91Ó°ÊÓ

Parametric equations give us a way to describe a geometric object using parameters rather than standard Cartesian coordinates. For a circle, such equations express the coordinates \((x, y)\) as functions of a parameter, often denoted by \(t\). This method is incredibly useful for describing curves and motions.In our exercise, the circle is described by the parametric equations:

• \(x = \cos t\)
• \(y = \sin t\)

These equations parameterize the circle with a radius of 1 centered at the origin, and with \(t\) ranging from \(0\) to \(2\pi\), they sweep out the entire circle once. Understanding parametric equations helps visualize motions and paths, such as the temperature changes along the circle in this exercise. They show how every point on the circle, described by \((x, y)\), changes as \(t\) changes from \(0\) to \(2\pi\).

Partial derivatives are a fundamental tool in calculus used to study changes in a multivariable function with respect to one of its variables while holding others constant. For the temperature function \(T = f(x, y)\), the partial derivatives are the rates of change of temperature with respect to \(x\) and \(y\).In this problem, we have:

• \(\frac{\partial T}{\partial x} = 8x - 4y\)
• \(\frac{\partial T}{\partial y} = 8y - 4x\)

These equations tell us how the temperature changes as we move along the x-axis and y-axis, respectively. They're important for understanding how temperature varies over the circle. Partial derivatives help us determine the direction in which the temperature increases or decreases most quickly, making it essential for finding maximum and minimum values on the surface described by the temperature function.

Critical points in the context of calculus are values of \(t\) where a function's derivative is zero or undefined. They're important because these are often the points where a function reaches its maximum, minimum, or saddle points, indicating where the change in the function transitions from increasing to decreasing or vice versa.For the exercise, to find the critical points of \(T = 4 - 2\sin(2t)\), we calculated the derivative \(\frac{dT}{dt} = -4\cos(2t)\). Setting this derivative equal to zero \(\frac{dT}{dt} = 0\) helps us find the critical points:-\(\cos(2t) = 0\) leads to solutions for \(t\) as \(\frac{\pi}{4} + \frac{k\pi}{2}\), where \(k\) is an integer. Within the given domain \(0 \leq t \leq 2\pi\), this yields critical points at \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \text{and} \frac{7\pi}{4}\). Evaluating \(T\) at these points helps identify where the maximum and minimum temperatures occur on the circle.

Trigonometric identities simplify the expressions involving trigonometric functions, making complex calculations more manageable. They are vital for transforming and simplifying parametric equations and derivatives.In the parameterization of the circle, we used the identity:

• \(\cos^2 t + \sin^2 t = 1\)

Also, we used the double angle identity for sine to transform the expression.For this problem, \(T = 4\cos^2 t - 4\cos t \sin t + 4\sin^2 t\) becomes \(T = 4 - 2\sin(2t)\) by the identity \(\sin(2t) = 2\sin t\cos t\). This simplification is essential for easily finding derivatives and identifying critical points.Trigonometric identities are tools that make it possible to turn potentially overwhelming calculations into something much more accessible, paving the way for clear understanding and solution of trigonometric puzzles.