Question

Arc length calculations with respect to \(y\) Find the arc length of the following curves by integrating with respect to \(y.\) $$x=2 e^{\sqrt{2} y}+\frac{1}{16} e^{-\sqrt{2} y}, \text { for } 0 \leq y \leq \frac{\ln 2}{\sqrt{2}}$$

Short answer

In order to calculate the arc length of the given curve with respect to \(y\), we first need to calculate the derivative of \(x\) with respect to \(y\), which is as follows: $$\frac{dx}{dy} = 2\sqrt{2} e^{\sqrt{2} y}-\frac{1}{8}\sqrt{2} e^{-\sqrt{2} y}.$$ Next, we compute the integrand, which is: $$\sqrt{1+ \left(\frac{dx}{dy}\right)^2} = \sqrt{1 + 8 e^{2\sqrt{2} y}+\frac{1}{32} e^{-2\sqrt{2} y}-4}$$ Finally, the arc length of the curve can be found by integrating the integrand over the given range of \(y\): $$ L = \int_{0}^{\frac{\ln 2}{\sqrt{2}}} \sqrt{1 + 8 e^{2\sqrt{2} y}+\frac{1}{32} e^{-2\sqrt{2} y}-4} dy$$ Numerical methods, such as Simpson's rule or a calculator, are needed to find an approximate value for \(L\).

Step-by-step solution

Calculate the derivative \(\frac{dx}{dy}\)

Calculate the derivative of \(x\) with respect to \(y:\) $$\frac{dx}{dy} = \frac{d}{dy}\left(2 e^{\sqrt{2} y}+\frac{1}{16} e^{-\sqrt{2} y}\right).$$ By applying the chain rule, we get: $$\frac{dx}{dy} = 2\sqrt{2} e^{\sqrt{2} y}-\frac{1}{8}\sqrt{2} e^{-\sqrt{2} y}.$$

Compute the integrand

Compute the integrand \(\sqrt{1+ \left(\frac{dx}{dy}\right)^2}\): $$\sqrt{1+ \left(\frac{dx}{dy}\right)^2} = \sqrt{1+ \left(2\sqrt{2} e^{\sqrt{2} y}-\frac{1}{8}\sqrt{2} e^{-\sqrt{2} y}\right)^2}.$$ Now squaring the term inside the square root: $$1 + \left(2\sqrt{2} e^{\sqrt{2} y}-\frac{1}{8}\sqrt{2} e^{-\sqrt{2} y}\right)^2 = 1 + 8 e^{2\sqrt{2} y}+\frac{1}{32} e^{-2\sqrt{2} y}-4 e^0$$ So, $$\sqrt{1+ \left(\frac{dx}{dy}\right)^2} = \sqrt{1 + 8 e^{2\sqrt{2} y}+\frac{1}{32} e^{-2\sqrt{2} y}-4}$$

Integrate the integrand

Integrate the integrand \(\sqrt{1+ \left(\frac{dx}{dy}\right)^2}\) over the given range of \(y:\) $$ L = \int_{0}^{\frac{\ln 2}{\sqrt{2}}} \sqrt{1 + 8 e^{2\sqrt{2} y}+\frac{1}{32} e^{-2\sqrt{2} y}-4} dy$$ This integral does not have a simple closed-form solution. To find an approximate value for \(L\), you can use numerical methods, such as Simpson's rule, or a calculator or software with the capability to numerically integrate functions.

Key concepts

Integration with respect to y

When we talk about integrating with respect to \(y\), it means we are thinking about changes happening along the vertical axis. This is quite different from the more frequent integrations with respect to \(x\), which consider changes along the horizontal axis. For our specific problem, we aimed to find the length of the curve described by the function \(x = 2 e^{\sqrt{2} y} + \frac{1}{16} e^{-\sqrt{2} y}\).

Integrating with respect to \(y\) involves calculating the length of the curve by considering tiny changes along the \(y\)-axis between 0 and its endpoint, \(\frac{\ln 2}{\sqrt{2}}\). You can think of this like using a ruler to measure a curved path going upwards instead of side-to-side. It's especially useful when the function is more naturally expressed in terms of \(y\) rather than \(x\).

Derivative calculation

Calculating derivatives is a crucial part of solving calculus problems, particularly when dealing with arc length. The derivative \( \frac{dx}{dy} \) tells us how \(x\) changes as \(y\) changes and is essential for integrating with respect to \(y\).

In our example, the derivative of \(x = 2 e^{\sqrt{2} y} + \frac{1}{16} e^{-\sqrt{2} y}\) was calculated using basic rules of differentiation. Each term was processed separately, applying the chain rule when necessary, to yield \( \frac{dx}{dy} = 2\sqrt{2} e^{\sqrt{2} y} - \frac{1}{8}\sqrt{2} e^{-\sqrt{2} y}\).

This derivative tells us the slope of the curve at any given point \(y\). It's used later to find the integrand, \( \sqrt{1+ \left( \frac{dx}{dy} \right)^2} \), which relates directly to the formula for arc length.

Chain rule in calculus

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. In essence, it helps us differentiate functions within functions.

When you have a function like \( e^{\sqrt{2} y} \), you notice that it's not just a simple \( y \) but \(\sqrt{2}y\) inside the exponential. The chain rule allows us to differentiate this by multiplying the derivative of the outer function by the derivative of the inside function. That’s how we ended up with the term \(2\sqrt{2} e^{\sqrt{2} y}\).

Similarly, for \( e^{-\sqrt{2} y} \), applying the chain rule gives us \(-\frac{1}{8}\sqrt{2} e^{-\sqrt{2} y}\). It's all about addressing each layer of the function, ensuring we correctly calculate how it changes with respect to the variable.

Numerical integration methods

Sometimes, integrals are too complex to solve exactly, especially when involving functions leading to expressions without simple antiderivatives. In our case, the integral of the arc length function did not yield a simple solution.

This is when numerical integration methods step in. Methods like Simpson’s rule or using numerical integration software are incredibly handy. These techniques approximate the area under the curve by breaking the problem into smaller, manageable parts and summing those areas. By increasing the number of sections, they can get very accurate results.

In our example, this approach allows us to compute the arc length by approximating the integral of \( \sqrt{1 + 8 e^{2\sqrt{2} y} + \frac{1}{32} e^{-2\sqrt{2} y} - 4} \). Using numerical tools or calculators capable of handling such calculations provides a solution where algebra falls short.