Derivada-parametrica-corregida1.pdf

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Derivadas en coordenadas paramétricas Sergio Yansen Núñez 1.

Determine la ecuación de la recta tangente a la curva:

x(t ) = −t cos(t ) y (t ) = π 2 + 2t 2

0 ≤ t ≤ 2π

,

en el punto P = (0, π ) 2.

Determine la ecuación de la recta normal (perpendicular a la tangente) a la curva

( ) ( )

 x(t ) = sen t 2 C:  y (t ) = cos 2 t 2 en t = 3.

π 2

π . 2

Sea x(t ) = 3 cos(t ) ; y (t ) = 4sen(t ) , t ∈ IR .

Calcule 4.

0
,

π d2y cuando t = 2 3 dx

Dadas las ecuaciones paramétricas:   t  x(t ) = ln tg    + cos(t )   2  y (t ) = sen(t )

¿Existe

0
,

dy para todos los valores de t ∈ ]0, π [ ? dx

Para los que exista determine

d2y . dx 2

Sergio Yansen Núñez

Derivadas en coordenadas paramétricas Sergio Yansen Núñez Solución

1. dy y ' (t ) = dx x' (t ) 1

⋅ 4t dy 2 π 2 + t2 = dx − cos(t ) − t (− sen(t ))

dy dx

=0 t =0

y =π ⇒

t=0

y − π = 0 ⋅ ( x − 0)

y =π

x' (t ) =

2.

1

⋅ cos(t 2 ) ⋅ 2t

2

2 sen(t )

y ' (t ) = 2 cos(t 2 ) ⋅ (− sen(t 2 ) ⋅ 2t

dy dx

π t= 2

=

y ' (t ) x' (t ) t =

π 2

= −24 2

La pendiente de la recta normal en el punto es

t=

y−

π 2 1 1 = 4 2 2 2



x=

4

2 2

,

1 4

2 2

y=

1 2

4  2 x−    2  

Sergio Yansen Núñez

Derivadas en coordenadas paramétricas Sergio Yansen Núñez

3.

x' (t ) = −3sen(t )

y ' (t ) = 4 cos(t )

dy y ' (t ) = dx x' (t ) dy 4 cos(t ) 4 = − cot(t ) =− dx 3sen(t ) 3

d  dy    d y dt  dx  = dx dx 2 dt 2

d2y = dx 2

d2y dx 2

t=



π 3

(

)

(

)

4 4 ⋅ − cos ec 2 (t ) − ⋅ − cos ec 2 (t ) 4 3 = 3 =− dx 3sen(t ) 9 sen 3 (t ) dt

=−

4 32 =− 3 9 sen (t ) t = π 27 3 3

Sergio Yansen Núñez

Derivadas en coordenadas paramétricas Sergio Yansen Núñez

x' (t ) =

4.

x' (t ) =

1 t 1 ⋅ sec 2   ⋅ − sen(t ) t 2 2 tg   2 1 1 cos 2 (t ) − sen(t ) = − sen(t ) = sen(t ) sen(t ) t t 2 sen  cos  2 2

y ' (t ) = cos(t ) dy y ' (t ) = dx x' (t ) dy cos(t ) = = tg (t ) dx cos 2 (t ) sen(t )

Para 0 < t < π ,

dy no existe cuando cos(t ) = 0 dx

d  dy    d y dt  dx  = dx dx 2 dt 2

d 2 y sec 2 (t ) sen(t ) π = = para t ≠ 2 2 4 2 dx cos (t ) cos (t ) sen(t )

Sergio Yansen Núñez



t=

π 2

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