Problem 68 - Algebra and Trigonometry

Question

Sketch the function represented by the given parametric equations. Then use the graph to determine each of the following: a. intervals, if any, on which the function is increasing and intervals, if any, on which the function is decreasing. b. the number, if any, at which the function has a maximum and this maximum value, or the number, if any, at which the function has a minimum and this minimum value. \(x=3(t-\sin t), y=3(1-\cos t) ; 0 \leq t \leq 2 \pi\)

Step-by-Step Solution

Answer

After applying the method described above, the positions, the intervals of increase or decrease, as well as the maximum or minimum values can be obtained. Due to the complexity of this function and its parametric representation, one can not easily provide explicit intervals or points without performing the algebra and graphing the function either by hand or with a computer.

Step 1: Sketching the function

The function is given by two parametric equations: \(x=3(t-\sin t)\) and \(y=3(1-\cos t)\) for \(0 \leq t \leq 2\pi\). Plot these equations on a graph to visualize the function. Use a graphing tool or plot points by substituting different values of \(t\) in the equation.

Step 2: Determine Intervals of Increase and Decrease

To find the intervals where the function is increasing or decreasing, we need to look at the derivative of \(y\) with respect to \(x\). In case of parametric equations, this derivative is given by \(\frac{dy}{dt} / \frac{dx}{dt}\). Compute these derivatives and look where this ratio is positive (the function is increasing) and where it is negative (the function is decreasing).

Step 3: Identify maximum or minimum points

To find the maximum or minimum values of a function, we need to take a look at the second derivative, \(\frac{d^2y}{dx^2}\) which is the derivative of \(\frac{dy}{dx}\). When this value changes sign, we have a maximum or minimum. The maximum/minimum values are the y-values at these points.

Step 4: Apply bounded interval

Remember that the task gives a bounded interval for \(t\) from \(0\) to \(2\pi\). Therefore, the increasing, decreasing and maximum/minimum points should be found within this interval.

Key Concepts

Increasing and Decreasing Intervals

In any parametric equation, finding where a function is increasing or decreasing helps us understand its behavior. For the functions defined by parametric equations, such as our example with \(x=3(t-\sin t)\) and \(y=3(1-\cos t)\), we determine these intervals by examining the derivative of \(y\) with respect to \(x\). To calculate the derivative in parametric form, we use the formula:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.\]

• Compute \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) by differentiating both \(y\) and \(x\) with respect to \(t\).
• Plug these derivatives into the formula to find \(\frac{dy}{dx}\).
• If \(\frac{dy}{dx} > 0\), the function is increasing; if \(\frac{dy}{dx} < 0\), the function is decreasing.

To solve our specific problem, carefully calculate over the interval \(0 \leq t \leq 2\pi\) and identify sections of \(t\) where \(\frac{dy}{dx}\) switches from positive to negative or vice versa. This gives you clear sections where the function's behavior shifts between increasing and decreasing.

Maximum and Minimum Values

Maximum and minimum values of a function tell us the highest or lowest points. To find these in a parametric context, after finding \(\frac{dy}{dx}\), we must examine the second derivative, \(\frac{d^2y}{dx^2}\). This second derivative helps determine concavity, which is crucial for spotting maxima and minima.Begin by computing \(\frac{dy}{dx}\), then differentiate it with respect to \(t\) to obtain \(\frac{d^2y}{dx^2}\).

• If \(\frac{d^2y}{dx^2} > 0\), it suggests a local minimum.
• If \(\frac{d^2y}{dx^2} < 0\), it indicates a local maximum.

Check these points within the interval \(0 \leq t \leq 2\pi\) to pinpoint t-values where these maxima or minima occur. Also note the corresponding \(y\)-values at these points, as they represent the actual maximum or minimum values of the function.

Derivatives of Parametric Equations

Understanding derivatives for parametric equations is integral when analyzing curves. Parametric equations define a set of coordinates on a plane using a third variable, typically \(t\), which doesn't appear in the traditional Cartesian equation format.For parametric forms:

• The first derivative, \(\frac{dy}{dx}\), provides the slope of the tangent to the curve at each point, helping us understand increasing or decreasing behavior.
• The second derivative, \(\frac{d^2y}{dx^2}\), reveals concavity and helps identify local maxima and minima points on the graph.

To master these concepts, practice computing these derivatives with various parametric equations.
This will enhance your grasp of how parametric curves behave and how to translate parametric information into graphical insights. Always keep an eye on the variable \(t\), as it provides a crucial link between the equations and the resulting shape of the graph.