Related Rates: Position, Velocity, and Acceleration Series

Other Related Rates

  • Most derivatives in this class are typically represented as ""dy/dx"", which describes the rate of change of y with respect to x. These are the most common in practice problems.

  • However, derivatives can relate the rate of change of any variable with respect to any other variable, provided there's an equation connecting them.

    • Examples include:

      • Rate of change of area with respect to radius (dA/dr).

      • Rate of change of volume with respect to height (dV/dh).

      • Temperature with respect to pressure (dT/dP).

      • Profit with respect to units (dP/dU).

Example 1: Spherical Balloon Inflation (Volume and Radius)

  • Problem: Find the rate of change of the volume of a spherical balloon with respect to its radius (dV/dr) as it's being inflated.

  • Formula: The volume of a sphere is given by V = \frac{4}{3}\pi r^3.

  • Derivative Calculation:

    • To find dV/dr, differentiate the volume formula with respect to r:
      dV/dr = \frac{d}{dr}(\frac{4}{3}\pi r^3)
      dV/dr = \frac{4}{3}\pi(3r^{3-1})
      dV/dr = 4\pi r^2

  • Evaluating dV/dr at a Specific Radius:

    • Question: What is the rate the volume is changing when the radius (r) is 1 inch?

    • Notation: dV/dr \Big|_{r=1}

    • Calculation:
      dV/dr \Big|_{r=1} = 4\pi (1)^2 = 4\pi \text{ inches}^3/\text{inch}

    • Interpretation: When the radius is 1 inch, the volume is increasing at a rate of 4\pi cubic inches for every inch the radius increases. It's crucial not to simplify units to inches squared, as this represents a rate of change of volume per unit of radius, not an area.

  • Finding Radius from a Given Rate:

    • Question: What is the radius (r) when the volume is changing at 780\pi inches^3 per inch?

    • Given: dV/dr = 780\pi

    • Set up the equation: 4\pi r^2 = 780\pi

    • Solve for r:

      • Divide both sides by 4\pi:
        r^2 = \frac{780\pi}{4\pi} = 195

      • Take the square root of both sides:
        r = \pm\sqrt{195}

      • Since radius cannot be negative in a real-world context:
        r = \sqrt{195} \approx 13.964 \text{ inches}

Example 2: Cylinder Volume (Derivative with respect to Radius or Height)

  • Formula: The volume of a cylinder is given by V = \pi r^2 h.

    • This formula involves two variables, r (radius) and h (height).

  • Finding dV/dr (Rate of change of volume with respect to radius):

    • When differentiating with respect to r, we assume h is a constant.

    • Rewrite as: V = (\pi h) r^2

    • Differentiate:
      dV/dr = \frac{d}{dr}((\pi h) r^2)
      dV/dr = (\pi h)(2r^{2-1})
      dV/dr = 2\pi rh

  • Finding dV/dh (Rate of change of volume with respect to height):

    • When differentiating with respect to h, we assume r is a constant.

    • Rewrite as: V = (\pi r^2) h

    • Differentiate:
      dV/dh = \frac{d}{dh}((\pi r^2) h)
      dV/dh = (\pi r^2)(1h^{1-1})
      dV/dh = \pi r^2

  • Key Takeaway: The choice of variable with respect to which the derivative is taken significantly impacts the result. Other variables are treated as constants during differentiation.

Example 3: Spherical Surface Area

  • Formula: The surface area of a sphere is given by SA = 4\pi r^2.

Part 1: Finding dSA/dr at a Specific Radius
  • Problem: Find the instantaneous rate of change of the surface area with respect to the radius (dSA/dr) when the radius is 400 miles.

  • Derivative Calculation:

    • Differentiate SA with respect to r:
      dSA/dr = \frac{d}{dr}(4\pi r^2)
      dSA/dr = 4\pi(2r^{2-1})
      dSA/dr = 8\pi r

  • Evaluation:

    • Plug in r = 400 miles:
      dSA/dr \Big|_{r=400} = 8\pi (400) = 3200\pi \text{ miles}^2/\text{mile}

Part 2: Finding dR/dSA at a Specific Surface Area
  • Problem: Find the instantaneous rate of change of the radius with respect to the surface area (dR/dSA) when the surface area is 4900\pi miles^2.

Method 1: Solve for r first, then differentiate
  1. Start with the surface area formula and solve for r: SA = 4\pi r^2 r^2 = \frac{SA}{4\pi} r = \sqrt{\frac{SA}{4\pi}}

    • Assuming positive radius in a real-world context.

  2. Rewrite in a power form for differentiation:
    r = \frac{1}{\sqrt{4\pi}} SA^{1/2}

  3. Differentiate r with respect to SA: dR/dSA = \frac{d}{dSA}(\frac{1}{\sqrt{4\pi}} SA^{1/2}) dR/dSA = \frac{1}{\sqrt{4\pi}} (\frac{1}{2} SA^{1/2 - 1}) dR/dSA = \frac{1}{2\sqrt{4\pi}} SA^{-1/2} = \frac{1}{2\sqrt{4\pi SA}}

    • This can also be written as dR/dSA = \frac{1}{2\sqrt{4\pi}\sqrt{SA}}

  4. Evaluate at SA = 4900\pi miles^2: dR/dSA \Big|{SA=4900\pi} = \frac{1}{2\sqrt{4\pi(4900\pi)}} = \frac{1}{2\sqrt{19600\pi^2}} dR/dSA \Big|{SA=4900\pi} = \frac{1}{2(140\pi)} = \frac{1}{280\pi}

    • As a decimal: 1.299934 \times 10^{-4} \text{ miles/mile}^2

Method 2: Flipping the Rate (Reciprocal Rule)
  • We previously found that dSA/dr = 8\pi r.

  • A relationship exists where dR/dSA = \frac{1}{dSA/dr}.

  • Therefore, dR/dSA = \frac{1}{8\pi r}.

  • To use this formula, we first need to find the radius (r) corresponding to the given surface area (SA = 4900\pi miles^2).

    • SA = 4\pi r^2

    • 4900\pi = 4\pi r^2

    • r^2 = \frac{4900\pi}{4\pi} = 1225

    • r = \sqrt{1225} = 35 \text{ miles}

  • Now plug this value of r into the flipped rate formula:
    dR/dSA = \frac{1}{8\pi (35)} = \frac{1}{280\pi}

  • Conclusion: Both methods yield the exact same answer. The ability to flip rates will be a useful trick revisited later in the course.