Question

In Exercises \(11-14,\) a planar curve \(C\) is given along with a surface \(f\) that is defined over \(C\). Set up the line integral \(\int_{C} f(s) d s,\) then approximate its value using technology. \(C\) is the portion of the parabola \(y=2 x^{2}+x+1\) on [0,1]\(;\) the surface is \(f(x, y)=x^{2}+2 y\).

Short answer

The line integral is approximated by numerically evaluating \( \int_{0}^{1} (5t^2 + 2t + 2) \sqrt{1 + (4t+1)^2} \: dt \).

Step-by-step solution

Parametrize the curve

The curve given is a portion of the parabola described by \( y = 2x^2 + x + 1 \) from \( x = 0 \) to \( x = 1 \). Therefore, we can parametrize the curve \( C \) as \( \mathbf{r}(t) = (t, 2t^2 + t + 1) \), where \( t \) ranges from \( 0 \) to \( 1 \).

Define the function over the curve

The function given is \( f(x, y) = x^2 + 2y \). After substituting the parametric equations, the function becomes \( f(t) = t^2 + 2(2t^2 + t + 1) \). Simplifying this gives \( f(t) = 5t^2 + 2t + 2 \).

Calculate the differential arc length

The differential arc length element \( ds \) on the curve can be calculated using the derivative \( \frac{dy}{dx} \) of the function \( y = 2x^2 + x + 1 \). So, \( \frac{dy}{dx} = 4x + 1 \). Thus, \( ds = \sqrt{1 + (4x+1)^2} \: dx \). For the parametric form, \( ds = \sqrt{1 + (4t+1)^2} \: dt \).

Set up the line integral

The line integral is defined as \( \int_{C} f(s) \: ds \), which becomes \( \int_{0}^{1} f(t) \sqrt{1 + (4t+1)^2} \: dt \) after substituting for \( f(t) \) and \( ds \).

Approximate the line integral numerically

Use numerical integration technology (e.g., a calculator or a software) to approximate the value of the integral \( \int_{0}^{1} (5t^2 + 2t + 2) \sqrt{1 + (4t+1)^2} \: dt \). Running this computation provides an approximate value for the integral.

Key concepts

Parametrization of Curves

Parametrizing curves is a fundamental concept in understanding how to represent a geometric figure analytically. For curves given by equations like parabolas, it's often useful to express them in parametric form. Here, we have the parabola described by the equation \( y = 2x^2 + x + 1 \) within the interval \([0, 1]\).

• The curve can be parametrized using a parameter \( t \), where \( x = t \) and accordingly, \( y = 2t^2 + t + 1 \).
• This transforms our 2D curve into parametric equations: \( \mathbf{r}(t) = (t, 2t^2 + t + 1) \).
• The parameter \( t \) varies from 0 to 1, covering the part of the parabola we're interested in.

These parametric forms let us handle the curve within a specific range and prepare it for further mathematical analysis, such as integration.

Differential Arc Length

Calculating the differential arc length \( ds \) along a curve is crucial for evaluating integrals over curves, known as line integrals. The arc length gives us the infinitesimal piece of the curve's length that we sum over.

• For a curve defined parametrically by \( (x(t), y(t)) \), \( ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \).
• In our specific problem, the curve is defined with \( \, \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 4t + 1 \).
• Consequently, \( ds = \sqrt{1 + (4t+1)^2} \, dt \) which simplifies the complexity in handling the curve's length.

This gives us the essential building block for setting up our line integral, reflecting the curve's geometry.

Numerical Integration

Numerical integration is a powerful tool for approximating the value of integrals that are difficult to calculate by hand. It involves using technology like calculators or software to give an estimated result.

• In scenarios involving complex functions or curves, as with our integral \( \int_{0}^{1} (5t^2 + 2t + 2) \sqrt{1 + (4t+1)^2} \, dt \), exact solutions manually can be tedious.
• Methods include the trapezoidal rule, Simpson's rule, and numerical integration software that can efficiently approximate the values.
• This approach allows students to get tangible results and focus on understanding the interplay of variables in the integral.

By providing flexibility and accessibility, numerical integration aids in solving real-world mathematical problems with high precision.

Parabola

A parabola is a symmetrical, open plane curve formed by the intersection of a cone with a plane parallel to its side. It is a specific quadratic curve that plays a significant role in mathematics.

• The general form for parabolas is \( y = ax^2 + bx + c \), making them easy to recognize and analyze.
• Our exercise focuses on the parabola \( y = 2x^2 + x + 1 \) illuminated within \( x \) values from 0 to 1.
• Parametrization over this domain allows precise calculation of lengths and areas related to the parabola.

Parabolas appear in physics, engineering, and various scientific fields due to their unique reflective and symmetrical properties.

Surface Function

Surface functions involve defining a mathematical function over a surface or a region, often using a 3D representation. It matches height values or other measures to coordinates on a 2D plane.

• In this exercise, the surface function is given by \( f(x, y) = x^2 + 2y \), which can be thought of as a surface over the xy-plane.
• This surface function is crucial since we're integrating over it with a specific parametric curve.
• The representation allows us to evaluate the way dimensions interact, particularly when combining it with differential arc length for integrals.

Surface functions are commonly used in calculus for applications involving physics, graphics, and optimization problems.