Question

For the following exercises, find the arc length of the curve on the indicated interval of the parameter.\(x=e^{t} \cos t, \quad y=e^{t} \sin t, \quad 0 \leq t \leq \frac{\pi}{2}\) (express answer as a decimal rounded to three places)

Short answer

The arc length is approximately 5.389.

Step-by-step solution

Understand the Formula for Arc Length

The arc length \( L \) of a parametric curve \( x = f(t) \) and \( y = g(t) \) on an interval \( a \leq t \leq b \) is given by the formula: \[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]. Here, \( a = 0 \) and \( b = \frac{\pi}{2} \), \( x = e^t \cos t \), and \( y = e^t \sin t \).

Calculate Derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)

Find the derivative \( \frac{dx}{dt} \) for \( x = e^t \cos t \) using the product rule: \[\frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t\].Similarly, find \( \frac{dy}{dt} \) for \( y = e^t \sin t \):\[\frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t\].

Substitute Derivatives into the Arc Length Formula

Substitute \( \frac{dx}{dt} = e^t \cos t - e^t \sin t \) and \( \frac{dy}{dt} = e^t \sin t + e^t \cos t \) into the formula for \( L \):\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{(e^t \cos t - e^t \sin t)^2 + (e^t \sin t + e^t \cos t)^2} \, dt\].

Simplify the Expression Under the Square Root

Simplify the expression inside the square root:\[(e^t \cos t - e^t \sin t)^2 + (e^t \sin t + e^t \cos t)^2 = e^{2t} ((\cos t - \sin t)^2 + (\sin t + \cos t)^2)\].Further simplifying, \[ (\cos t - \sin t)^2 + (\sin t + \cos t)^2 = 2\], leading to:\[L = \int_{0}^{\frac{\pi}{2}} \sqrt{2} \, e^t \, dt\].

Compute the Definite Integral

The integral becomes:\[L = \sqrt{2} \int_{0}^{\frac{\pi}{2}} e^t \, dt\].Therefore, integrate:\[\int e^t \, dt = e^t \bigg|_0^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} - 1\].Thus,\[L = \sqrt{2}(e^{\frac{\pi}{2}} - 1)\].

Compute the Final Result

Calculate \(L = \sqrt{2}(e^{\frac{\pi}{2}} - 1) \). First, approximate \( e^{\frac{\pi}{2}} \) and then simplify the expression as a decimal:- \( e^{\frac{\pi}{2}} \approx 4.810478\),- then \( e^{\frac{\pi}{2}} - 1 \approx 3.810478 \).Finally, \( L \approx \sqrt{2} \times 3.810478 \approx 5.389 \). Hence, the arc length is approximately 5.389 rounded to three decimal places.

Key concepts

Parametric Equations

When plotting curves, instead of relying on the Cartesian coordinates with formulas like \( y = f(x) \), parametric equations provide a different approach. They define both \( x \) and \( y \) in terms of a third variable, usually \( t \). This can be particularly useful for describing complex curves that might loop over themselves or be difficult to represent with a single function. In our exercise, we have:

• \( x = e^t \cos t \)
• \( y = e^t \sin t \)

These equations describe the path of a point moving around in space. One way to visualize this is to imagine that \( t \) is time, and as time passes, the point moves along the curve, tracing out its shape. The interval \( 0 \leq t \leq \frac{\pi}{2} \) specifies the portion of the curve we're interested in; \( t = 0 \) is the starting point, and \( t = \frac{\pi}{2} \) is the ending point. This combination of \( x \) and \( y \) values results in a spiral-like curve starting at the origin and moving outward.

Integral Calculus

Integral calculus arises when seeking to calculate quantities related to accumulation. In the context of arc length, it is used to find the stretch of curve over an interval. For parametric curves, the arc length formula is given by:\[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]In this integral, \( \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \) represents the length of a tiny segment of the curve. Summing up all these segments over the interval \( t \) allows us to find the total arc length. It's important to notice that the function under the integral must be simplified appropriately to make the calculation feasible, as seen in the provided solution. Here, the interval is \( t = 0 \) to \( t = \frac{\pi}{2} \), meaning we're calculating the arc length of the part of the curve that passes through this range.

Derivatives

Derivatives are a fundamental tool in calculus, providing a way to measure how a function changes. In the context of parametric equations like those in our exercise, we need to calculate derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to use in the arc length formula. Using the product rule, which is necessary here due to the products of functions in \( e^t \cos t \) and \( e^t \sin t \), we arrive at:

• \( \frac{dx}{dt} = e^t \cos t - e^t \sin t \)
• \( \frac{dy}{dt} = e^t \sin t + e^t \cos t \)

These derivatives tell us how the \( x \) and \( y \) coordinates change as \( t \) changes. Substituting these into the arc length integral, we find the contribution of these changes to the total arc length. Simplifying and integrating this expression forms the essence of calculating the arc length from a parametric function, an approach that requires a strong grasp of differentiation techniques.