Ｃａｌｃｕｌｕｓ　１　ｌｅｃｔｕｒｅ　２．８　Rｅｌａｔｅｄ　Rａｔｅｓ

Ｉｎｔｒｏｄｕｃｔｉｏｎ　ｔｏ　ｔｈｅ　ｃｏｎｃｅｐｔ　ｏｆ　ｒｅｌａｔｅｄ　ｒａｔｅｓ

• [0:00]. Definition and examples.
  • Definition of related rates: how a formula relates to its change over time.
  • Examples:
    • Cost as a function of time.
    • Profit as a function of time.
    • Volume in expansion or contraction as a function of time.
  • Objective of related rates: relating a formula to time.

Ｉｎｔｒｏｄｕｃｔｏｒｙ　ｅｘａｍｐｌｅ：　ｃｏｎｅ　ｗｉｔｈ　ｗａｔｅｒ　ｌｅａｋ

• [📷image]
• [1:05]. Description and variables.
  • Description of the problem:
    • A cone with a water leak at the bottom.
    • The water level in the cone changes over time.
  • Variables to consider:
    • Height of the water ($𝒉$).
    • Radius of the water surface ($𝒓$).
• [2:45]. Objective: Find the rate of change of the volume of water with respect to time ($\dfrac{d𝓥}{d𝓉}$).
• [3:34]. Considerations:
  • The height ($𝒉$) and radius ($𝒓$) of the water change as the water drains.
  • The volume of water ($𝓥$) depends on $𝒉$ and $𝒓$, and thus also changes over time.
• [4:13]. Steps to Solve the Problem:
  • Step 1: Find a formula that relates the involved variables.
    • In this case, the formula for the volume of a cone: $𝓥=\dfrac{\pi}{3}𝒓^{2}𝒉$.
  • [5:35] Step 2: Differentiate the formula implicitly with respect to time ($𝓉$).
    • Justification: Volume ($𝓥$), radius ($𝒓$), and height ($𝒉$) are functions of time.
    • Use the chain rule to differentiate each variable with respect to $𝓉$, resulting in
      terms like $\dfrac{d𝓥}{d𝓉}$, $\dfrac{d𝒓}{d𝓉}$, and $\dfrac{d𝒉}{d𝓉}$.
    • Take advantage of Implicit Differentiation
      • Why Use Implicit Differentiation in Related Rates.
        • Many related rates problems present relationships where variables cannot be easily isolated.
        • Implicit differentiation allows us to handle these complexities by:
          • Treating dependent variables (like $𝐲$) as functions of independent variables (like $𝒙$ or $𝓉$).
          • Applying the chain rule to account for the interdependence of variables.
            • Example Insight: In the cone with a water leak example, the volume $𝓥$ depends on both the radius $𝒓$ and the height $𝒉$, which are themselves changing over time. Implicit differentiation facilitates finding how $𝓥$ changes with time without needing to isolate $𝒓$ or $𝒉$.
• [8:27]. Interpretation of the Derivative Terms:
  • $\dfrac{d}{d𝓉}[𝓥]=\dfrac{d}{d𝓉}\!\Big[\dfrac{\pi}{3}𝒓^{2}𝒉\Big]$
    • $\dfrac{d𝓥}{d𝓉}$: Rate of change of volume with respect to time; $𝓥$ is a function of $𝓉$; describes how the volume is changing with respect to time.
    • $\dfrac{d𝒓}{d𝓉}$: Rate of change of radius with respect to time; $𝒓$ is a function of $𝓉$; describes how the radius is changing with respect to time.
    • $\dfrac{d𝒉}{d𝓉}$: Rate of change of height with respect to time; $𝒉$ is a function of $𝓉$; describes how the height is changing with respect to time.
• [10:00]. Use of the product rule when differentiating the volume formula.
  • $\dfrac{d}{d𝓉}[𝓥]=\dfrac{d}{d𝓉}\!\Big[\dfrac{\pi}{3}𝒓^{2}𝒉\Big]$
    • Differentiate implicitly, as $r$ and $h$ depend on $t$
    • $\dfrac{d𝓥}{d𝓉}=\dfrac{\pi}{3}\cdot\dfrac{d}{d𝓉}[𝒓^{2}𝒉]$ # Constants can be factored out
    • $\dfrac{d𝓥}{d𝓉}=\dfrac{\pi}{3}\cdot\Big[\,\dfrac{d}{d𝓉}(𝒓^{2})\cdot 𝒉+𝒓^{2}\cdot\dfrac{d𝒉}{d𝓉}\,\Big]$ # Product rule: $\dfrac{d}{d𝓉}[uv]=\dfrac{d}{d𝓉}[u]\cdot v+u\cdot\dfrac{d}{d𝓉}[v]$
      • $\dfrac{d}{d𝓉}(𝒓^{2})=2𝒓\cdot\dfrac{d𝒓}{d𝓉}$ # Chain rule, since $𝒓=𝒓(𝓉)$
    • $\dfrac{d𝓥}{d𝓉}=\dfrac{\pi}{3}\cdot\Big[\,2𝒓\cdot\dfrac{d𝒓}{d𝓉}\cdot 𝒉+𝒓^{2}\cdot\dfrac{d𝒉}{d𝓉}\,\Big]$ # Substitute the result above
    • $\dfrac{d𝓥}{d𝓉}=\dfrac{\pi}{3}\Big[\,2𝒓𝒉\cdot\dfrac{d𝒓}{d𝓉}+𝒓^{2}\cdot\dfrac{d𝒉}{d𝓉}\,\Big]$ # This gives the rate of change of volume in terms of rates of $r$ and $h$
• [13:33]. Information Needed to Solve the Problem:
  • Value of the radius ($𝒓$) at a given moment.
  • Value of the height ($𝒉$) at a given moment.
  • Rate of change of the radius ($\dfrac{d𝒓}{d𝓉}$).
  • Rate of change of the height ($\dfrac{d𝒉}{d𝓉}$).
• [14:54]. Transition to Simpler Examples to Illustrate the Concept of Related Rates. – [📷image]
  • [15:26]. 🧩 Example –: Find $\dfrac{d𝑦}{d𝓉}$ when $𝑦=𝒙^{3}$ at $𝓉=1$
    • Assume that $𝑦$ and $𝒙$ are functions of time ($𝓉$).
    • Differentiate the equation implicitly with respect to $𝓉$, resulting in $\dfrac{d𝑦}{d𝓉}=3𝒙^{2}\cdot\dfrac{d𝒙}{d𝓉}$
      • [17:34]. Additional Information Needed!:
        • Value of $𝒙$ at a given moment (e.g., $𝓉=1$) and value of $\dfrac{d𝒙}{d𝓉}$ at the same moment (e.g., $𝓉=1$).
          • Find this out: If $𝒙=2$ and $\dfrac{d𝒙}{d𝓉}=4$ at $𝓉=1$.
            • What is happening at $𝓉=1$?
              • $\dfrac{d𝒙}{d𝓉}=4$ "The variable $𝒙$ is changing with respect to time $𝓉$ at a rate of 4 units per unit of time."
              • The actual position is $𝒙=2$
        • $\dfrac{d𝑦}{d𝓉}=3\cdot(2)^{2}\cdot 4=3\cdot 4\cdot 4=48$
          • So, at $𝓉=1$, $\dfrac{d𝑦}{d𝓉}=48$

