91Ó°ÊÓ

Implicit differentiation is a powerful tool used when dealing with equations that are not easily solved for one variable in terms of another.
Unlike explicit differentiation where a function is written as \(y = f(x)\), implicit differentiation handles relationships like \(x^2 + y^2 = 25\). This allows us to differentiate both sides of the equation with respect to \(x\), treating \(y\) as an implicit function of \(x\).

• Start by differentiating each term of the equation with respect to \(x\).
• Remember to apply the chain rule on terms involving \(y\), as \(y\) is a function of \(x\), leading to derivatives like \(\frac{dy}{dx}\).
• After differentiation, you solve for \(\frac{dy}{dx}\), which gives the slope of the tangent line to the curve at any point \((x, y)\) on the curve.

In this problem, implicit differentiation of \(x^2 + y^2 = 25\) leads to \(2x + 2y\frac{dy}{dx} = 0\), and solving for \(\frac{dy}{dx}\) gives \(\frac{dy}{dx} = -\frac{x}{y}\). This tells us that the slope of the tangent line at any point \((x, y)\) on the circle is perpendicular to the radial line through \((x, y)\).

The equation of a circle provides a beautiful symmetry that often simplifies analysis in geometry and calculus.
Among its characteristics is the constant radius from the center, making the equation \(x^2 + y^2 = r^2\) a circle centered at the origin \((0, 0)\) with radius \(r\).

• For this form, substituting specific \((x, y)\) values checks if those points lie on the circle by ensuring the equation is satisfied.
• If they satisfy \(x^2 + y^2 = 25\), these coordinates lie exactly on the circle with radius 5.

This was verified when plugging in \((3, -4)\) into \(3^2 + (-4)^2 = 25\). Since both sides are equal, the point is indeed on the circle. To find the tangent to this circle at any point like \((3, -4)\), we ultimately need the slope through implicit differentiation. This approach neatly interconnects these mathematical expressions and the geometry of the circle.

The point-slope form of a line is essential when finding equations of tangent and normal lines. It is written as \(y - y_1 = m(x - x_1)\), a convenient formula when you know:

• A specific point \((x_1, y_1)\) through which the line passes.
• The slope \(m\) of the line.

For the tangent line at \((3, -4)\) and slope \(\frac{3}{4}\), the equation becomes \(y + 4 = \frac{3}{4}(x - 3)\). Simplifying gives \(y = \frac{3}{4}x - \frac{25}{4}\), which is the tangent line's equation.
The normal line, whose slope is the negative reciprocal of the tangent, has a slope of \(-\frac{4}{3}\). Using the same point \((3, -4)\) results in \(y + 4 = -\frac{4}{3}(x - 3)\), simplifying to \(y = -\frac{4}{3}x - \frac{14}{3}\).
Point-slope form effortlessly converts these slopes and points into equations, making it indispensable in calculus.