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Exam (elaborations)

EXAMEN CALCULO INGENIERIA AEROSPACIAL

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Examen de practica de calculo de primero de ingeniería aerospacial.

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  • April 7, 2021
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  • 2020/2021
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Càlcul. Graus d’Enginyeria Aeronàutica (EETAC)
Examen de Mig quadrimestre. 19 de novembre de 2012


1. (2 punts) Calculeu la recta perpendicular a la corba d’equació y 2 sin(πx y) + x ln2 (y) = 0
en el punt (−1, 1).

Resolució:

Observem que el punt P pertany a la corba, donat que en substituir x = −1 i y = 1 se’n
satisfà l’equació.
Suposem que, prop del punt P , la corba es pot expressar de la forma y = f (x) i derivem
implı́citament per calcular el pendent de la recta tangent en P :
[ ]

√ √ √ xy (x) ln(y(x))
2y(x)y ′ (x) sin(πx y)+y 2 cos(πx y)· π y + π √ +ln2 (y(x))+2xy ′ (x) =0
2 y(x) y(x)

Ara substituı̈m en P , és a dir x = −1 i y(−1) = 1. ( )

Tenint en compte que ln 1 = sin(−π) = 0 i cos(−π) = −1, obtenim −π 1 − y (−1) 2
= 0.
Aixı́ doncs, el pendent de la recta tangent és y ′ (−1) = 2, i per tant la recta perpendicular
a la corba en el punt P té pendent y′−1
(−1)
= − 12 .
Conclusió: la recta perpendicular demanada té equació y − 1 = − 12 (x + 1).


2. (5 punts)
4+x
a) Raoneu l’existència d’extrems absoluts de la funció g(x) = en cadascun
(1 + x)4
dels intervals [−6, 0], [− 13 , 0] i [ 23 , +∞). Calculeu-los, quan existeixin.
b) Calculeu el polinomi de Taylor de grau 3 en a = 0 de la funció
( 2 ) f (x) = x ln(1 + x) i
utilitzeu-lo per calcular un valor aproximat de V = − 3 ln 3 .
1
∑ k+1
Recordeu: PT (ln(1 + x), 0, n) = nk=1 (−1)k xk .
c) Sabent que f (iv) (x) = 2g(x), useu l’apartat (a) per donar una fita de l’error comès
en l’aproximació de V calculada a l’apartat (b).
x ln(1 + x)
d ) Useu l’apartat (b) per calcular lı́m .
x→0 2x + 3x3 + 4x4
2


Resolució:

a) Per l’estudi dels extrems absoluts de la funció g cal primer observar que és continua
i derivable a tot el seu domini, R \ {−1}. Al punt x = −1 g té una ası́mptota vertical.
Els punts crı́tics seran, a part de x = −1 on g no és derivable, els punts que anul·len
la seva primera derivada
15 + 3x
g ′ (x) = − = 0 ⇔ x = −5.
(1 + x)5

Estudiem el signe de g ′ a tot el domini de g, tenint en compte els punts crı́tics trobats
x = −5 i x = −1:

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