🧩Ｅｘａｍｐｌｅ：　ｏｉｌ　ｓｐｉｌｌ

• [📷image]
• [19:46]. Description ad objective of the problem:
  • An oil spill is expanding in a circular shape.
  • The radius of the spill is increasing at a constant rate of 3 feet (0.91 m) per second.
  • Objective: Determine how quickly the area of the spill is increasing when the radius is 30 feet (9.14 m).
• [23:19]. Steps to Solve the Problem:
  • Step 1: Assign letters to the variables:
    • $𝓉$ for time.
    • $𝓐$ for area.
    • $𝒓$ for radius.
  • Step 2: Identify the formula that relates the variables:
    • In this case, the area of a circle: $𝓐=\pi 𝒓^{2}$.
  • Step 3: Identify the given rates in the problem:
    • $\dfrac{d𝒓}{d𝓉}=3$ feet per second (rate of change of the radius).
    • $\dfrac{d𝓐}{d𝓉}$ rate of change of area
  • [27:12]. Step 6: Differentiate the formula implicitly with respect to time ($𝓉$):
    • $\dfrac{d𝓐}{d𝓉}=2\pi 𝒓\cdot\dfrac{d𝒓}{d𝓉}$
  • Step 5: Substitute the known values into the differentiated equation and solve for
    the unknown rate of change ($\dfrac{d𝓐}{d𝓉}$):
    • $\dfrac{d𝓐}{d𝓉}=2\pi\cdot (30\ \text{feet})\cdot\big(3\ \text{feet/second}\big)=180\pi\ \text{square feet per second}.$

