Problem 75 - Algebra and Trigonometry

Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. The normal line to a graph at a point is the line perpendicular to the tangent line of the graph at the point. If the tangent line is \(y=\frac{2}{3} x-1\) when \(f(3)=1,\) find an equation of the normal line.

Step-by-Step Solution

Answer

\( y = -\frac{3}{2}x + \frac{11}{2} \)

Step 1 - Identify the slope of the tangent line

The tangent line equation is given as \[ y = \frac{2}{3}x - 1 \] From this equation, it can be seen that the slope of the tangent line (m) is \( \frac{2}{3} \).

Step 2 - Find the slope of the normal line

The normal line is perpendicular to the tangent line, so its slope will be the negative reciprocal of the tangent line's slope. The slope of the tangent line is \( \frac{2}{3} \), so the slope of the normal line (m_{normal}) will be: \[ m_{normal} = -\frac{1}{(\frac{2}{3})} = -\frac{3}{2} \]

Step 3 - Use the point-slope form of the line equation

Since we know the slope of the normal line and a point through which it passes (3,1), we can use the point-slope form equation: \[ y - y_1 = m(x - x_1) \] Plugging in the values, we get: \[ y - 1 = -\frac{3}{2}(x - 3) \]

Step 4 - Simplify the equation

Distribute and simplify the equation: \[ y - 1 = -\frac{3}{2}x + \frac{9}{2} \] Add 1 to both sides: \[ y = -\frac{3}{2}x + \frac{9}{2} + 1 \] Convert 1 to halves: \[ y = -\frac{3}{2}x + \frac{9}{2} + \frac{2}{2} \] Combine terms: \[ y = -\frac{3}{2}x + \frac{11}{2} \].

Key Concepts

tangent line

A tangent line is a straight line that touches a curve at a single point, without crossing it. This point is known as the point of tangency. A tangent line has the same slope as the curve it touches at this specific point. In the given exercise, the tangent line to the graph at the point is given by the equation: y = \frac{2}{3}x - 1 This means the curve and the tangent line share the same slope, \( \frac{2}{3} \), at the point where \( x = 3 \) and \( y = 1 \). The tangent line helps us understand the instantaneous rate of change of the curve at that specific point. It is useful in various applications in calculus and physics, such as finding derivatives or predicting motion.

slope

The slope of a line measures its steepness and direction. It is often represented by the letter 'm'. The formula to calculate slope between two points \((x_1, y_1)\) and \((x_2, y_2)\) is: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] In the context of the given exercise, the slope of the tangent line is \( \frac{2}{3} \). To find the slope of the normal line, which is perpendicular to the tangent line, we take the negative reciprocal of the tangent slope. If the tangent slope is \( \frac{2}{3} \), the normal slope is: \[ m_{normal} = -\frac{1}{(\frac{2}{3})} = -\frac{3}{2} \] This negative reciprocal relationship ensures the normal line is perpendicular to the tangent line.

point-slope form

The point-slope form is a way to write the equation of a line when you know its slope and one point it passes through. The formula for the point-slope form is: \[ y - y_1 = m(x - x_1) \] Here, \( m \) is the slope of the line, let's say the slope of the normal line \( m_{normal} = -\frac{3}{2} \) from our exercise, and \( (x_1, y_1) \) is the point through which the line passes, which is \( (3, 1) \). With these values, plugging into the point-slope form gives us: \[ y - 1 = -\frac{3}{2}(x - 3) \] We can then simplify this to get the final equation of the normal line: \[ y - 1 = -\frac{3}{2}x + \frac{9}{2} \] Adding 1 to both sides and simplifying further: \[ y = -\frac{3}{2}x + \frac{11}{2} \] This form is useful for easily converting a linear equation to a more familiar slope-intercept form \( y = mx + b \).