Problem 46 - Precalculus with Limits

Question

In Exercises 37-46, find the angle \(\theta\) (in radians and degrees)between the lines. \(0.02x - 0.05y = -0.19\) \(0.03x + 0.04y = 0.52\)

Step-by-Step Solution

Answer

The angle between the two lines is -0.85 radians or -48.86 degrees.

Step 1: Convert the given equations into slope-intercept form.

This can be done by isolating \(y\) on one side of the equation. The equations are as follows: From \(0.02x - 0.05y = -0.19\), we get \(y = 0.4x + 3.8\)From \(0.03x + 0.04y = 0.52\), we get \(y = -0.75x + 13\)

Step 2: Calculate the gradients of the lines

The gradient of a line in the form \(y = mx + c\) is given by \(m\). Therefore, from the first equation, \(m1 = 0.4\) and from the second equation, \(m2 = -0.75\)

Step 3: Calculate the angle between the two lines

We use the formula \(\tan(\theta) = \frac{m2 - m1}{1 + m1*m2}\). Plugging in the values we got for \(m1\) and \(m2\), we get \(\tan(\theta) = \frac{-0.75 - 0.4}{1 + 0.4*(-0.75)} = -1.125\)Now, the value of \(\theta\) is then given by the arctangent of -1.125. Therefore, \(\theta = \arctan(-1.125) = -0.85\) radians.To convert this into degrees, multiply by \(\frac{180}{\pi}\). Therefore, \(\theta = -0.85 * \frac{180}{\pi} = -48.86^o\)

Key Concepts

Slope-Intercept Form

The slope-intercept form is one of the most common ways to express the equation of a straight line. This form is written as \( y = mx + c \), where \( m \) represents the slope (or gradient) of the line, and \( c \) is the y-intercept, the point where the line crosses the y-axis.

To convert a linear equation like \( Ax + By = C \) into the slope-intercept form, solve for \( y \) by isolating it on one side of the equation. This involves manipulating the equation to get \( y \) by itself:

• First, move \(x\) to the other side of the equation by subtraction or addition.
• Next, divide every term by the coefficient of \(y\) to solve for \(y\).

For example, converting \(0.02x - 0.05y = -0.19\) into this form involves adding \(0.02x\) to both sides and then dividing through by \(-0.05\) to end up with \(y = 0.4x + 3.8 \). This expressively shows us how the line behaves graphically.

Gradient of Lines

The gradient of a line, often referred to as the slope, is a measure of how steep the line is. In the slope-intercept form \( y = mx + c \), the gradient is denoted by \( m \), indicating how much \( y \) changes per unit increase in \( x \). This is a crucial feature in understanding the relationship between two variables.

When you visualize a line, the gradient tells you:

• If the line slopes upward from left to right (positive slope).
• If the line slopes downward from left to right (negative slope).
• If the line is horizontal, indicating zero slope.

In the provided equations, \( y = 0.4x + 3.8 \) and \( y = -0.75x + 13 \), the gradients are \( m_1 = 0.4 \) and \( m_2 = -0.75 \), respectively. These values give us immediate insight into how the lines would look graphically and help determine the angle between them.

Arctangent Function

The arctangent function, often written as \( \arctan \), is used in trigonometry to find an angle whose tangent is a given number. In the context of finding the angle between two lines, it enables us to calculate the angle using the formula:

\[ \tan(\theta) = \frac{m_2 - m_1}{1 + m_1m_2} \]

Here, \(m_1\) and \(m_2\) are the gradients of the two lines. The result of this formula gives us \(\tan(\theta)\), which we then convert into an angle \( \theta \) by calculating \( \arctan(\tan(\theta)) \).

For instance, using the provided line gradients \(m_1 = 0.4\) and \(m_2 = -0.75\), the computation would be \( \tan(\theta) = -1.125 \). The angle is the inverse tangent of this result, approximating to \(\theta = -0.85\) radians.

To convert this to degrees from radians, multiply by \( \frac{180}{\pi} \), resulting in \(-48.86^\circ\). The arctangent function thus provides a precise method to find the angle which tells us about the intersectional behavior of the two lines.