🧩Ｅｘａｍｐｌｅ：　ｃａｍｅｒａ　ｆｏｌｌｏｗｉｎｇ　ａ　ｒｏｃｋｅｔ

• [📷image-1] – [📷image-2]
• [30:18]. Description ad objective of the problem.
  • Description of the problem:
    • A rocket is launched from the ground vertically.
    • A camera on the ground must follow the rocket by adjusting its elevation angle.
      It is a distance of 3000 feet (0.91 km) from the rocket.
    • The rocket ascends at a speed of 600 feet (0.18 km) per second when the rocket is at 4000 feet (1.22 km).
  • Objective: Determine how quickly the elevation angle of the camera must change to keep the rocket centered in the frame when the rocket is 4000 feet high.
• [31:43]. Considerations:
  • The height of the rocket ($𝒉$) changes over time.
  • The elevation angle of the camera ($\theta$) must change to follow the rocket.
• [32:29]. Steps to Solve the Problem:
  • Step 1: Assign letters to the variables:
    • $𝓉$ for time.
    • $𝒉$ for the height of the rocket.
    • $\theta$ for the elevation angle of the camera.
  • Step 2: Find a formula that relates the variables:
    • Use the tangent function: $\tan(\theta)=\dfrac{𝒉}{3000}$
      (where $3000$ is the constant horizontal distance between the camera and the launch point).
  • [36:30]. Step 3: Identify the given rates in the problem:
    • $\dfrac{d𝒉}{d𝓉}=600$ feet (0.18 km) per second (rate of change of the rocket's height) when $𝒉=4000$ feet (1.22 km).
    • $\dfrac{d\theta}{d𝓉}$ rate of change of angle of the camera.
  • [42:32]. Step 4: Differentiate the formula implicitly with respect to time ($𝓉$):
    • $\dfrac{d}{d𝓉}\big[\tan(\theta)\big]=\dfrac{d}{d𝓉}\!\Big[\dfrac{1}{3000}\cdot 𝒉\Big]$
      • $\sec^{2}(\theta)\cdot\dfrac{d\theta}{d𝓉}=\dfrac{1}{3000}\cdot\dfrac{d𝒉}{d𝓉}$
  • [44:40]. Step 5: Determine the value of $\sec^{2}(\theta)$ when the rocket's height is 4000 feet (1.22 km):
    • Use the Pythagorean theorem to calculate the hypotenuse of the right triangle formed by the rocket, the camera, and the ground.
      • The rocket is $4{,}000$ feet high.
      • $\text{hypotenuse}^{2}=4000^{2}+3000^{2}\;\Rightarrow\;\text{hypotenuse}=5000$ feet
    • Use the definition of secant ($\sec(\theta)=1/\cos(\theta)$) to calculate $\sec(\theta)$.
      • $\cos(\theta)=\dfrac{3000}{5000}=\dfrac{3}{5}\;\Rightarrow\; \sec(\theta)=\dfrac{5}{3}$
  • Step 6: Substitute the known values into the differentiated equation and solve for the unknown rate of change ($\dfrac{d\theta}{d𝓉}$):
    • $\dfrac{d\theta}{d𝓉}=\dfrac{\big(\dfrac{1}{3000}\big)\big(600\ \text{feet/second}\big)}{\big(\dfrac{5}{3}\big)^{2}}=\dfrac{9}{125}\ \text{radians/second}$ (when the height is $4000$ feet)
    • Convert the rate of change to degrees per second: $\big(\dfrac{9}{125}\ \text{radians/second}\big)\cdot\dfrac{180^\circ}{\pi}\approx 4.13^\circ/\text{second}$