Question

Find the lengths of the following curves: a. \(y(x)=x\) for \(x \in[0,2]\). b. \((x, y, z)=(t, \ln t, 2 \sqrt{2} t)\) for \(1 \leq t \leq 2\). c. \(y(x)=\cosh x, x \in[-2,2]\). (Recall the hanging chain example from classical dynamics.)

Short answer

The lengths of the curves are: a) \(2\sqrt{2}\), b) \(\int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + 8} dt\), c) \(2\sinh 2\).

Step-by-step solution

Finding length of y=x

Find the derivative of \(y=x\), which is 1. Apply the arc length formula on interval [0,2] \(\int_{0}^{2} \sqrt{1+(1)^{2}} dx = \int_{0}^{2} \sqrt{2} dx = \sqrt{2}x \mid_{0}^{2} = 2\sqrt{2}\)

Finding length of vector curve

Find the derivative of the vector function \((t, \ln t, 2\sqrt{2} t)\) which is \((1, \frac{1}{t}, 2\sqrt{2})\). Next, calculate the magnitude: \(\sqrt{(1)^{2}+(\frac{1}{t})^{2}+(2\sqrt{2})^{2}} = \sqrt{1 + \frac{1}{t^{2}} + 8}\). Then the length from \(t=1\) to \(t=2\) is \(\int_{1}^{2} \sqrt{1 + \frac{1}{t^{2}} + 8} dt\). This is an integral calculation problem.

Finding length of y=cosh(x)

Observe that the derivative of \(y=\cosh x\) is \(\sinh x\) and so the square is \(\sinh^{2}x\). Use the identity \(1 + \sinh^{2}x = \cosh^{2}x\) to simplify the expression under the square root in the integral: \(\sqrt{1+(\sinh x)^{2}} = \sqrt{\cosh^{2}x} = \cosh x\). So the length from \(x=-2\) to \(x=2\) is \(\int_{-2}^{2}\cosh x dx = \sinh x \mid_{-2}^{2}=\sinh 2 - \sinh(-2) = 2\sinh 2\).

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that deals with the accumulation of quantities and the area under curves. It provides tools for solving problems related to the total size, value, or length, often where the immediate values are not straightforward to calculate but can be described as functions.

For example, in the context of curve length calculation, integral calculus lets us add up an infinite number of infinitesimally small line segments to find the total length of the curve. When we look at the step-by-step solutions for finding the lengths of curves in the exercise, we rely on integral calculus to handle continuous curves. Whether we're working with a simple straight line or a more complex vector function, the arc length can be determined through definite integration between two points.

Even in the specific problem of calculating the length of a curve described by vector functions, it is the integral with respect to the parameter that provides the actual length. Thus, integral calculus is the key to unlocking a variety of applications, including the computation of curve lengths.

Arc Length Formula

The arc length formula is a powerful tool provided by integral calculus that allows us to calculate the distance along a smooth, continuous curve. It's given by the integral of the square root of the sum of the squares of the function's derivative and one, with respect to the independent variable.

In our exercise, for part a, the application of the arc length formula involves integrating the square root of the sum of the square of the derivative of the function, which is 1, and 1, resulting in the calculation of a definite integral that determines the length of the curve from point 0 to point 2.

Essentially, the arc length formula embodies the principle that a curve can be approximated by a series of infinitesimally short straight lines, and by summing the lengths of these lines, we obtain the total length of the curve. Therefore, understanding the arc length formula is essential for solving problems like the ones presented in homework exercises where we calculate the length of various curves.

Vector Functions Derivative

When dealing with curves in space, we often describe them using vector functions. The derivative of a vector function represents the rate of change of the function's components and is crucial for finding the length of the curve represented by the vector.

In the context of our problem, for part b, we first find the derivative of the vector function given. This derivative vector represents the change in the curve's direction and magnitude at each point along the curve. Once we have the derivative, we can determine the infinitesimally small line segments that make up the curve by calculating the magnitude of the derivative.

The length of the curve is then the integral of the magnitude of the derivative over the interval in question. In the presented exercise, we find the length of the curve by integrating the magnitude of the derivative from t=1 to t=2. This process highlights the importance of understanding how to derive vector functions and utilize their derivatives to solve problems involving spatial curves.