when the curve is horizontal dy/dx = 0
at that point if d^2y/dx^2 is NEGATIVE, the slope gets smaller than zero as you move up in x
That means it starts down, sheds water, maximum
if d^2y/dx^2 is POSITIVE
then the slope gets greater than zero as you move up in x from that horizontal point.
That means it starts up, bottom of bowl, holds water, minimum
can someone please explain to me
1. When doing parametric curves how do you find concavity (concave up or concave down)
2. When doing polar coordinates how do you graph them and how to find different polar points that satisfy a condition like theta<0 or r<0 or r>0 or theta>0
Have an exam and these two sections are not clicking thank you in advance
3 answers
r < 0 does not make sense. r is the distance from the origin.
theta is the distance counterclockwise from the x axis.
theta = 0 is +x direction from the origin
theta = pi/2 or 90 deg is +y from origin
theta = pi or 180 deg is -x from origin
thera = 3 pi/2 or 270 deg is -y from origin
theta is the distance counterclockwise from the x axis.
theta = 0 is +x direction from the origin
theta = pi/2 or 90 deg is +y from origin
theta = pi or 180 deg is -x from origin
thera = 3 pi/2 or 270 deg is -y from origin
If r<0 you move in the direction opposite to θ
(1,π) = (-1,0)
for concavity in parametric equations, it gets a little more complicated finding d^2y/dx^2
y' = dy/dx = (dy/dt) / (dx/dt)
So you have to find all that as a messy rational function of t.
Then d^y/dx^2 = (dy'/dt) / (dx/dt) and you need the quotient rule to come up with dy'/dt.
Or, you may be able to eliminate t as a parameter, getting an equation in just x and y, and then find y" as usual, (though that can also be involved if you get stuck using implicit derivatives)
(1,π) = (-1,0)
for concavity in parametric equations, it gets a little more complicated finding d^2y/dx^2
y' = dy/dx = (dy/dt) / (dx/dt)
So you have to find all that as a messy rational function of t.
Then d^y/dx^2 = (dy'/dt) / (dx/dt) and you need the quotient rule to come up with dy'/dt.
Or, you may be able to eliminate t as a parameter, getting an equation in just x and y, and then find y" as usual, (though that can also be involved if you get stuck using implicit derivatives)