Tangent Lines

I.  Quiz

II.  Homework

III.  The Derivative of Parametric Equations

Suppose that x = x(t) and y = y(t) then

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

as long as dx/dt is nonzero

Example:  Find dy/dx for x(t) = 2cos(t) and y(t) = 2sin(t)

Solution:  We have dx/dt = -2sin(t) and dy/dt = 2cos(t) hence

dy/dx =  (dy/dt)/(dx/dt) = 2cos(t)/-2sin(t) = cot(t)

IV.  The Second Derivative of Parametric Equations

To calculate the second derivative we use the chain rule twice.  

d²y/dx² = d/dx(dy/dx) = d/dx(dy/dt / dx/dt) =  d/dt(dy/dt / dx/dt) dt/dx

= [d/dt(dy/dx / dx/dt]/dx/dt

Hence to find the second derivative, we find the derivative with respect to t of the first derivative and then divide by the derivative of x with respect to t.

Example:

Let x(t) = t³ y(t) = t⁴ then dy/dx = 4t³/3t² = 4/3 t

Hence       

d²y/dx² = [d/dt(4/3 t)] / dx/dt = 4/3 / 3t² = 4/3t²

IV.  Derivatives in Polar Coordinates

Theorem Let r = r(theta) represent a polar curve, then

dy/dx = (dy/dtheta)/(dx/dtheta) = [r'(theta)cos(theta) - r(theta)sin(theta)]/[r'(theta)sin(theta) + r(theta)cos(theta)]

Proof:  Since x = rcos(theta), and y = rsin(theta),

we can substitute in r = r(theta) to get

x = r(theta)cos(theta)

x' = r'(theta)cos(theta) - r(theta)sin(theta) and

y' = r'(theta)sin(theta) + r(theta)cos(theta).

Since dy/dx = (dy/dtheta)/(dx/dtheta), dividing gives the result.