### Geometry: Parametric curves

### Derivatives of parametric curves

Given a parametric equation #\ivcc{\blue{x(t)}}{\green{y(t)}}#, we can use our notion of a vector to describe the speed and direction of the curve in a point.

Recall the parametric equation describing a throwing a ball. They are given by

\[\orange P \colon \phantom{x}\begin{cases}\blue{x(t)}&=10 \cdot \cos( \theta ) t\\ \green{y(t)}&= 10 \cdot \sin(\theta) t - \frac{1}{2}g \cdot t^2 \end{cases}\]

Here is a picture where we have added a vector #\vec{v}# representing the direction of the ball. The length #\| \vec{v} \|# of #\vec{v}# is equal to the total speed. Note that the ball travels fastest when it is has just been thrown or right at the moment when it hits the ground.

For a real valued function the derivative makes the notion of the direction of a function precise. We generalize this notion to parametric equations. Having related vectors to parametric equations we have the added benefit that we also acquire the definition of the "speed" of a parametric curve.

Let #\orange C# be a parametric curve described by #\cv {\blue{x(t)} \\ \green{y(t)} }#.

The **direction vector** **at** #t_0# of #\orange C# is defined as the vector #\cv{\blue{x'(t)}\\ \green{y'(t)}}#.

The **speed in #t_0#** is given by the length of the direction vector. Explicitely it is given by\[\sqrt{\blue{x'(t_0)}^2+\green{y'(t_0)}^2}\]

**Example**

The direction vector in #t=2# of the curve described by \[\cv{\blue{x(t)}\\ \green{y(t)}} = \cv{\blue{2t-1} \\ \green{t^2}}\] is given by \[\cv{\blue{x'(2)}\\ \green{y'(2)} }=\cv{ \blue{2}\\ \green{4} }\] The speed is therefore given by #\sqrt{\blue{2}^2+\green{4}^2}=2\sqrt{5}#.

As an application of all of the previous theory we can now determine the tangent line at a point to a curve.

Tangent line to a curve

Let #\orange{C}# be the curve represented by #\ivcc{ \blue{ x(t) }}{ \green{ y(t)} }#. Set #\blue{a} = \blue{x'(t_0)}# and # \green{b} = \green{ y'(t_0)}#. Then the derivative of #\orange C# at time #t = t_0# is given by \[ \cv{ \blue{x'(t_0) } \\ \green{y'(t_0)}} = \cv{ \blue{a} \\ \green{b}}\]

We have seen in the expandable block converse statement that any line of the form \[ - \green{b} x + \blue{a} y = c \] has direction vector #\cv{ \blue{a} \\ \green{b}}#.

We can find #c# by substituting #x = \blue{x(t_0)}# and #y = \green{y(t_0)}# in the equation.

**Example**

Suppose #\orange{C}# is represented by #\ivcc{\blue{ \cos (2t) }}{ \green{ \sin(t) + \cos(t)} }# and #t = \pi#.

The derivative at #t = \pi# is given by \[\begin{array}{rcl}\cv{\blue{x'(\pi)} \\ \green{y'(\pi)

}}& = & \cv{ \blue{- 2\sin(2\pi)} \\ \green{\cos(\pi) - \sin(\pi)}} \\ & = & \cv{ \blue{0} \\ \green{-1} }\end{array} \]

So the line #l \colon \green{1} x + \blue{0} y = c# has the correct slope.

Substituting #x = \blue{x(\pi)} = 1# and #y = \green{y(\pi)} = -1# we find #c = 1#.

So vertical line #l \colon x = 1# is the tangent line.

The speed of a curve at #t# is equal to #\sqrt{ x'(t)^2 + y'(t)^2}#. We get

\[\begin{array}{rcl}

\sqrt{ x'(t)^2 + y'(t)^2 } & = & \sqrt{\left(2\cdot t+1\right)^2+9}\\

&& \blue{ \text { determined the derivative of } x(t) \text{ and } y(t) } \\

& = & \sqrt{4\cdot t^2+4\cdot t+10}\\

&& \blue{ \text { expanded the brackets } } \\

& = & \sqrt{4\cdot (-4)^2+4\cdot (-4)+10}\\

&& \blue{ \text { substituted } t = -4 } \\

& = & \sqrt{58}\\

&& \blue{ \text { simplified the expression } }

\end{array}\]

We find the speed at #t = -4# is equal to #\sqrt{58}#.

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