2008-01-11

Today's nonsense
- Due Monday 01-14: pg 725 #1-14.

ex 1) Find dy/dx for the curve given by x = sinT and y = cosT. So you take dx/dt and dy/dt and then you do (dy/dt) / (dx/dt).
So that's (-sin(T)) / (cosT) = - tan(T)

2) x=T^{^(2)} - 4T + 1
y = T^{^(3)} - 3
dx/dt = 2T - 4
dy/dt = 3T^{^(2)}
dy/dx = 3T^{^(2)} / (2T - 4)

Higher order derivatives for Parametric Equations

d^{^(2)}y / dx^{^(2)} = d / dx of [ dy/dx ] = (d/dt of [ dy/dx] ) / (dx/dt) So every time you take another derivative, you take the derivative of the previous divided by dx/dt

Find the second derivative with respect to x where x=(1/2)T^{^(2)} and y=T^{^(2)} + T
dx/dt = T
dy/dt = 2T + 1
dy/dx = (2T+1)/T
(2T+1)/T --> 2 + (1/T) --> (-1/T^{^(2)}) d^{^(2)}y / dx^{^(2)} = (-1/T^{^(2)}) / T = -1/T^{^(3)}

2) Find the second derivative of x = 2 cos(theta) and y = sin(theta)
dx/dt = -2sin(theta)
dy/dt = cos(theta)
dy/dx = (cos(theta)) / (-2sin(theta)) = (-1/2)cot(theta)
d2y/dx2 = (1/2)csc^{^(2)}(theta) all over (-2sin(theta)) (aka dx d theta)
= -(1/4) csc^{^(3)}(theta)

3) For the curve given by x = T^(1/2) and y = (1/4)(T^2 - 4) where T > = 0, find the slope and concavity at (2,3)
dx/dt = (1/2)T
dy/dt = (1/4)(2T)
dy/dx = T^{^(3/2)}
@(2,3) dy/dx = 4^(3/2) = 8
(but you need to go find T first)
sqrt(2) = T
3 = (1/4)(T^2 - 4)
T = 4

d2y/dx2 = (3/2)T^{^(1/2)} / (dx/dt or just 1/(2T^{^(1/2)}))
d2y/dx2 = 3T
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