Calculus/Parametric Differentiation

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Taking Derivatives of Parametric Systems[edit]

Just as we are able to differentiate functions of x, we are able to differentiate x and y, which are functions of t. Consider:

x = \sin t \

y = t \

We would find the derivative of x with respect to t, and the derivative of y with respect to t:

x' = \cos t \

y' = 1 \

In general, we say that if

x = f(t) \ and y = g(t) \ then:

x' = f'(t) \ and y' = g'(t) \

It's that simple.

This process works for any amount of variables.

Slope of Parametric Equations[edit]

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

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

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

{y' \over x'} = {{dy \over dt} \over {dx \over dt}} = {dy \over dx}

So, for the example in section 1, the slope at any time t:

{1 \over \cos t} = \sec t

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

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

If there is a time when both x' and y' are 0, that point is called a singular point.

Concavity of Parametric Equations[edit]

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 {dy \over dx} in terms of t, you are left with {{d^2y \over dx}\over dt}:

{d\over dt}[{dy \over dx}] = {{d^2y \over dx}\over dt}.

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

{{d^2y \over dx}\over dt} \times {dt \over dx} = {d^2y\over dx^2}.

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

{d\over dt}[{dy \over dx}] \times {dt \over dx}

So for the example in sections 1 and 2, the concavity at any time t:

{d\over dt}[\csc t] \times \cos t = - \csc^2 t \times \cos t