# Calculus/Parametric Differentiation

 ← Parametric Introduction Calculus Parametric Integration → Parametric Differentiation

## Taking Derivatives of Parametric Systems

Just as we are able to differentiate functions of ${\displaystyle x}$  , we are able to differentiate ${\displaystyle x}$  and ${\displaystyle y}$  , which are functions of ${\displaystyle t}$  . Consider:

{\displaystyle {\begin{aligned}x&=\sin(t)\\y&=t\end{aligned}}}

We would find the derivative of ${\displaystyle x}$  with respect to ${\displaystyle t}$  , and the derivative of ${\displaystyle y}$  with respect to ${\displaystyle t}$  :

{\displaystyle {\begin{aligned}x'&=\cos(t)\\y'&=1\end{aligned}}}

In general, we say that if

{\displaystyle {\begin{aligned}x&=x(t)\\y&=y(t)\end{aligned}}}

then:

{\displaystyle {\begin{aligned}x'&=x'(t)\\y'&=y'(t)\end{aligned}}}

It's that simple.

This process works for any amount of variables.

## Slope of Parametric Equations

In the above process, ${\displaystyle x'}$  has told us only the rate at which ${\displaystyle x}$  is changing, not the rate for ${\displaystyle y}$  , and vice versa. Neither is the slope.

In order to find the slope, we need something of the form ${\displaystyle {\frac {dy}{dx}}}$  .

We can discover a way to do this by simple algebraic manipulation:

${\displaystyle {\frac {y'}{x'}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}={\frac {dy}{dx}}}$

So, for the example in section 1, the slope at any time ${\displaystyle t}$  :

${\displaystyle {\frac {1}{\cos(t)}}=\sec(t)}$

In order to find a vertical tangent line, set the horizontal change, or ${\displaystyle x'}$  , equal to 0 and solve.

In order to find a horizontal tangent line, set the vertical change, or ${\displaystyle y'}$  , equal to 0 and solve.

If there is a time when both ${\displaystyle x',y'}$  are 0, that point is called a singular point.

## Concavity of Parametric Equations

Solving for the second derivative of a parametric equation can be more complex than it may seem at first glance.

When you have take the derivative of ${\displaystyle {\frac {dy}{dx}}}$  in terms of ${\displaystyle t}$  , you are left with ${\displaystyle {\frac {\frac {d^{2}y}{dx}}{dt}}}$  :

${\displaystyle {\frac {d}{dt}}\left[{\frac {dy}{dx}}\right]={\frac {\frac {d^{2}y}{dx}}{dt}}}$  .

By multiplying this expression by ${\displaystyle {\frac {dt}{dx}}}$  , we are able to solve for the second derivative of the parametric equation:

${\displaystyle {\frac {\frac {d^{2}y}{dx}}{dt}}\times {\frac {dt}{dx}}={\frac {d^{2}y}{dx^{2}}}}$  .

Thus, the concavity of a parametric equation can be described as:

${\displaystyle {\frac {d}{dt}}\left[{\frac {dy}{dx}}\right]\times {\frac {dt}{dx}}}$

So for the example in sections 1 and 2, the concavity at any time ${\displaystyle t}$  :

${\displaystyle {\frac {d}{dt}}[\csc(t)]\times \cos(t)=-\csc ^{2}(t)\times \cos(t)}$

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