Related Rates: Position, Velocity, and Acceleration Series

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 ( $dP/dU$ ).

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:
    $r = \sqrt{195} \approx 13.964 \text{ inches}$

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.

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

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 $^2$ .

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.
Rewrite in a power form for differentiation:
$r = \frac{1}{\sqrt{4\pi}} SA^{1/2}$
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 $dR/dSA = \frac{1}{2\sqrt{4\pi}\sqrt{SA}}$
Evaluate at $SA = 4900\pi$ miles $^2$ : $dR/dSA \Big|<em>{SA=4900\pi} = \frac{1}{2\sqrt{4\pi(4900\pi)}} = \frac{1}{2\sqrt{19600\pi^2}}$ $dR/dSA \Big|</em>{SA=4900\pi} = \frac{1}{2(140\pi)} = \frac{1}{280\pi}$
- As a decimal: $1.299934 \times 10^{-4} \text{ miles/mile}^2$

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.