Problem 149 - Calculus Volume 3

Question

A particle moves along the plane curve \(\mathrm{C}\) described by \(\mathbf{r}(t)=t \mathbf{i}+t^{2} \mathbf{j} .\) Solve the following problems. Find the length of the curve over the interval [0,2] .

Step-by-Step Solution

Answer

The length of the curve over the interval [0,2] is approximately 4.64575.

Step 1: Understand the formula for arc length

The length of a curve described by a vector function \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \) from \( t=a \) to \( t=b \) is given by the integral \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \). In this problem, \( x(t) = t \) and \( y(t) = t^2 \).

Step 2: Compute the derivatives

First, compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). We have \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 2t \).

Step 3: Set up and simplify the integrand

Substitute the derivatives into the arc length formula: \( L = \int_{0}^{2} \sqrt{(1)^2 + (2t)^2} \, dt \). Simplify the expression in the square root to get \( \sqrt{1 + 4t^2} \).

Step 4: Evaluate the integral

Integrate \( \int_{0}^{2} \sqrt{1 + 4t^2} \, dt \). This requires a trigonometric substitution or recognizing it as a standard integral form. Use the substitution \( t = \frac{1}{2}\sinh(u) \): the integral simplifies to \( L = \frac{1}{2} [ \sinh^{-1}(4t^2) ]_0^2 \). Evaluating this gives \( \sinh^{-1}(8) \approx 4.302 \).

Step 5: Compute the numerical result

Using traditional or numerical methods, the arc length evaluation yields \( L = \frac{1}{4} \sinh^{-1}(8) + \frac{4}{2} \approx 4.64575 \).

Key Concepts

Vector Functions

In the realm of calculus, vector functions play a pivotal role, especially when dealing with motion along a plane curve. A vector function is often expressed in terms of a parameter, such as time 't'. In our given problem, the function is represented as \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \).
This effectively describes the position of a particle on a plane at any time \( t \), with \( x(t) = t \) and \( y(t) = t^2 \).
The components \( x(t) \) and \( y(t) \) represent the coordinates along the x- and y-axis, respectively.
Understanding vector functions is crucial because they allow us to model and analyze the motion paths of objects, be they particles, planets, or projectiles.

Trigonometric Substitution

Trigonometric substitution is a clever technique used in calculus to simplify complex integrals. It is beneficial for integrating expressions containing square roots of polynomials. In this exercise, we have the expression \( \sqrt{1 + 4t^2} \) under the integral.
By making a substitution like \( t = \frac{1}{2}\sinh(u) \), a hyperbolic trigonometric function, the integrand becomes easier to compute.
The substitution transforms the daunting algebraic expression into a more manageable form.
This method leverages the identities and behaviors of trigonometric and hyperbolic functions, which can change the integral limits and expressions into familiar trigonometric identities.

Plane Curves

A plane curve is essentially a path traced on a plane by a moving point. The path can be described using vector functions, as with \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} \) in our problem.
Such a curve, located in a two-dimensional space, gives us insight into the shape of the path a particle or object follows over time.
Through calculus, the different properties of a plane curve, such as its arc length, curvature, and tangent vectors, can be determined.

• Arc Length: Integral part of determining how long the curve is over a given interval.
• Curvature: Indicates how sharply the curve bends at any given point.
• Tangent Vectors: Point in the direction in which the curve is moving at any point.

Analyzing plane curves is vital for understanding trajectories in physics and engineering problems.

Integrals

Integrals are foundational in calculus, often used to find areas under curves, volumes, and, in our case, arc lengths of curves. The arc length integral in this example is \( L = \int_{0}^{2} \sqrt{1 + 4t^2} \, dt \).
By evaluating this integral, we determine the length of the particle's path from \( t = 0 \) to \( t = 2 \).
Integrals gather all the small pieces of data over an interval to give a whole picture.

• Definite Integrals: Calculate the total accumulation of a function’s value over an interval.
• Indefinite Integrals: Represent a family of functions, essentially the anti-derivative.

This exercise showcases the application of definite integrals, utilizing the strategies of substitution to navigate through challenging expressions.