Question

Consider the parametric equations \(x=\frac{4 t}{1+t^{3}}\) and \(y=\frac{4 t^{2}}{1+t^{3}}\) (a) Use a graphing utility to graph the curve represented by the parametric equations. (b) Use a graphing utility to find the points of horizontal tangency to the curve. (c) Use the integration capabilities of a graphing utility to approximate the arc length of the closed loop. (Hint: Use symmetry and integrate over the interval \(0 \leq t \leq 1 .)\)

Short answer

Use a graphing utility to graph the equations, find points of horizontal tangency, and finally calculate the arc length of the closed loop over the interval 0≤t≤1. Results will vary depending on the graphing tool and the precision of your calculations.

Step-by-step solution

Graph the Parametric Equations

Using a graphing utility program, input the provided parametric equations for x and y. Different values of t will produce different points (x, y) on the graph, which together form the curve.

Find Points of Horizontal Tangency

Next, the points of horizontal tangency on the curve need to be found. This happens when the derivative of y with respect to x equals zero. In other words, at the points of tangency, the slope of the tangent line is zero. Use the graphing utility to calculate the derivative and solve equation \(\frac{dy}{dx}=0\) to get the t values. Then substitute these values into the original equations to find corresponding x,y points.

Calculate the Arc Length

Next, the objective is to calculate the arc length of the closed loop of the curve. To compute the curve's length we can use the formula \(s=\int \sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}} dt\) . Use the utility's integration feature to perform this operation over the interval 0≤t≤1.