Question

The length of one arch of the curve \(y=3 \sin (2 x)\) is given by \(L=\int_{0}^{\pi / 2} \sqrt{1+36 \cos ^{2}(2 x)} d x\). Estimate \(L\) using the trapezoidal rule with \(n=6\).

Short answer

The estimated length \(L\) is approximately 6.87.

Step-by-step solution

Understand the Problem

The task is to estimate the integral \(L=\int_{0}^{\pi / 2} \sqrt{1+36 \cos ^{2}(2 x)} dx\) using the trapezoidal rule with \(n=6\) subintervals. This is an application of numerical integration to find the length of a curve.

Determine Step Size

For the trapezoidal rule, first calculate the width of each subinterval. Given \(n=6\) subintervals over \([0, \pi/2]\), the step size \(h\) is \(h = \frac{\pi/2 - 0}{6} = \frac{\pi}{12}\).

Compute Subinterval Points

Compute the \(x\) values at which function evaluations are needed: \(x_0 = 0, x_1 = \frac{\pi}{12}, x_2 = \frac{2\pi}{12}, x_3 = \frac{3\pi}{12}, x_4 = \frac{4\pi}{12}, x_5 = \frac{5\pi}{12}, x_6 = \frac{6\pi}{12} = \frac{\pi}{2}\).

Evaluate Function at Subinterval Points

Evaluate the function \(f(x) = \sqrt{1 + 36 \cos^2(2x)}\) at each \(x_i\):- \(f(x_0) = \sqrt{1 + 36 \cos^2(0)} = \sqrt{37}\)- \(f(x_1) = \sqrt{1 + 36 \cos^2(\frac{\pi}{6})} = \sqrt{28}\)- \(f(x_2) = \sqrt{1 + 36 \cos^2(\frac{\pi}{3})} = \sqrt{10}\)- \(f(x_3) = \sqrt{1 + 36 \cos^2(\frac{\pi}{2})} = 1\)- \(f(x_4) = \sqrt{1 + 36 \cos^2(\frac{2\pi}{3})} = \sqrt{10}\)- \(f(x_5) = \sqrt{1 + 36 \cos^2(\frac{5\pi}{6})} = \sqrt{28}\)- \(f(x_6) = \sqrt{1 + 36 \cos^2(\pi)} = \sqrt{37}\)

Apply Trapezoidal Rule

Apply the trapezoidal rule formula:\[L \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + 2f(x_4) + 2f(x_5) + f(x_6) \right]\]Substituting the values, we get:\(L \approx \frac{\pi}{12} \times \frac{1}{2} \left[ \sqrt{37} + 2\sqrt{28} + 2\sqrt{10} + 2(1) + 2\sqrt{10} + 2\sqrt{28} + \sqrt{37} \right]\) Calculate this to get an approximate numerical value.

Calculate Numerical Value

Compute the numerical value:1. \(f(x_0) + f(x_6) = 2\sqrt{37}\)2. \(2f(x_1) + 2f(x_5) = 4\sqrt{28}\)3. \(2f(x_2) + 2f(x_4) = 4\sqrt{10}\)4. \(2f(x_3) = 2\)Thus, the expression becomes:\(L \approx \frac{\pi}{24} [2 \sqrt{37} + 4 \sqrt{28} + 4 \sqrt{10} + 2]\).Evaluating that gives an approximate value for \(L\).

Final Answer Calculation

By simplifying the numerical expressions:Approximate \(L = \frac{\pi}{24} [(2 \times 6.08) + (4 \times 5.29) + (4 \times 3.16) + 2]\).This simplifies to approximately \((0.1309 \times 52.5) \approx 6.87\).

Key concepts

Trapezoidal Rule

The trapezoidal rule is a method used in numerical integration to approximate the definite integral of a function. Unlike methods that require finding the exact antiderivative, this rule allows us to find the area under a curve by dividing it into small, adjacent trapezoids rather than rectangles.

Here's how it works: we first divide the interval of integration into smaller subintervals. In the particular exercise case, the domain y = 3 sin (2x) was segmented into six parts because n = 6 for the trapezoidal rule setup. With these segments, each subinterval spans a height of h = (π/12). Each segment end shares interval boundaries with its neighbor, creating a series of trapezoidal slices that collectively approximate the total area.

After determining the subintervals, we evaluate the function at each point (like evaluating f(x) at x0, x1, etc.) to find the sum. Crucially, remember that the formula for the trapezoidal rule is: \[ L \approx \frac{h}{2} \left[ f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n) \right] \]. This formula is key because it multiplies all inner terms by 2, as these points (except for the endpoints) contribute to two trapezoids each. This step ensures our approximate integral, represented by L, is more accurate, balancing between summation of function evaluations and weighting of the interval steps.

Curve Length Estimation

Estimating the length of a curve can present challenges, especially when dealing with transcendent functions like sine and cosine. When you deal with non-linear expressions like \y = 3 sin(2x)\, the task becomes trickier. To find the length of a curve, we ideally solve \( L = \int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^2} \, dx \).

In the given exercise, the formula is already squared in the integral: \( L = \int_{0}^{\pi / 2} \sqrt{1+36 \cos ^{2}(2 x)} \, dx \). This results from applying the derivative of the function \y = 3 sin(2x), simplifying the expression, then applying the arc length formula. The new integral contains the function to be evaluated numerically.

One practical technique is to simplify the differentiable part: \3 sin(2x)\ to bi-products like \dy/dx\, which can transform the given formula. By transforming the integral into a squareroot-based function, with some constants, this prepares numerical methods such as trapezoidal rule. These techniques offer estimations for curve lengths without solving tricky integration analytic equations straight away.

Definite Integrals

The idea of a definite integral revolves around determining the accumulated "net change" or total area under a curve between two limits. A simple definition could be thinking of it as the sum up of infinitely small product areas along a curvy path.

In the example integral \( \int_{0}^{\pi / 2} \sqrt{1+36 \cos ^{2}(2 x)} \, dx \), the limits define the range on the x-axis: from \(0\) to \(\pi/2\). This segment implies we are only concerned with data points within this bound, sampling these to calculate the sum of vertical slices. The function within the integral will modify to output particular values particular to each resulting point along this interval.

Definite integrals help in finding meaningful measurements, such as area captured below curve(s), average, or total changes. When assessing curves, particularly ones displaying non-standard patterns, definite integrals often direct us towards numerical methods like trapezoidal rule or Simpson's for estimations, ensuring the calculation precision without relying solely on abstract algebraic solutions.