Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given parametric equation and find its length. \(x=\sqrt{5} \sin 2 t-2, y=\sqrt{5} \cos 2 t-\sqrt{3} ; 0 \leq t \leq \pi / 4\)

Step-by-Step Solution

Answer

The arc length is \(\frac{\pi \sqrt{5}}{2}\).

Step 1: Analyze Parametric Equations

The given parametric equations are \(x = \sqrt{5} \sin 2t - 2\) and \(y = \sqrt{5} \cos 2t - \sqrt{3}\). These equations describe a circular path since they are based on sine and cosine functions. Specifically, if we set \(x' = \sqrt{5} \sin \theta - 2\) and \(y' = \sqrt{5} \cos \theta - \sqrt{3}\), this represents a circle centered at \((-2, -\sqrt{3})\) with a radius of \(\sqrt{5}\).

Step 2: Sketch the Circle

To sketch the graph of the parametric equations, plot a circle centered at \((-2, -\sqrt{3})\). The radius is \(\sqrt{5}\). The range of \(t\) from \(0\) to \(\pi/4\) implies this section of the circle is in the first quadrant where the sine and cosine are positive/reducing to zero.

Step 3: Compute Derivatives for Arc Length

To find the length of the curve between \(0\) and \(\pi/4\), compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). We have \(\frac{dx}{dt} = 2\sqrt{5} \cos 2t\) and \(\frac{dy}{dt} = -2\sqrt{5} \sin 2t\).

Step 4: Set Up Arc Length Integral

The arc length \(L\) of a parametric curve is given by the integral \(L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into this formula.

Step 5: Calculate the Integral

Simplify inside the square root: \(\sqrt{(2\sqrt{5} \cos 2t)^2 + (-2\sqrt{5} \sin 2t)^2} = \sqrt{20}\). This simplifies the integral to \(L = \int_0^{\pi/4} \sqrt{20} \, dt = \sqrt{20} \left[ t \right]_0^{\pi/4}\). Calculate the definite integral yields \(L = \sqrt{20}\cdot \frac{\pi}{4}\).

Step 6: Simplify the Arc Length

Since \(\sqrt{20} = 2\sqrt{5}\), the arc length \(L\) becomes \(2\sqrt{5} \cdot \frac{\pi}{4}\), which simplifies to \(\frac{\pi \sqrt{5}}{2}\).

Key Concepts

Arc Length

Arc length is a measure of the distance along a curve. For parametric equations, it is essential to have derivatives of both the x and y components with respect to a parameter, such as t, which in our case is time. To find the arc length of a parametric curve, use the formula:

• \( L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)

Here:

• \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are the derivatives with respect to t.
• \([a, b]\) is the interval over which the curve is measured.

The integral evaluates the length for each increasing value of t which, when summed, gives the total arc length. In the given problem, we evaluated from \(0\) to \(\pi/4\). This approach captures the accumulation of tiny segments of straight line approximations, resembling the whole curve.

Derivative Calculations

Derivatives tell us the rate at which one variable changes with respect to another. In parametric equations, it’s important to find the derivatives of both x and y with respect to the parameter t. This helps in assessing how the components of the curve behave individually over time.

• The derivative \(\frac{dx}{dt}\) provides the rate of change of \(x\) with respect to \(t\).
• Similarly, \(\frac{dy}{dt}\) gives the rate of change of \(y\) with respect to \(t\).

In the context of the exercise, these derivatives were used to set up the integral for calculating the arc length. Here, derivative calculations offer insight into the instantaneous direction and speed of points as they trace the curve.

Integral Evaluation

Integral evaluation is the process of finding the value of an integral, essentially adding up an infinite number of infinitesimal data points to find a whole value. When dealing with arc length, integral evaluation collects the bits of length across the curve's t-interval. We began with computing the derivatives as the first key step, followed by substituting these into the arc length integral formula to facilitate evaluation:

• Integrating over the bounds aims to compute the total path traveled by a point along the curve.
• In our specific case, we simplified the integral to \(\sqrt{20} \left[ t \right]_{0}^{\pi/4}\). This meant we evaluated the length of the curve over a specified segment.

The simplicity of numerical outcomes during evaluation often relies on effective algebraic simplification before integration, aiding in more straightforward solutions.

Trigonometric Functions

Trigonometric functions like sine and cosine, often appear in parametric equations due to their periodic nature. They effectively describe oscillatory motions, such as circles or waves. In our parametric equations:

• \(\sin(2t)\) and \(\cos(2t)\) dictate the paths through their inherent properties of oscillation between -1 and 1.
• Their usage allows the equation to represent a circular motion, which is centered at \((-2, -\sqrt{3})\) with a radius of \(\sqrt{5}\).

Understanding these functions within the context of parametric forms aids in determining changes in angle and radius simultaneously over the defined interval. Trigonometric identities also help simplify equations, enhancing our understanding and efficiency in finding required solutions.