Related Rates for MATH 139

Step 1: Identify what we know and what we want.

• Given: $\frac{dr}{dt} = 2$ m/s
• Find: $\frac{dA}{dt}$ when $r = 5$ m

Step 2: Write the equation relating the variables.

The area and radius of a circle are related by: $$A = \pi r^2$$

Step 3: Differentiate with respect to $t$ to see how the rates are related.

$$\frac{dA}{dt} = \pi \cdot 2r \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt}$$

Now we can see: the rate of area change depends on both the current radius AND the rate of radius change.

Step 4: Substitute known values.

$$\frac{dA}{dt} = 2\pi (5)(2) = 20\pi$$

$$\boxed{\frac{dA}{dt} = 20\pi \text{ m}^2/\text{s} \approx 62.8 \text{ m}^2/\text{s}}$$

Step 1: Identify variables and rates.

• Given: $\frac{dV}{dt} = 100$ cm³/s
• Find: $\frac{dr}{dt}$ when $r = 10$ cm

Step 2: How are the variables related?

$$V = \frac{4}{3}\pi r^3$$

Step 3: Differentiate to see how the rates are related.

$$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$$

Step 4: Solve for $\frac{dr}{dt}$.

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

Step 5: Substitute.

$$\frac{dr}{dt} = \frac{100}{4\pi (10)^2} = \frac{100}{400\pi} = \frac{1}{4\pi}$$

$$\boxed{\frac{dr}{dt} = \frac{1}{4\pi} \text{ cm/s} \approx 0.08 \text{ cm/s}}$$

Step 1: Draw and label.

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

Step 2: How are $x$ and $y$ related?

By the Pythagorean theorem: $$x^2 + y^2 = 10^2 = 100$$

Step 3: Differentiate to relate the rates.

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

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

This tells us: when $x$ increases (bottom moves out), $y$ must decrease (top slides down) — the rates are connected!

Step 4: Find $y$ when $x = 6$.

$$6^2 + y^2 = 100 \Rightarrow y^2 = 64 \Rightarrow y = 8$$

Step 5: Substitute and solve.

$$6(2) + 8\frac{dy}{dt} = 0$$ $$12 + 8\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = -\frac{12}{8} = -\frac{3}{2}$$

$$\boxed{\frac{dy}{dt} = -\frac{3}{2} \text{ ft/s}}$$

The negative sign means the top is moving down (decreasing $y$).

Step 1: Variables and rates.

• Given: $\frac{dV}{dt} = 2$ m³/min
• Find: $\frac{dh}{dt}$ when $h = 4$ m

Step 2: Volume of a cone: $V = \frac{1}{3}\pi r^2 h$

But we have two variables ($r$ and $h$). We need another relationship!

Similar triangles: The ratio of radius to height is constant. $$\frac{r}{h} = \frac{5}{10} = \frac{1}{2} \Rightarrow r = \frac{h}{2}$$

Step 3: Substitute to get $V$ in terms of $h$ only.

$$V = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{4} \cdot h = \frac{\pi h^3}{12}$$

Now we have one equation relating $V$ and $h$.

Step 4: Differentiate.

$$\frac{dV}{dt} = \frac{\pi}{12} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{4} \frac{dh}{dt}$$

Step 5: Solve and substitute.

$$\frac{dh}{dt} = \frac{4}{\pi h^2} \cdot \frac{dV}{dt} = \frac{4}{\pi (4)^2} \cdot 2 = \frac{8}{16\pi} = \frac{1}{2\pi}$$

$$\boxed{\frac{dh}{dt} = \frac{1}{2\pi} \text{ m/min} \approx 0.159 \text{ m/min}}$$

Step 1: Set up coordinates with the intersection at the origin.

Let $x$ = A's distance from intersection, $y$ = B's distance, $z$ = distance between cars.

Step 2: How are the variables related?

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

Step 3: Differentiate.

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

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

Step 4: Find $z$ and set up signs.

When $x = 3$ and $y = 4$: $z = \sqrt{9 + 16} = 5$ km

Since both cars are approaching the intersection:

• $\frac{dx}{dt} = -60$ km/h (x is decreasing)
• $\frac{dy}{dt} = -80$ km/h (y is decreasing)

Step 5: Substitute.

$$5\frac{dz}{dt} = 3(-60) + 4(-80) = -180 - 320 = -500$$

$$\frac{dz}{dt} = -100$$

$$\boxed{\frac{dz}{dt} = -100 \text{ km/h}}$$

The distance is decreasing at $100$ km/h.