Related Rates Problem Solving

Setup:

• Let $s$ = side length, $A$ = area
• Given: $\frac{ds}{dt} = 4$ cm/s
• Find: $\frac{dA}{dt}$ when $s = 12$ cm

Relationship: $A = s^2$

Differentiate: $$\frac{dA}{dt} = 2s \frac{ds}{dt}$$

Substitute: $$\frac{dA}{dt} = 2(12)(4) = 96 \text{ cm}^2/\text{s}$$

Answer: The area is increasing at 96 cm$^2$/s.

Setup:

      |
      |___  y (height)
      |   \
      |    \ 13 ft
      |     \
      |______\
         x (distance from wall)

• Let $x$ = distance from bottom to wall
• Let $y$ = height of top
• Given: $\frac{dx}{dt} = 3$ ft/s
• Find: $\frac{dy}{dt}$ when $x = 5$ ft

Find $y$ when $x = 5$: $$x^2 + y^2 = 169$$ $$25 + y^2 = 169$$ $$y^2 = 144$$ $$y = 12 \text{ ft}$$

Relationship: $x^2 + y^2 = 169$

Differentiate: $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$$

Solve for $\frac{dy}{dt}$: $$\frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt}$$

Substitute: $$\frac{dy}{dt} = -\frac{5}{12}(3) = -\frac{15}{12} = -\frac{5}{4} \text{ ft/s}$$

Answer: The top is sliding down at $\frac{5}{4}$ = 1.25 ft/s.

(Negative sign confirms downward motion.)

Setup:

• Given: $\frac{dV}{dt} = -2$ m$^3$/min (negative because draining)
• Find: $\frac{dh}{dt}$ when $h = 4$ m

Similar triangles: $$\frac{r}{h} = \frac{3}{8} \implies r = \frac{3h}{8}$$

Volume in terms of $h$ only: $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{3h}{8}\right)^2 h = \frac{1}{3}\pi \cdot \frac{9h^2}{64} \cdot h = \frac{3\pi h^3}{64}$$

Differentiate: $$\frac{dV}{dt} = \frac{3\pi}{64} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{9\pi h^2}{64}\frac{dh}{dt}$$

Substitute $\frac{dV}{dt} = -2$ and $h = 4$: $$-2 = \frac{9\pi (4)^2}{64}\frac{dh}{dt}$$ $$-2 = \frac{9\pi (16)}{64}\frac{dh}{dt}$$ $$-2 = \frac{144\pi}{64}\frac{dh}{dt}$$ $$-2 = \frac{9\pi}{4}\frac{dh}{dt}$$ $$\frac{dh}{dt} = -\frac{8}{9\pi} \approx -0.283 \text{ m/min}$$

Answer: The water level is falling at $\frac{8}{9\pi} \approx 0.283$ m/min.

Setup:

        N
        |
   W----+----E
        |   Car → x (west)
        |
        ↓ y (south)
     Truck

        z = distance between them

• Let $x$ = car's distance west
• Let $y$ = truck's distance south
• Let $z$ = distance between them
• $\frac{dx}{dt} = 60$ km/h, $\frac{dy}{dt} = 80$ km/h

At $t = 0.5$ hours:

• $x = 30$ km
• $y = 40$ km
• $z = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$ km

Relationship: $z^2 = x^2 + y^2$

Differentiate: $$2z\frac{dz}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}$$ $$z\frac{dz}{dt} = x\frac{dx}{dt} + y\frac{dy}{dt}$$

Substitute: $$50\frac{dz}{dt} = 30(60) + 40(80)$$ $$50\frac{dz}{dt} = 1800 + 3200$$ $$50\frac{dz}{dt} = 5000$$ $$\frac{dz}{dt} = 100 \text{ km/h}$$

Answer: The distance between them is increasing at 100 km/h.

Setup:

     Wall
      |
      |
      x
      |
      +----θ----O (searchlight)
              50 m

• Let $\theta$ = angle from perpendicular
• Let $x$ = distance along wall from closest point
• Given: $\frac{d\theta}{dt} = 2 \text{ rev/min} = 4\pi \text{ rad/min}$
• Find: $\frac{dx}{dt}$ when $\theta = 45° = \frac{\pi}{4}$

Relationship: $$\tan\theta = \frac{x}{50}$$ $$x = 50\tan\theta$$

Differentiate: $$\frac{dx}{dt} = 50\sec^2\theta \cdot \frac{d\theta}{dt}$$

When $\theta = \frac{\pi}{4}$: $$\sec^2\left(\frac{\pi}{4}\right) = \frac{1}{\cos^2(45°)} = \frac{1}{(\frac{\sqrt{2}}{2})^2} = \frac{1}{\frac{1}{2}} = 2$$

Substitute: $$\frac{dx}{dt} = 50 \cdot 2 \cdot 4\pi = 400\pi \text{ m/min}$$

Convert to m/s: $$\frac{dx}{dt} = \frac{400\pi}{60} = \frac{20\pi}{3} \approx 20.9 \text{ m/s}$$

Answer: The beam is moving along the wall at $400\pi$ m/min ≈ 1257 m/min (or about 21 m/s).

(B) Numbers should be substituted AFTER differentiating

By substituting $r = 5$ first, the student turned $V$ into a constant $\frac{500\pi}{3}$. The derivative of a constant is zero, which is meaningless here.

Correct approach:

1. Differentiate: $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
2. Then substitute: $\frac{dV}{dt} = 4\pi(5)^2 \frac{dr}{dt} = 100\pi \frac{dr}{dt}$

(B) Decrease (radius grows slower)

From $\frac{dr}{dt} = \frac{1}{4\pi r^2} \cdot \frac{dV}{dt} = \frac{k}{4\pi r^2}$

As $r$ increases, $r^2$ increases, so $\frac{1}{r^2}$ decreases, making $\frac{dr}{dt}$ smaller.

Physical intuition: The same amount of air ($k$ per second) has to spread over a larger surface area, so the radius increases more slowly.