Question

Consider the segment of the line \(y=m x+c\) on the interval \([a, b] .\) Use the arc length formula to show that the length of the line segment is \((b-a) \sqrt{1+m^{2}}\) Verify this result by computing the length of the line segment using the distance formula.

Short answer

Answer: The length of the line segment is \((b - a)\sqrt{1 + m^2}\).

Step-by-step solution

Use arc length formula to find the length of line segment

To find the length of the line segment, we will use the arc length formula which is given by: \( L = \int_{a}^b \sqrt{1 + [f'(x)]^2} dx\) Where \(f'(x)\) is the derivative of the function \(f(x) = mx + c\). Let's first find the derivative of the given function: \(f'(x) = \frac{d}{dx}(mx + c) = m\) Now, we can plug the derivative back into the arc length formula: \( L = \int_{a}^b \sqrt{1 + m^2} dx\) Since \(\sqrt{1 + m^2}\) does not depend on x, we can take it out of the integral: \( L = \sqrt{1 + m^2} \int_{a}^b dx\) Now, integrate with respect to x: \( L = \sqrt{1 + m^2} [x]_{a}^b\) Plug in the limits of integration: \( L = \sqrt{1 + m^2} (b - a)\) So, the length of the line segment is \((b - a)\sqrt{1 + m^2}\).

Verify result using distance formula

To verify the obtained result, let's use the distance formula to compute the length of the line segment. For this, we need the coordinates of the endpoints of the segment on the line \(y = mx + c\). The coordinates of the endpoints are: - Point A: \((a, ma + c)\) - Point B: \((b, mb + c)\) The distance formula is given by: \( L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) Let's plug in the coordinates of points A and B and calculate the length L: \( L = \sqrt{(b - a)^2 + (mb + c - ma - c)^2}\) \( L = \sqrt{(b - a)^2 + (m(b - a))^2}\) Factor out \((b - a)^2\): \( L = \sqrt{(b - a)^2(1 + m^2)}\) Take the square root of \((b - a)^2\): \( L = (b - a)\sqrt{1 + m^2}\) This confirms that the length of the line segment computed using the distance formula is equal to the length obtained using the arc length formula, which is \((b - a)\sqrt{1 + m^2}\).

Key concepts

Derivative

To understand the concept of a derivative, imagine a function that represents a curve. The derivative provides us with the slope of the tangent line at any point on that curve. It's a way to measure how a function is changing at any given point, and it's a fundamental tool in calculus.

For a linear function like \(y = mx + c\), which represents a straight line, the derivative is simpler to find because the slope is constant. The derivative \(f'(x)\) of this function is simply \(m\), the coefficient of \(x\). This tells us that the slope of the line is \(m\) at any point along it, consistently.

• The derivative signifies the rate of change of a function.
• For a line \(y = mx + c\), the rate of change (or slope) is constant and equals \(m\).

In our exercise, finding the derivative of \(y = mx + c\) helped us determine part of the arc length formula by giving us the constant slope \(m\) to work with. This is particularly useful for calculating the arc length of straight segments.

Integral Calculus

Integral Calculus is like the reverse of taking a derivative. Instead of finding the slope, we use it to calculate the total accumulation of quantities, like area under a curve or, as in our case, length of a line segment. The arc length of a curve between two points can be found using integration, which is where Integral Calculus plays its crucial role.

The arc length formula for a curve \(y = f(x)\) between two points \(x = a\) and \(x = b\) is given by:\[ L = \int_{a}^b \sqrt{1 + [f'(x)]^2} \, dx \]This equation integrates the values determined by the derivative across the interval \([a, b]\).

• Integration sums up small pieces (infinitesimals) to find total values like length or area.
• In this problem, we integrate over the interval \([a, b]\) with a constant factor due to the linear nature of the function \(f(x) = mx + c\).

This process helps convert a complicated problem of finding a line's length into manageable integration with a straightforward application, as linear functions simplify the integral considerably.

Distance Formula

The Distance Formula is a reliable method to calculate the distance between two points in a plane. For two points \((x_1, y_1)\) and \((x_2, y_2)\), the distance \(L\) is defined by:\[ L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]This takes into account both the horizontal and vertical differences between two points.

In our problem, this formula is used to verify the length of the line segment found using the arc length formula. The endpoints of the line \(y = mx + c\) on the interval \([a, b]\) are \((a, ma + c)\) and \((b, mb + c)\), so the distance formula validates our result.

• The Distance Formula provides a straightforward way to calculate the length of a line segment.
• It uses the Pythagorean Theorem to account for changes in both x and y directions.

By applying the Distance Formula, we were able to confirm that the derived length equation \((b - a)\sqrt{1 + m^2}\) was correct. It simplifies to the known result, showing how calculus and geometry come together in solving real-world problems.