Set derivative to zero: $-0.03 \sin(0.1\pi t) + 0.1\pi \cos(0.1\pi t) = 0$.

["Set Derivative to Zero: Solving $-0.03 \sin(0.1\pi t) + 0.1\pi \cos(0.1\pi t) = 0$", "Finding critical points in mathematical models is essential in fields like physics, engineering, and economics. One common technique is setting the derivative of a function to zero, which helps identify maxima, minima, and inflection points. In this article, we explore how to solve the equation:", "$$\n-0.03 \sin(0.1\pi t) + 0.1\pi \cos(0.1\pi t) = 0\n$$", "by setting the derivative of the underlying function to zero.", "---", "### Step 1: Understand the Function", "Let’s define the function of interest as:", "$$\nf(t) = -0.03 \sin(0.1\pi t) + 0.1\pi \cos(0.1\pi t)\n$$", "To locate critical points where the rate of change stops—potential turning points—we compute the derivative $ f'(t) $ and set it equal to zero:", "$$\nf'(t) = 0\n$$", "Differentiating term by term:", "- The derivative of $ -0.03 \sin(0.1\pi t) $ is $ -0.03 \cdot 0.1\pi \cos(0.1\pi t) = -0.003\pi \cos(0.1\pi t) $\n- The derivative of $ 0.1\pi \cos(0.1\pi t) $ is $ 0.1\pi \cdot (-0.1\pi \sin(0.1\pi t)) = -0.01\pi^2 \sin(0.1\pi t) $", "So,", "$$\nf'(t) = -0.003\pi \cos(0.1\pi t) - 0.01\pi^2 \sin(0.1\pi t)\n$$", "Set this equal to zero:", "$$\n-0.003\pi \cos(0.1\pi t) - 0.01\pi^2 \sin(0.1\pi t) = 0\n$$", "---", "### Step 2: Simplify the Equation", "Factor out $ -\pi $:", "$$\n-\pi \left( 0.003 \cos(0.1\pi t) + 0.01\pi \sin(0.1\pi t) \right) = 0\n$$", "Since $ \pi <br/>\ne 0 $, divide both sides:", "$$\n0.003 \cos(0.1\pi t) + 0.01\pi \sin(0.1\pi t) = 0\n$$", "Rewriting:", "$$\n\frac{\cos(0.1\pi t)}{\sin(0.1\pi t)} = -\frac{0.01\pi}{0.003}\n$$", "$$\n\cot(0.1\pi t) = -\frac{10\pi}{3}\n$$", "---", "### Step 3: Solve for $ t $", "We solve:", "$$\n\cot(0.1\pi t) = -\frac{10\pi}{3}\n\quad \Rightarrow \quad\n\ an(0.1\pi t) = -\frac{3}{10\pi}\n$$", "Let $ \ heta = 0.1\pi t $. Then:", "$$\n\ an\ heta = -\frac{3}{10\pi}\n$$", "The general solution for $ \ heta $ is:", "$$\n\ heta = \ an^{-1}\left(-\frac{3}{10\pi}\right) + n\pi, \quad n \in \mathbb{Z}\n$$", "Recall $ \ heta = 0.1\pi t $, so:", "$$\n0.1\pi t = \ an^{-1}\left(-\frac{3}{10\pi}\right) + n\pi\n$$", "Solve for $ t $:", "$$\nt = \frac{1}{0.1\pi} \left( \ an^{-1}\left(-\frac{3}{10\pi}\right) + n\pi \right)\n= \frac{10}{\pi} \left( \ an^{-1}\left(-\frac{3}{10\pi}\right) + n\pi \right)\n$$", "---", "### Step 4: Interpret the Solution", "This expression gives all critical points $ t $ where the derivative of $ f(t) $ is zero—points where the function reaches local maxima or minima.", "Because the functions involved are periodic (with period $ \frac{2}{0.1\pi} = \frac{20}{\pi} $), these critical points repeat every $ \frac{20}{\pi} \approx 6.37 $ units.", "---", "### Step 5: Practical Applications", "In real-world scenarios—such as modeling damped oscillations, wave interference, or electrical signals—solving $ f'(t) = 0 $ helps determine peak performance, resonance conditions, or stability thresholds. This derivative-zero condition often marks significant events in dynamic systems.", "---", "### Summary", "Setting the derivative of $ f(t) = -0.03 \sin(0.1\pi t) + 0.1\pi \cos(0.1\pi t) $ to zero yields a key trigonometric equation, leading to:", "$$\nt = \frac{10}{\pi} \left( \ an^{-1}\left(-\frac{3}{10\pi}\right) + n\pi \right), \quad n \in \mathbb{Z}\n$$", "These solutions identify critical points where the function’s behavior changes, essential for analysis in science and engineering.", "---", "Keywords:\nderivative to zero, solve $ -0.03 \sin(0.1\pi t) + 0.1\pi \cos(0.1\pi t) = 0 $, critical points, derivative method, periodic functions, calculus applications, time derivatives, applied mathematics."]