Problem 77 - Calculus Early Transcendentals: Pearson New International Edition

Question

A wheel whose rim has equation \(x^{2}+(y-6)^{2}=25\) is rotating rapidly in the counterclockwise direction. A speck of dirt on the rim came loose at the point \((3,2)\) and flew toward the wall \(x=11\). About how high up on the wall did it hit? Hint: The speck of dirt flies off on a tangent so fast that the effects of gravity are negligible by the time it has hit the wall.

Step-by-Step Solution

Answer

The speck of dirt hits the wall at a height of 8 units.

Step 1: Identify the Center and Radius of the Circle

The given equation of the circle is \(x^{2} + (y-6)^{2} = 25\). This is of the form \((x-h)^{2}+(y-k)^{2}=r^{2}\), where \((h, k)\) is the center and \(r\) is the radius of the circle. Here, \(h = 0\), \(k = 6\), and \(r = 5\). Thus, the center of the circle is \((0, 6)\) and the radius is 5.

Step 2: Find the Tangent Slope at the Point (3, 2)

A tangent to a circle is perpendicular to the radius at the point of tangency. First, find the slope of the radius from the center \((0, 6)\) to the point \((3, 2)\). The slope is \(\frac{2-6}{3-0} = \frac{-4}{3}\). The slope of the tangent line is the negative reciprocal of the radius slope, which is \(\frac{3}{4}\).

Step 3: Write the Equation of the Tangent Line

Use point-slope form to write the equation of the tangent line at \((3, 2)\). The point-slope form is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (3, 2)\) and \(m = \frac{3}{4}\). Substituting these values gives us \(y - 2 = \frac{3}{4}(x - 3)\). Simplify to get \(y = \frac{3}{4}x - \frac{9}{4} + 2\) or \(y = \frac{3}{4}x + \frac{-1}{4}\).

Step 4: Determine the Point Where the Tangent Line Hits the Wall

The tangent line goes towards the wall at \(x = 11\). To find where it hits the wall, substitute \(x = 11\) into the tangent line equation: \(y = \frac{3}{4}(11) + \frac{-1}{4}\). Calculate this to find \(y = \frac{33}{4} + \frac{-1}{4} = \frac{32}{4} = 8\).

Key Concepts

Circle geometry

In circle geometry, understanding the equation of a circle is crucial. A standard circle equation takes the form

• \((x - h)^2 + (y - k)^2 = r^2\)

Here, \((h, k)\) represents the center of the circle and \(r\) is the radius. For the wheel problem, the circle's equation is given as

• \(x^2 + (y-6)^2 = 25\).

This means the center of this circle is at

• \((0, 6)\)

and the radius is 5 (since the square root of 25 is 5). Circle geometry helps us understand the position and size of a circular object based on its equation. Recognizing the circle's center and radius provides a foundational understanding of how to relate different points and lines associated with the circle, such as its tangents.
Understanding this allows us to tackle complex problems where a circle interacts with lines or other shapes.

Tangent line

A tangent line is a line that touches a circle at exactly one point. This means it just 'grazes' the circle without cutting through it. The tangent line is unique because it is always perpendicular to the radius at the point of tangency.
In our problem, the speck flies off from point

• \((3, 2)\)

This point is on the circle. The corresponding radius connects the center

• \((0, 6)\)

to the point

• \((3, 2)\)

The slope of this radius is critical because it helps find the tangent's slope. The slopes are related in that the slope of the tangent (denoted as \(m_t\)) is the negative reciprocal of the radius slope. For our radius slope of

• \(\frac{-4}{3}\)

The tangent's slope formulates as

• \(\frac{3}{4}\)

This principle is vital in identifying how points on a circle can influence the direction of tangent lines.

Coordinate geometry

Coordinate geometry involves using algebraic techniques to solve geometric problems. This branch of mathematics allows us to find exact points and slopes on graphs using coordinate points like

• \((x_1, y_1)\)

In our example, we found the equation of the tangent line using point-slope form, a common coordinate geometry technique.
The point-slope form is expressed as

• \(y - y_1 = m(x - x_1)\)

By combining the known tangent slope

• \(\frac{3}{4}\)

with the point

• \((3, 2)\)

we formulated the tangent equation. This resulted in

• \(y = \frac{3}{4}x + \frac{-1}{4}\)

Coordinate geometry is exceedingly useful for managing and analyzing the relationships between geometric figures and their algebraic representations. It provides a systematic way to find unknown values and solve for elements within geometric problems.

Slope calculation

Slope is a measure of the steepness or direction of a line. In simpler terms, it tells us how much a line rises or falls as it goes from left to right across a graph.
The formula to find the 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 given circle problem, we calculate the slope of the radius connecting the circle's center

• \((0, 6)\)

to

• \((3, 2)\)

This slope comes out as

• \(\frac{-4}{3}\)

The tangent line's slope, a negative reciprocal of the radius' slope, is

• \(\frac{3}{4}\)

These slopes help us understand the angles and directions of lines in relation to one another on a coordinate plane. Mastering slope calculation is key for working with linear equations.