Parametric Form of the Derivative
Let's say you have the parametric equation x = f(t), and y = g(t). dy/dx, or the combined equation (if you were to merge the two equations into one) equals (dy/dt)/(dx/dt) or g'(t)/f'(t).
If you let Δy = g(t + Δt) - g(t), and Δx = f(t + Δt), you can say
dy/dx = lim as Δx->0 (Δy/Δx) = lim[ (g(t + Δt) - g(t))/(f(t + Δt) - f(t))]
If you divide both the numerator and denominator by Δx, you get lim [(g(t + Δt) - g(t))/Δt] / lim [(f(t + Δt) - f(t)/Δt].
Recall that that's the definition of a derivative, so you get dy/dx = g'(t)/f'(t) = (dy/dt)/(dx/dt).
(Proof from textbook)
Second and Higher Degree Derivatives
The second derivative is the derivative of the derivative / derivative of x , so it's:
The 3rd,4th, 5th, etc. derivatives are almost the same as the second derivative. In the third derivative, you replace 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's not a linear change).
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 = t3/2
Derivative at (2,3): x = 2, t = 4, 43/2 = 8
Second derivative = d/dt [t3/2]/(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.
Sometimes, a problem might ask what the speed of a set of parametric equations is.
Sometimes, a problem might ask about the velocity.