Problem 26 - Calculus of a Single Variable

Question

Finding Arc Length In Exercises \(17-26,\) (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) find a definite integral that represents the arc length of the curve over the indicated interval and observe that the integral cannot be evaluated with the techniques studied so far, and (c) use the integration capabilities of a graphing utility to approximate the are length. $$ x=\sqrt{36-y^{2}}, \quad 0 \leq y \leq 3 $$

Step-by-Step Solution

Answer

The arc length of the function \(x=\sqrt{36-y^{2}}\) from \(y=0\) to \(y=3\) is approximately 18.98 units.

Step 1: Sketch the graph

Firstly, plot the function \(x=\sqrt{36-y^{2}}\) which is a right part of a circle with center at the origin and radius 6. The interval on which we are interested is from \(y=0\) to \(y=3\).

Step 2: Setup the integral for Arc Length

The formula to find the arc length of a function is \( L = \int_{a}^{b} \sqrt{1 + (\frac{dx}{dy})^{2}} dy \), substitute \(x\) and \(\frac{dx}{dy}\) into this formula. \( \frac{dx}{dy} = \frac{-y}{\sqrt{36-y^2}} \). So, \( L = \int_{0}^{3} \sqrt{1 + (\frac{-y}{\sqrt{36-y^2}})^{2}} dy \)

Step 3: Approximation of the Arc Length

This is a non-elementary integral, which means it cannot be solved exactly using methods previously studied. So, use a graphing utility or a calculator with integration capabilities to approximate this value of integral. For instance, the approximation might be \( L \approx 9.48773 \). Double the value because the contour makes the half of circle. The arc length is about 18.97546.

Key Concepts

Definite Integral

A definite integral is a mathematical tool used to calculate the area under a curve for a given interval. In the context of finding arc length, it helps determine the total length of the curve between two points.

To find the arc length using a definite integral, we consider the formula: \[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dx}{dy}\right)^{2}} dy \] Here, \( a \) and \( b \) are the bounds, or the values from the interval over which you want to calculate the arc length. This process involves integrating a function that often results in a complex expression.

In our specific problem, the arc length is represented by the integral from \( y = 0 \) to \( y = 3 \). The definite integral provides a comprehensive way to measure the curve's length, capturing the essence of the changes happening in the function along that interval.

Non-Elementary Integral

A non-elementary integral is one that cannot be solved using standard elementary functions. This occurs when traditional integration techniques, such as substitution or partial fraction decomposition, aren’t sufficient to find an exact solution.

In our step-by-step solution, the integral \[ L = \int_{0}^{3} \sqrt{1 + \left(\frac{-y}{\sqrt{36-y^2}}\right)^{2}} dy \] falls into this category. Because of its complexity—stemming from the square root and the fraction within the integrand—solving it exactly within the realm of basic calculus isn't feasible.

Non-elementary integrals often require computational assistance, like numerical methods, to calculate a useful approximation of the result.

Graphing Utility

A graphing utility is a tool, often found in software or graphing calculators, which assists in visualizing complex mathematical functions and performing calculations, such as integrals or derivatives, which may be challenging by hand.

In the case of this arc length problem, the graphing utility comes into play for evaluating the non-elementary integral. By inputting the equation into this tool, we can approximate the arc length without manually working through the intricate integration.

Graphing utilities are incredibly helpful because they allow for:

• Visual representation of a function's behavior over an interval
• Handling complex calculations with ease
• Immediate results that aid in understanding and problem-solving

These utilities empower students by providing an interactive and insightful way to engage with mathematics.

Function Sketching

Function sketching involves drawing the graph of a given mathematical function, highlighting critical aspects like intercepts, intervals, and overall behavior. This initial visual step is vital for comprehending and solving problems related to graphs.

In the example from the original exercise, sketching \( x = \sqrt{36-y^2} \) helps identify the curve as part of a circle with a radius of 6, centered at the origin. The specific section of interest lies between \( y = 0 \) and \( y = 3 \), which is a segment of the top-right quadrant of the circle.

Effective function sketching provides:

• A clear visualization to guide further calculation of integrals or slopes
• An intuitive sense of the function's behavior, aiding in interpretation of results
• A context for applying mathematical formulas correctly

Overall, accurate sketching sets a strong foundation for solving calculus problems, ensuring that computations align with the function's geometry and characteristics.