L31 Calculus with Parametric Curves - and C is travesed...

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Lecture 31: Parametric Curves Consider a parametric curve de ned by the equa- tions x = f ( t ) and y = g ( t ) : the slope of the tangent line to the curve is given dy dx = dy=dt dx=dt = g 0 ( t ) f 0 ( t ) Derivation: Let y = F ( x ) ; so, y = g ( t ) = F ( f ( t )) By chain rule, we have g 0 ( t ) = F 0 ( f ( t )) f 0 ( t ) F 0 ( f ( t )) = g 0 ( t ) f 0 ( t ) hence, y 0 = F 0 ( x ) = g 0 ( t ) f 0 ( t ) The second derivate is given by d 2 y dx 2 = d dx dy dx = d ( y 0 ) dx = dy 0 dt dx dt 1
ex. Consider the parametric curve x = p t; y = t 2 4 = 4 ; t > 0 Find the slope and concavity at the point (2 ; 3) : (8 ; up) 2
ex. Find equation of the tangent line to the parametric curve at the point corresponding to the given value of the parameter. x = 2 t 2 + 1 ; y = 1 3 t 3 t; t = 3 ( y = 2 3 x 20 3 ) 3
ex. Find the points on the curve where the tangent line is horizontal or vertical. x = 2 cos t; y = sin 2 t (HTL at :( p 2 ; 1) ; ( p 2 ; 1) : ( p 2 ; 1) ; ( p 2 ; 1) : VTL at:(2 ; ) ; ( 2 ; 0)) 4
Arc Length Consider a parametric curve C de ned by the equations x = f ( t ) and y = g ( t ), t : If f 0 and g 0 are continuous on [

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