1) = -7e^(-0.1t). The question asks for the rate at a specific temperature, not a specific time. First, find the time t when T(t) = 55. 55 = 20 + 70e^(-0.1t), which leads to 35 = 70e^(-0.1t), so 0.5 = e^(-0.1t). We don't need to solve for t. Notice that T(t) - 20 = 70e^(-0.1t). And T'(t) = -0.1 * (70e^(-0.1t)) = -0.1 * (T(t) - 20). When T(t) = 55, the rate is T'(t) = -0.1 * (55 - 20) = -0.1 * 35 = -3.5. The temperature is decreasing at a rate of 3.5 °C/min.
15. 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?
    Answer: V'(t_0) > 0 and V''(t_0) < 0 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.
16. 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?
    Answer: Decreasing by approximately 7.4 people/day 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 $N(t) = 100(1 - e^{-0.2t})$ represents people who have heard the rumor, the number who have not heard it is $100 - N(t) = 100 - 100(1 - e^{-0.2t}) = 100e^{-0.2t}$. To find how fast this is changing, take the derivative: $\frac{d}{dt}[100e^{-0.2t}] = 100 \cdot (-0.2) \cdot e^{-0.2t} = -20e^{-0.2t}$. At $t = 5$ days: $-20e^{-0.2(5)} = -20e^{-1} = -20(0.368) \approx -7.4$ 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).
17. 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?
    Answer: Decreasing at 1.6 cm/min 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.
18. A person 6 feet tall walks away from a streetlight that is 15 feet high at a rate of 3 feet per second. How fast is the length of the person's shadow increasing?
    Answer: 2 ft/s Let x be the distance of the person from the streetlight and s be the length of the shadow. We are given dx/dt = 3. We want to find ds/dt. By similar triangles, we have the ratio of heights equal to the ratio of bases: (height of person) / (height of light) = (length of shadow) / (distance from light to tip of shadow). This gives 6 / 15 = s / (x + s). Simplifying, 2 / 5 = s / (x + s). Cross-multiply: 2(x + s) = 5s, which gives 2x + 2s = 5s, so 2x = 3s. Now, differentiate with respect to time t: 2(dx/dt) = 3(ds/dt). Substitute the given dx/dt = 3: 2(3) = 3(ds/dt), so 6 = 3(ds/dt). Solving for ds/dt gives ds/dt = 2 ft/s.
19. The length $l$ of a rectangle is increasing at a rate of $3 \text{ cm/s}$ and its width $w$ is decreasing at a rate of $2 \text{ cm/s}$. At the moment when $l=10 \text{ cm}$ and $w=5 \text{ cm}$, what is the rate of change of the rate of change of the area?
    Answer: $-12 \text{ cm}^2/\text{s}^2$ The area of the rectangle is $A = lw$. We are given the rates of change of the sides: $dl/dt = 3$ and $dw/dt = -2$. These rates are constant. The rate of change of the area is the first derivative, $dA/dt$. Using the product rule: $dA/dt = l(dw/dt) + w(dl/dt)$. $dA/dt = l(-2) + w(3) = -2l + 3w$. The question asks for the 'rate of change of the rate of change of the area', which is the second derivative, $d^2A/dt^2$. We differentiate the expression for $dA/dt$ with respect to $t$: $d^2A/dt^2 = \frac{d}{dt}(-2l + 3w) = -2(dl/dt) + 3(dw/dt)$. Now we substitute the given constant rates for $dl/dt$ and $dw/dt$: $d^2A/dt^2 = -2(3) + 3(-2) = -6 - 6 = -12$. The rate of change of the rate of change of the area is $-12 \text{ cm}^2/\text{s}^2$. The specific values $l=10$ and $w=5$ are not needed for this calculation, but are needed to calculate the first derivative at that instant. Distractor (A) is the value of $dA/dt$ at that specific moment: $dA/dt = -2(10) + 3(5) = -20 + 15 = -5$. This is a common error of calculating the first derivative instead of the second.
20. The population of a species of fish in a lake is modeled by the logistic function P(t) = 2500 / (1 + 4e^(-0.2t)), where t is the time in years. At what rate is the fish population growing at the instant the population reaches 1250 fish?
    Answer: 125 fish per year The rate of growth is P'(t). The population grows fastest when it is at half the carrying capacity, which is L/2 = 2500/2 = 1250. So the question asks for the maximum growth rate. We can find P'(t) using the chain and quotient rules: P'(t) = -2500 * (1 + 4e^(-0.2t))^(-2) * (4e^(-0.2t) * -0.2) = (2000e^(-0.2t)) / (1 + 4e^(-0.2t))^2. To find the time when P(t) = 1250, we solve 1250 = 2500 / (1 + 4e^(-0.2t)), which gives 1 + 4e^(-0.2t) = 2, so 4e^(-0.2t) = 1, and e^(-0.2t) = 1/4. We can substitute 4e^(-0.2t) = 1 into the expression for P'(t). P'(t) = (500 * 4e^(-0.2t)) / (1 + 4e^(-0.2t))^2 = (500 * 1) / (1 + 1)^2 = 500 / 4 = 125. So the rate is 125 fish per year. An alternative method recognizes that for logistic growth P'(t) = kP(1 - P/L). Here k=0.2 and L=2500. At P=1250, P'(t) = 0.2 * 1250 * (1 - 1250/2500) = 250 * (1 - 0.5) = 125.

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