Parametric Equations and Derivatives

Parametric Form of the Derivative

First Derivative

Let's say you have the parametric equation $$\begin{cases}x = f(t)\\y = g(t)\end{cases}$$ $\dfrac{dy}{dx}$, or the combined derivative (if you were to merge the two equations into one), equals $$\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} \text{ or } \dfrac{g'(t)}{f'(t)}$$

Proof

If you let $\Delta y = g(t + \Delta t) - g(t)$ and $\Delta x = f(t + \Delta t)$, you can say $$\dfrac{dy}{dx} = \lim_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x} = \lim_{\Delta t \to 0} \dfrac{g(t + \Delta t) - g(t)}{f(t + \Delta t) - f(t)}$$ If you divide the numerator and denominator by $\Delta t$, you get an interesting result. $$\lim_{\Delta t \to 0} \dfrac{g(t + \Delta t) - g(t)}{f(t + \Delta t) - f(t)} = \lim_{\Delta t \to 0} \dfrac{\left(\dfrac{g(t + \Delta t) - g(t)}{\Delta t}\right)}{\left(\dfrac{f(t + \Delta t) - f(t)}{\Delta t}\right)}$$ That's the definition of a derivative! This equals $$\dfrac{g'(t)}{f'(t)}$$ So, we have $$\dfrac{dy}{dx} = \dfrac{g'(t)}{f'(t)} = \dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}$$

Second and Higher Degree Derivatives

The second derivative is the derivative of the first derivative / derivative of x , so it's:

$$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \boxed{\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}}$$ The 3rd,4th, 5th, etc. derivatives are almost the same as the second derivative. In the third derivative, you replace $\dfrac{dy}{dx}$ with the second derivative. The idea is the same, you want to find the derivative of that, but you must divide by the derivative of x, because x changes differently than usual (if $x = t^2$, it does not change linearly).

Example

$$\begin{cases}x = \sqrt{t}\\y = \frac{1}{4}\left(t^2 - 4\right)\\t \ge 0\end{cases}$$ Find the derivative and second derivative at (2,3) $$\dfrac{dy}{dx} = \dfrac{\dfrac{t}{2}}{\frac{1}{2\sqrt{t}}} = t^\frac{3}{2}$$ Derivative at (2,3): x = 2, t = 4, $(4)^\frac{3}{2} = \boxed{8}$
Second derivative = $$\dfrac{\frac{d}{dt}t^\frac{3}{2}}{\dfrac{1}{2\sqrt{t}}} = \dfrac{\frac{3}{2}\sqrt{t}}{\dfrac{1}{2\sqrt{t}}} = 3t = 3(4) = \boxed{12}$$ So, the graph is concave up at (2,3), and its slope is 8.

x = √t, y = 1/4(t² - 4), t≥0

Find the derivative and second derivative at (2,3)

dy/dx = (dy/dt)/(dx/dt) =

(t/2) /(1/(2√t) = t√t = t^3/2

Derivative at (2,3): x = 2, t = 4, 4^3/2 = 8

Second derivative = d/dt [t^3/2]/(dx/dt) = (3/2)√t/(1/(2√t)) = 3t

When t = 4, 3t = 12.

So, the graph is concave up at (2,3), and the slope is 8.

Speed

Sometimes, a problem might ask what the speed of a set of parametric equations is. 

Example

If x(t) = 2t, y(t) = sin(4t), what is the speed at t = 3? $$x'(t) = 2, y'(t) = 4\cos(4t)$$ We know the x and y speeds, so if you add the two vectors, you get the overall speed. $$\text{Speed(t)} = \sqrt{(x'(t))^2 + (y'(t))^2}$$

Velocity

Sometimes, a problem might ask about the velocity.

Example (Same as before, just with velocity)

If x(t) = 2t, y(t) = sin(4t), what is the velocity at t = 3? $$x'(t) = 2, y'(t) = 4\cos(4t)$$ So, you can write it as a vector $$<2, 4\cos{12}> = <2, 3.375>$$

Finding a tangent line

This is the same as finding the tangent to any other curve. You create a line containing that point and the slope. You use point-slope form, so if the point is (a,b), you find what t is and you do y = b + g'(t)/f'(t)(x-a).