Tangent Line Problems

[MarissaDelozier]

Find equation of tangent line in Cartesian coordinates of the curve given:
r= 3(1-cos 0) at pie/2 ( cos 0 is cos angle symbol)

 

[arthur ohlsten]

to start pie you eat pi is a Greek letter

slope = dy/dx at @=pi/2 or 90*
eq. R=3[1-cos@)
at @=pi/2 R=3

find dy/dx
Given; y=Rsin @ x=R cos @ by definition of polar coordinates
then by derivative:

dy = R d (sin@) + dR cos@
dy= Rcos@ + dR cos@
at @=pi/2 R=3
dy= 0

x= R cos@
dx= -R sin@ + dR cos@
at @=pi/2 R=3
dx= -3 +0

dy/dx = -0/3 or horizontal line

Please check for errors
Arthur

 

[skeeter]

let t = theta

x = r*cost
dx/dt = -r*sint + cost*(dr/dt)

y = r*sint
dy/dt = r*cost + sint*(dr/dt)

r = 3(1 - cost)
dr/dt = 3sint

at t = pi/2 ...
r = 3
dr/dt = 3
x = 0
dx/dt = -3
y = 3
dy/dt = 3

dy/dx = (dy/dt)/(dx/dt) = (3)/(-3) = -1

tangent line equation is ...
y - 3 = -1(x - 0)
y = -x + 3

 

[galactus]

Since \(\displaystyle \L\\x=rcos{\theta}\)

\(\displaystyle \L\\r=3(1-\frac{x}{r})\)

\(\displaystyle \L\\r'=\frac{-3}{r}\)

\(\displaystyle \L\\r=3(1-cos{\frac{\pi}{2}})=3\)

\(\displaystyle \L\\3(1-\frac{x}{r})=\frac{-3}{r}x+b\)

\(\displaystyle \L\\3(1-\frac{x}{3})=\frac{-3}{3}x+b\)

\(\displaystyle \L\\3-x=-x+b\). So, b=3

\(\displaystyle \L\\y=-x+3\)