Exam (elaborations)
EXAMEN CALCULO INGENIERIA AEROSPACIAL
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Examen de practica de calculo de primero de ingeniería aerospacial.
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Càlcul. Enginyeria Aeronàutica (EETAC).Examen de Mig Quadrimestre. 24 d’abril de 2013
Cal raonar i desenvolupar adequadament les respostes.
1. (3 punts)
a) Estudieu les superfı́cies de nivell de la funció
x2 + y 2 − 9
f (x, y, z) = .
3z 2
Indicant, a cada cas, de quina superfı́cie es tracta i fent un esbós de la seva gràfica.
b) Parametritzeu la corba obtinguda com a intersecció de les superfı́cies seguents
S1 = {(x, y, z) ∈ R3 | y = x2 }, S2 = {(x, y, z) ∈ R3 | z = x2 + xy − 1}.
Trobeu la recta tangent a l’esmentada corba en el punt (0, 0, −1) .
Solució:
a) La expresión del conjunto de nivel h es
x2 + y 2 − 9
h=
3z 2
o, equivalentemente
x2 + y 2 − 3hz 2 = 9.
Por lo tanto
q
h ∈ (−∞, 0)\{− 13 }, elipsoide de semiejes a = 3, b = 3, c = 3
|h|
.
h = − 13 , esfera de centro (0, 0, 0) y radio 3.
h = 0 cilindro circular de radio 3 y eje de simetrı́a OZ
h > 0 hiperboloide de 1 hoja y eje de simetrı́a OZ
En todos los casos el plano z = 0 está excluido de la superficie.
b) γ(t) = (t, t2 , t2 + t3 − 1), t ∈ R. Comprobamos que el punto (0, 0, −1) pertenece a la
curva y corresponde al valor del parámetro t = 0.
Su vector tangente en un punto arbitrario es
γ 0 (t) = (1, 2t, 2t + 3t2 )
Por lo tanto
γ 0 (0) = (1, 0, 0).
La recta tangente es por tanto
r(t) = (0, 0, −1) + t(1, 0, 0) = (t, 0, −1)
O alternativamente, la intersección de las dos superficies
y = 0 y z = −1
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