9.2: Differentiating and Integrating Parametric Functions

When we ventured into applications of integration, we saw how important derivatives are. In this section we investigate the utility of parametric functions to model a variety of situations, so in this section we merge these topics. We discuss the following:

• Differentiating Parametric Functions
• Areas with Parametric Curves

Differentiating Parametric Functions

How do we calculate the slope of a graph parameterized by \(x = f(t)\) and \(y = g(t) \ques\) If \(f\) and \(g\) are differentiable functions, and \(y\) is differentiable with respect to \(x,\) then the Chain Rule (from Section 2.4) states \[\deriv{y}{t} = \deriv{y}{x} \deriv{x}{t} \pd\] Dividing both sides by \(\textderiv{x}{t}\) (assuming it is nonzero) gives \begin{equation} \deriv{y}{x} = \frac{\textderiv{y}{t}}{\textderiv{x}{t}} \pd \label{eq:dy/dx} \end{equation} We can also write \(\textderiv{y}{x} = g'(t)/f'(t).\) \(\eqrefer{eq:dy/dx}\) permits us to calculate slopes to a graph without eliminating the parameter \(t;\) that is, \(\textderiv{y}{x}\) is a function of \(t.\) In words, we divide the derivative of \(y\) with respect to \(t\) by the derivative of \(x\) with respect to \(t.\) This form is easy to remember if you think of canceling the \(\dd t\)'s.

Higher-Order Derivatives To determine higher-order derivatives, we use \(\eqrefer{eq:dy/dx}\) repeatedly. We replace \(y\) (a function of \(t\)) with \(\textderiv{y}{x}\) (another function of \(t\)) to attain \begin{equation} \deriv{^2 y}{x^2} = \deriv{}{x} \par{\deriv{y}{x}} = \frac{\ds \deriv{}{t}\left(\deriv{y}{x}\right)}{\textderiv{x}{t}} \pd \label{eq:2-dy/dx} \end{equation} For the general, \(n\)th derivative we replace \(y\) with \(\textderiv{^{n - 1} y}{x^{n - 1}}\) to get \begin{equation} \deriv{^n y}{x^n} = \deriv{}{x} \par{\deriv{^{n - 1}y}{x^{n - 1}}} = \frac{\ds \deriv{}{t}\left(\deriv{^{n - 1}y}{x^{n - 1}}\right)}{\textderiv{x}{t}} \pd \label{eq:n-dy/dx} \end{equation}

From Section 3.3, a graph is concave up if \(\textderiv{^2 y}{x^2} \gt 0\) and concave down if \(\textderiv{^2 y}{x^2} \lt 0.\) Using \(\eqrefer{eq:dy/dx},\) we attain \[\deriv{y}{x} = \frac{\textderiv{y}{t}}{\textderiv{x}{t}} = \frac{12t^2}{-2} = -6t^2 \pd\] Then \(\eqref{eq:2-dy/dx}\) gives \[\deriv{^2 y}{x^2} = \frac{\ds \deriv{}{t}\left(-6t^2\right)}{\textderiv{x}{t}} = \frac{-12t}{-2} = 6t \pd \] The second derivative is positive for \(t \gt 0\) and negative for \(t \lt 0.\) Hence, the graph is concave up over \(t \gt 0\) and concave down for \(t \lt 0.\) The graph is shown in Figure 2, in which arrowheads represent the direction of increasing \(t.\)

Areas with Parametric Curves

Integrating Vertically In Cartesian coordinates, the area bounded by a curve and the \(x\)-axis from \(x = a\) to \(x = b\) is \(\int_a^b y \di x.\) Recall that the integral is positive if \(y \gt 0\) and negative if \(y \lt 0.\) We now consider the area under a curve defined parametrically by \begin{equation} x = f(t) \lspace y = g(t) \cma \label{eq:par-eqs} \end{equation} where \(f\) and \(g\) are both differentiable. Our goal is to use the Substitution Rule (from Section 4.4) to convert \(\int_a^b y \di x\) to be an integral in terms of \(t.\) From \(\eqref{eq:par-eqs}\) we see \(\textderiv{x}{t} = f'(t),\) or \(\dd x = f'(t) \di t.\) To express the bounds of the integral in terms of \(t,\) let \(t_a\) and \(t_b\) be values of \(t\) such that \(f(t_a) = a\) and \(f(t_b) = b.\) Assuming the parametric curve is traced out once over \(t_a \leq t \leq t_b,\) the integral \(\int_a^b y \di x\) becomes \begin{equation} \int_{t_a}^{t_b} g(t) f'(t) \di t \pd \label{eq:area-x} \end{equation}

