\(12(1) + 16 \frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{12}{16} = -\frac{3}{4}\) ft/s.

["# Solving (12(1) + 16 \frac{dy}{dt} = 0): Deriving ( \frac{dy}{dt} = -\frac{3}{4} ) ft/s", "Understanding how to solve differential equations is fundamental in physics, engineering, and applied mathematics. One common type encountered is a first-order linear equation involving a rate of change, such as (12(1) + 16 \frac{dy}{dt} = 0). In this article, we'll walk through solving this equation step-by-step to find ( \frac{dy}{dt} = -\frac{3}{4} ) ft/s, explaining both the math and its practical significance.", "---", "## The Equation: A Practical Rate Equation", "The expression\n$$\n12(1) + 16 \frac{dy}{dt} = 0\n$$\noften arises in scenarios involving steady rates of change, such as the speed of an object moving constant acceleration or a fluid flowing at a linear rate. Here, ( y(t) ) typically represents a quantity changing over time — for example, position, displacement, or flow rate — and ( \frac{dy}{dt} ) denotes its instantaneous rate of change.", "---", "## Step-by-Step Solution", "### Step 1: Simplify the Constant", "First, (12(1)) simplifies to 12:\n$$\n12 + 16 \frac{dy}{dt} = 0\n$$", "### Step 2: Isolate the Derivative Term", "Subtract 12 from both sides:\n$$\n16 \frac{dy}{dt} = -12\n$$", "### Step 3: Solve for ( \frac{dy}{dt} )", "Divide both sides by 16:\n$$\n\frac{dy}{dt} = -\frac{12}{16}\n$$", "Reduce the fraction:\n$$\n\frac{dy}{dt} = -\frac{3}{4}\n$$", "---", "## Interpretation: What Does ( \frac{dy}{dt} = -\frac{3}{4} ) Mean?", "The result ( \frac{dy}{dt} = -\frac{3}{4} ) implies that quantity ( y ) decreases at a constant rate of ( \frac{3}{4} ) units per second. The negative sign indicates a loss or decay. In real-world contexts:", "- If ( y ) represents distance moved along a straight line, ( y ) is decreasing at ( 0.75 ) ft/s — moving backward at 0.75 feet per second.\n- If ( y ) is flow rate, a negative rate means outflow exceeds inflow, causing a drop in volume.", "This steady rate is ideal for modeling linear motion with constant deceleration or substances flowing under regulated conditions.", "---", "## Why This Format Matters in Calculus", "This equation is a simplified version of a first-order differential equation — useful in modeling continuous change. Solving such equations helps predict future values, optimize systems, and analyze trends. For instance, knowing ( \frac{dy}{dt} = -\frac{3}{4} ) lets engineers compute ( y(t) ) by integrating:\n$$\ny(t) = y(0) - \frac{3}{4}t + C \quad \ ext{(where } C = y(0) \ ext{)}\n$$", "---", "## Final Summary", "- The equation ( 12 + 16 \frac{dy}{dt} = 0 ) models a rate of change\n- Isolating ( \frac{dy}{dt} ) yields ( -\frac{3}{4} ) ft/s\n- This represents a consistent decrease at 0.75 units per second\n- Such rates are foundational in mechanics, optimization, and dynamic modeling", "Understanding how to manipulate and interpret these expressions equips you with a powerful tool for analyzing real-world change — whether tracking motion, energy flow, or environmental trends.", "---", "Keywords:\n( \frac{dy}{dt} = -\frac{3}{4} ), differential equation, linear decay, rate of change, calculus application, physics modeling, integration, velocity, flow rate, math problem solution"]