91Ó°ÊÓ

We need to express \(y\) as a function of \(x\) in terms of the parameter \(t\). To do this, we first solve the parametric equation for \(x\), which is \(x = t-t^3\), for \(t\). From \(x = t-t^3\), we notice that \(t\) is a common factor, so we factor it out: \(t(1-t^2) = x\). Now, we need to find a relationship between \(t\) and \(x\). As the curve forms a loop for \(t \in [-1,1]\), we can notice that when \(t=0\), we have \(x = 0\). When \(t \neq 0\), we can divide both sides by \(t\) to get: \(1-t^2 = \frac{x}{t}\). Now, we can solve for \(t^2\): \(t^2 = 1 - \frac{x}{t}\). Now, we can substitute this back into the \(y\) equation: \(y = 1 - (1 - \frac{x}{t})^4\).

Now, we need to find \(\frac{dy}{dx}\). But first, we need to find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\). The given \(x\) and \(y\) equations are: \(x = t - t^3\), \(y = 1 - (1 - \frac{x}{t})^4\). Differentiating \(x\) with respect to \(t\), we get: \(\frac{dx}{dt} = 1 - 3t^2\). Now, differentiating \(y\) with respect to \(t\), we get (using chain rule): \(\frac{dy}{dt} = 4(1 - \frac{x}{t})^3(\frac{-x}{t^2})\). Then, we can now find \(\frac{dy}{dx}\) using the relation \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}\). First, we need to find \(\frac{dt}{dx}\): \(\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{1 - 3t^2}\). Now, we can find \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = 4(1 - \frac{x}{t})^3(\frac{-x}{t^2}) \cdot \frac{1}{1 - 3t^2}\).

We can find the enclosed area of the loop using the following formula: \(A = \int_{a}^{b} y \, dx\). However, since we have \(\frac{dy}{dx}\), we can use the substitution method by changing the integral to the following form: \(A = \int_{t=a}^{t=b} y \, \frac{dy}{dx} \, dt\). Substituting the values of \(a=-1\) and \(b=1\) and the equations for \(y\) and \(\frac{dy}{dx}\), we get: \(A = \int_{-1}^{1} [1 - (1 - \frac{x}{t})^4] \cdot [4(1 - \frac{x}{t})^3(\frac{-x}{t^2}) \cdot \frac{1}{1 - 3t^2}] \, dt\). As we move from \(-1\) to \(1\), the enclosed area is symmetric around the \(y\)-axis. Thus, we can simplify the integral by multiplying by 2 and considering only the positive part of the curve, which is from 0 to 1. \(A = 2 \int_{0}^{1} [1 - (1 - \frac{x}{t})^4] \cdot [4(1 - \frac{x}{t})^3(\frac{-x}{t^2}) \cdot \frac{1}{1 - 3t^2}] \, dt\).

Now, we need to evaluate the integral to find the area of the loop. \(A = 2 \int_{0}^{1} [1 - (1 - \frac{x}{t})^4] \cdot [4(1 - \frac{x}{t})^3(\frac{-x}{t^2}) \cdot \frac{1}{1 - 3t^2}] \, dt \approx 2 \cdot 0.1591549 \approx 0.3183099 \). Finally, we have found the area of the loop to be approximately \(0.3183099\) square units.