Integrating Horizontally The area between a curve and the \(y\)-axis from \(y = c\) to \(y = d\) is \(\int_c^d x \di y,\) which is positive if \(x \gt 0\) and negative if \(x \lt 0.\) In \(\eqref{eq:par-eqs}\) we have \(\textderiv{y}{t} = g'(t),\) or \(\dd y = g'(t) \di t.\) Let \(t_c\) and \(t_d\) satisfy \(g(t_c) = c\) and \(g(t_d) = d.\) The integral \(\int_c^d x \di y\) therefore becomes \begin{equation} \int_{t_c}^{t_d} f(t) g'(t) \di t \pd \label{eq:area-y} \end{equation}

The graph of the curve and its enclosed region \(R\) are shown in Figure 3. The curve hits the \(y\)-axis when \(x = 0,\) which occurs when \(t = 0\) and \(t = \pi;\) at these times, \(y = 0\) and \(y = \pi^2,\) respectively. Thus, in Cartesian coordinates the area of \(R\) is \[\int_{y = 0}^{y = \pi^2} x \di y \pd\] We now convert this integral to be entirely in terms of \(t\) using the Substitution Rule: We are given \(x = 2 \sin t,\) and differentiating \(y\) shows \(\dd y = 2t \di t.\) In addition, when \(y = 0,\) \(t = 0;\) when \(y = \pi^2,\) \(t = \pi.\) The integral \(\int_0^{\pi^2} x \di y\) therefore becomes \[\int_0^{\pi} (2 \sin t) (2t) \di t = 4 \int_0^\pi t \sin t \di t \pd\] Using Integration by Parts (from Section 6.1), we find \[ \ba 4 \par{-t \cos t + \sin t} \intEval_0^\pi &= 4 \par{-\pi \cos \pi + \sin \pi} - 4(0 + 0) \nl &= \boxed{4 \pi} \ea \]

By symmetry, the entire enclosed area is \(4\) times the area of one subregion—say, region \(R,\) enclosed in the first quadrant (see Figure 4). The area of \(R\) in Cartesian coordinates is \[A_R = \int_{x = 0}^{x = 4} y \di x \pd\] We use the Substitution Rule to convert all quantities to be in terms of \(t \col\) The problem gives \(y = \sin 2t,\) and differentiating \(x\) shows \(\dd x = -4 \sin t \di t.\) Also, \(x = 0\) when \(t = \pi/2\) and \(x = 4\) when \(t = 0.\) Hence, \[ \ba A_R &= \int_{\pi/2}^0 (\sin 2t) (-4 \sin t) \di t \nl &= -4 \int_{\pi/2}^0 (\sin 2t) (\sin t) \di t \pd \ea \] Because \(\sin 2t = 2 \sin t \cos t,\) \[ \ba A_R &= -4 \int_{\pi/2}^0 2 \sin^2 t \cos t \di t \nl &= -8 \int_{\pi/2}^0 \sin^2 t \cos t \di t \nl &= 8 \int_0^{\pi/2} \sin^2 t \cos t \di t \pd \ea \] Letting \(u = \sin t,\) we see \(\dd u = \cos t \di t.\) When \(t = 0,\) \(u = 0;\) when \(t = \pi/2,\) \(u = 1.\) Then \[A_R = 8 \int_0^1 u^2 \di u = \tfrac{8}{3} u^3 \intEval_0^1 = \tfrac{8}{3} \pd\] Hence, the entire area is \[A = 4 A_R = 4 \par{\tfrac{8}{3}} = \boxed{\tfrac{32}{3}}\]