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
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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