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To understand how to find a normal line to a curve, it's vital to first grasp the concept of a derivative. The derivative of a function, denoted as \(y'(x)\), provides us with the slope of the tangent line at any point along the curve. In simpler terms, it tells us how steep the curve is at a specific location.
In the provided example, the function is \(y = \sqrt{x}\). When we find the derivative, we are calculating the instantaneous rate of change or the slope of the curve at any point \(x\). The derivative of \(y = \sqrt{x}\) is \(y'(x) = \frac{1}{2\sqrt{x}}\), which illustrates the slope formula based on \(x\).
Knowing the slope of the tangent line is the first step towards finding both the tangent and normal lines. This leads us to the next core concept, the tangent line, which further builds on this understanding.

A tangent line is a straight line that touches a curve at a single point without crossing it at that vicinity. It represents the instantaneous slope of the function at that point. To find this line, we use the derivative, which we previously discussed.
In the exercise, once the derivative \(y'(x) = \frac{1}{2\sqrt{x}}\) is evaluated at \(x = 4\), it results in the slope \(\frac{1}{4}\). This tells us that at the point \(P(4,2)\), the tangent line is relatively gentle with an upward slope.

• The tangent line equation is crucial for constructing the normal line.
• Understanding the slope of the tangent line helps identify how steep or flat the curve is at that point.

Having this slope, the tangent line can be calculated from any given point on the curve. We now use this slope to derive the equation of the normal line.

Once we have the slope of the perpendicular line, the point-slope form is instrumental in crafting the equation of a normal line. The point-slope formula is a simple yet powerful tool: \(y - y_1 = m(x - x_1)\). It requires a single point \((x_1, y_1)\) and the slope \(m\) to create a line equation.
In our instance, we've identified the slope of the tangent as \(\frac{1}{4}\). The perpendicular line should have a slope that is the negative reciprocal of this, resulting in \(m = -4\).
Utilizing the point-slope form with our point \(P(4,2)\) gives us:

• Substitute \(m = -4\), \(x_1 = 4\), and \(y_1 = 2\) into the equation.
• Simplify to find \(y = -4x + 18\).

The point-slope form simplifies constructing line equations, making it easier to understand and derive both tangent and normal lines swiftly. Thus, this foundational concept ties together derivatives, slopes, and line equations into a cohesive method for tackling such exercises.