Question

Find the arc length of the graph of the parametric equations on the given interval(s). \(x=5 t+2, \quad y=1-3 t\) on [-1,1]

Short answer

The arc length is \(2\sqrt{34}\).

Step-by-step solution

Differentiate Parametric Equations

To find the arc length of a parametric curve, we first need to find the derivatives of the given parametric equations. For the given equations, we have:- For \(x(t) = 5t + 2\), differentiate with respect to \(t\) to get \(\frac{dx}{dt} = 5\).- For \(y(t) = 1 - 3t\), differentiate with respect to \(t\) to get \(\frac{dy}{dt} = -3\).

Apply the Arc Length Formula

The formula for the arc length \(L\) of a parametric curve \(x(t)\), \(y(t)\) from \(t = a\) to \(t = b\) is given by:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]Substitute the derivatives obtained:\[ L = \int_{-1}^{1} \sqrt{5^2 + (-3)^2} \, dt \]

Simplify the Expression Inside the Integral

Simplify the expression inside the integral:\(\sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}\).Thus, the integral becomes:\[ L = \int_{-1}^{1} \sqrt{34} \, dt \]

Evaluate the Integral

The integral \(\int_{-1}^{1} \sqrt{34} \, dt\) is evaluated by recognizing \(\sqrt{34}\) as a constant with respect to \(t\):\[ L = \sqrt{34} \int_{-1}^{1} dt = \sqrt{34} \cdot [t]_{-1}^{1} = \sqrt{34} \cdot (1 - (-1)) = \sqrt{34} \cdot 2 = 2\sqrt{34} \].

Key concepts

Arc Length

Arc length is a crucial concept in mathematics, especially when dealing with curves defined by parametric equations. The arc length of a curve is essentially the distance you would travel if you followed the path of the curve from one point to another.
For parametric equations like \(x(t)\) and \(y(t)\), the arc length \(L\) between the parameters \(t = a\) and \(t = b\) can be calculated using the formula:
\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]
In this formula:

• \(\frac{dx}{dt}\) is the derivative of \(x\) with respect to \(t\).
• \(\frac{dy}{dt}\) is the derivative of \(y\) with respect to \(t\).
• The integral sums up the infinitesimal elements of the curve's length between \(t = a\) and \(t = b\).

This formula essentially accounts for both the horizontal and vertical changes in the curve, allowing us to calculate the precise length of potentially complex paths.

Differentiation

Differentiation is a fundamental aspect of calculus used to determine the rate at which a quantity changes. In the context of parametric equations, differentiation helps us find the slopes of the tangent lines along the curve.
For a parametric equation \(x(t) = 5t + 2\) and \(y(t) = 1 - 3t\), we differentiate each component with respect to \(t\):
- The derivative of \(x(t)\) with respect to \(t\) is \(\frac{dx}{dt} = 5\). This result tells us that for each unit increase in \(t\), \(x\) increases by 5 units.- Similarly, the derivative for \(y(t)\) is \(\frac{dy}{dt} = -3\), meaning \(y\) decreases by 3 units for every unit increase in \(t\).
Understanding these derivatives is integral when using the arc length formula because they depict how fast and in what direction the curve moves at each point \(t\). This knowledge allows us to sum up the subtle changes along the curve to establish the total distance traveled through the parameter interval.

Integral Calculus

Integral calculus allows us to find the total accumulation of quantities, such as area under a curve or the total length of a curve. When calculating the arc length of parametric curves, we use integral calculus to sum the small segments of the curve determined by differentiation.
The arc length is found by evaluating the integral:
\[ L = \int_{-1}^{1} \sqrt{34} \, dt \]
In this scenario:

• \(\sqrt{34}\) is treated as a constant since it doesn't depend on \(t\).
• The integral then becomes simply \(\sqrt{34} \cdot [t]_{-1}^{1}\), representing the total change in \(t\) over the interval.
• Solving gives \(\sqrt{34} \times 2\), since the total change in \(t\) from -1 to 1 is 2.

Integral calculus not only allows us to handle abstract mathematical concepts but also to pin down real-world measurements, providing valuable insight into the nature of parametric curves and their characteristics.