Derivatives of parametric curves

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}#.

Relation to ordinary derivative

Earlier, we gave the example of the graph of a function as a parametric curve. Let us now compare our ordinary concept of derivative with the concept of derivative mentioned above. We can describe the graph as the parametric curve #c(t)=\cv{t\\f(t)}#, whose derivative is given by #c'(t)=\cv{1\\f'(t)}#. The speed in a point #t_0# is therefore given by \[\sqrt{1+f'(t_0)^2}\]

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.

Picture

\[\phantom{x}\]

We shown above that the vertical line #l \colon x = 1# is the tangent line to the curve #\orange{C}# at #t = \pi#. In the picture this is demonstrated. It is possible to adjust the value of #t = t_0# using the slider. The line #l# is the tangent line at the time #t#.

Let #P# be a point whose motion is described by
\[\begin{cases}
x(t) & = -2\cdot t^2+4\cdot t+4, \\
y(t) & = 4-4\cdot t.
\end{cases}\]
Determine the speed #v_t# of #P# at #t = 5#.

#v_t = 4\cdot \sqrt{17}#

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

\[\begin{array}{rcl}
\sqrt{ x'(t)^2 + y'(t)^2 } & = & \sqrt{\left(4-4\cdot t\right)^2+16}\\
&& \blue{ \text { determined the derivative of } x(t) \text{ and } y(t) } \\
& = & \sqrt{16\cdot t^2-32\cdot t+32}\\
&& \blue{ \text { expanded the brackets } } \\
& = & \sqrt{16\cdot (5)^2-32\cdot (5)+32}\\
&& \blue{ \text { substituted } t = 5 } \\
& = & 4\cdot \sqrt{17}\\
&& \blue{ \text { simplified the expression } }
\end{array}\]

We find the speed at #t = 5# is equal to #4\cdot \sqrt{17}#.

New example