r = 5 - 4sinθ
find the tangents at the pole. create a table of polar points of interest.
please write it out. i need help to check my answer.
x = r cos t
and r = f ( theta )
dx / dt = r ( - sin t ) + cos t ( dr / d theta )
dx / dt = (dr / d theta ) cos t - r sin t
y = r sin theta
dy dt = r cos theta + sin theta ( dr / d theta )
dy / dt = (dr / d theta ) sin theta + r cos theta
dy / dx = ( dy / dt ) / (dx / dt )
now to do the substituting.
dy / dt = (- 4 cos theta ) sin theta + ( 5 - 4 sin theta ) cos theta
dy / dt = - 8 sin theta cos thtea + 5
dy / dt = - 4 sin ( 2 theta ) + 5
dx / dt = ( - 4 cos theta ) cos theta - ( 5 - 4 sin theta ) sin theta
dx / dt = - 4 cos (2 theta ) - 5
dy / dx = ( - 4 sin ( 2 theta ) + 5 ) / ( - 4 cos (2 theta ) - 5 )
do what you like with the equations.
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x = r cos t
and r = f ( theta )
dx / dt = r ( - sin t ) + cos t ( dr / d theta )
dx / dt = (dr / d theta ) cos t - r sin t
y = r sin theta
dy dt = r cos theta + sin theta ( dr / d theta )
dy / dt = (dr / d theta ) sin theta + r cos theta
dy / dx = ( dy / dt ) / (dx / dt )
now to do the substituting.
dy / dt = (- 4 cos theta ) sin theta + ( 5 - 4 sin theta ) cos theta
dy / dt = - 8 sin theta cos thtea + 5
dy / dt = - 4 sin ( 2 theta ) + 5
dx / dt = ( - 4 cos theta ) cos theta - ( 5 - 4 sin theta ) sin theta
dx / dt = - 4 cos (2 theta ) - 5
dy / dx = ( - 4 sin ( 2 theta ) + 5 ) / ( - 4 cos (2 theta ) - 5 )
do what you like with the equations.