(a) Changing quantities:

• $A$ = area of the oil spill
• $r$ = radius of the oil spill

(b) Given rate: $$\frac{dA}{dt} = 50 \text{ m}^2/\text{min}$$

(c) Reasonable unknown: $$\frac{dr}{dt} = ? \text{ (rate at which radius is increasing)}$$

Let $x$ = distance from bottom of ladder to wall Let $y$ = height of top of ladder

Given rate: The bottom slides away, so $x$ is increasing: $$\frac{dx}{dt} = 2 \text{ ft/s}$$

Unknown rate: The top slides down, so $y$ is decreasing: $$\frac{dy}{dt} = ? \quad \text{(will be negative)}$$

Instant: When $x = 5$ ft

Diagram:

      |← 4 m →|
       \     /
        \   /    ← full tank
         \ / 10 m
          |

      |← r →|
       \   /
        \ /  h    ← water level
         |

Changing quantities:

• $V$ = volume of water
• $h$ = depth (height) of water
• $r$ = radius of water surface

Given rate: Water flows out, so volume decreases: $$\frac{dV}{dt} = -3 \text{ m}^3/\text{min}$$

Unknown rate: Typically we want: $$\frac{dh}{dt} = ? \text{ (rate at which depth is changing)}$$

Note: We expect $\frac{dh}{dt}$ to be negative since water is draining.

Diagram (3D):

                  Drone •←─ altitude 50 m
                       /|
                      / |
                   s /  | 50
                    /   |
               ────+────•──→ x (east)
                  /|   /
                 / |  /
            Car •  | y (north)

Variables:

• $x$ = horizontal distance of drone from origin (east)
• $y$ = distance of car from origin (north)
• $s$ = distance between drone and car

Given rates: $$\frac{dx}{dt} = 8 \text{ m/s} \quad \text{(drone's eastward speed)}$$ $$\frac{dy}{dt} = 12 \text{ m/s} \quad \text{(car's northward speed)}$$

Unknown rate: $$\frac{ds}{dt} = ? \quad \text{when } t = 10 \text{ s}$$

At $t = 10$ s:

• $x = 8(10) = 80$ m
• $y = 12(10) = 120$ m
• Altitude = 50 m (constant)

Relationship: By the 3D distance formula: $$s^2 = x^2 + y^2 + 50^2$$

Variables and rates:

• $\ell$ = length, $\frac{d\ell}{dt} = 3$ cm/s
• $w$ = width, $\frac{dw}{dt} = -2$ cm/s (negative because decreasing)

(a) Perimeter rate: $$P = 2\ell + 2w$$ $$\frac{dP}{dt} = 2\frac{d\ell}{dt} + 2\frac{dw}{dt} = 2(3) + 2(-2) = 2 \text{ cm/s}$$

Area rate: Using the product rule: $$A = \ell w$$ $$\frac{dA}{dt} = \frac{d\ell}{dt} \cdot w + \ell \cdot \frac{dw}{dt} = 3w + \ell(-2) = 3w - 2\ell$$

(b) Area neither increasing nor decreasing:

Set $\frac{dA}{dt} = 0$: $$3w - 2\ell = 0$$ $$w = \frac{2\ell}{3}$$

Answer: The area is momentarily constant when the width equals $\frac{2}{3}$ of the length.

For example, when $\ell = 6$ cm and $w = 4$ cm.

(B) $\frac{dr}{dt} < 0$ because the radius is getting smaller

The derivative tells us the rate of change. A quantity that is decreasing has a negative rate of change, regardless of its current value.

(B) Yes, but only if we also know the current radius $r$

Differentiating: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$

Solving: $\frac{dr}{dt} = \frac{1}{4\pi r^2} \frac{dV}{dt}$

We need to know both $\frac{dV}{dt}$ and the current value of $r$ to compute $\frac{dr}{dt}$.