1.2 Calculus of parametric curves  (Page 2/6)

Finding a tangent line

Find the equation of the tangent line to the curve defined by the equations

$x\left(t\right)=t^2-3,y\left(t\right)=2t-1,\mathrm{-3}\le t\le 4\text{when}t=2.$

First find the slope of the tangent line using [link] , which means calculating $x^{\prime }\left(t\right)$ and $y^{\prime }(t)\text{:}$

$\begin{array}{c}{x}^{\prime }\left(t\right)=2t\hfill \\ y^{\prime }\left(t\right)=2.\hfill \end{array}$

Next substitute these into the equation:

$\begin{array}{c}\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt}\hfill \\ \frac{dy}{dx}=\frac{2}{2t}\hfill \\ \frac{dy}{dx}=\frac{1}{t}.\hfill \end{array}$

When $t=2,$ $\frac{dy}{dx}=\frac{1}{2},$ so this is the slope of the tangent line. Calculating $x\left(2\right)$ and $y\left(2\right)$ gives

$x\left(2\right)={\left(2\right)}^2-3=1\text{and}y\left(2\right)=2\left(2\right)-1=3,$

which corresponds to the point $\left(1,3\right)$ on the graph ( [link] ). Now use the point-slope form of the equation of a line to find the equation of the tangent line:

$\begin{array}{ccc}\hfill y-y_0& =\hfill & m\left(x-x_0\right)\hfill \\ \hfill y-3& =\hfill & \frac{1}{2}\left(x-1\right)\hfill \\ \hfill y-3& =\hfill & \frac{1}{2}x-\frac{1}{2}\hfill \\ \hfill y& =\hfill & \frac{1}{2}x+\frac{5}{2}.\hfill \end{array}$

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Second-order derivatives

Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y=f\left(x\right)$ is defined to be the derivative of the first derivative; that is,

Since $\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt},$ we can replace the $y$ on both sides of this equation with $\frac{dy}{dx}.$ This gives us

If we know $dy\text{/}dx$ as a function of t, then this formula is straightforward to apply.

Integrals involving parametric equations

Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x\left(t\right)=t-\text{sin}t,y\left(t\right)=1-\text{cos}t.$ Suppose we want to find the area of the shaded region in the following graph.

To derive a formula for the area under the curve defined by the functions

we assume that $x\left(t\right)$ is differentiable and start with an equal partition of the interval $a\le t\le b.$ Suppose $t_0=a<t_1<t_2<\text{⋯}<t_n=b$ and consider the following graph.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is $y\left(x\left({\bar{t}}_i\right)\right)$ for some value ${\bar{t}}_i$ in the i th subinterval, and the width can be calculated as $x\left(t_i\right)-x\left(t_{i-1}\right).$ Thus the area of the i th rectangle is given by