CALCULO 3 Resolucion

④.sean f(t) = t²+1 ; 0 ;t³ ) y g(t) = (sent ; cost ; 0 )
A.f (a+b) = ( (a+b)++1 ;0 ; (a+b)³ )
B.g (t-3) = ( sen (t-3) ; -cos ( t-3) ; 0 )
C.f (sent) = ( (sent)²+ 1 ; 0 ; (sent)³ )
g( t²+1) = ( sen (t²+1) ; - cos (t²+1) ; 0)
→ f ( sent) x g (t²+1) = î ĵ k̂sen²t+1 0 sen³t
sen (t²+1) -cos (t²+1) 0
Resolviendo por método de la estrella
= [ (î) (0)(0)+( ĵ ) (sen (t²+1) ) + (sen²t+1 )(-cos (t²+1) ) (k̂) ]
= [ (k̂) (0) (sem (t²+1) ) + ( î) ( -cos ( t²+1)) (sen³t) + (sen²t+1)(0) ]
sen³x sen (t²+1) ĵ - (sen²t+1)) k̂+ sen³t x cos(t²+1) î→f(sent) x g(t²+1) = ( sen³cos (t²+1) ; sen³x sen(t²+1) ; sen²t cos(t²+1) + cos(t²+19
⑤.Define una f del intervlo [a;b] ssobre segmento dee recta de extremos Po y P1de
Supongamos :P₀ (X˳ , Y˳, Z˳) y P1 (X₁ , Y₁ , Z ₁)
A (ax , ay , az ) →vector unitarion paralelo al vecto P
→ De la ecuación dde la recta |f= P₀ +t a |Donde t €[a; b]
P = (x₀ , y₀, z₀)+ t ( aₓ , ay , az )
( x,y,z ) = (x₀+ t ax ; Y₀ + t ay;z₀ + t az )
→ x = x₀+ t aₓ
Y = y₀+ t ay ● f (t) = ( x₀+ t aₓ; y₀+ t ay ; z₀+ t az
Z = z₀+ t az
⑦sean f (t)= (2t¯, g(t) = ( ln (t+1) ;
Calcule f ± g , f ● g , f x g , 4f-2 g
A. f + g = (2t¯¹ ± ln (t+1) ; ± )
Hallando dan :
X = 2t¯¹ ± lm (t+1) → t ≠ 0 ^ t+1 >0 → > - 1 , t – {0}
Y = ± → 4 - t² ≥ 0 ^ t² +2t – 8 ≥ 0
t² - 4 ≤ 0 ^ t² + 2t – 8 ≥ 0
→Dominio : t ( - 1 , + ) [- 2 , 2 ] ( - , - 4 ] [ 2 + )
Dominio f ( t) = { ( x , y ) IR² / t IR ( t > - 1)} – {0}
B. f●g = ( 2t¯¹ ) (ln (t+1)) + ( ) ( )
= + )
t + 1 > 0 ^ t ≠ 0 ^ ( 4 - t²) ( t² + 2t – 8 ) ≥ 0
t > - 1 ^ t≠ 0 ^ ( t² - 4 ) ( t² + 2t – 8 ) ≤ 0
t > - 1 ^ t ≠ 0 ^ ( t – 2 ) ( t + 2 ) ( t + 4 ) ( t – 2 ) ≤ 0
t > - 1 ^ t ≠ 0 ^ ( t -2 )² ( t + 2 ) ( t + 4 ) ≤ 0
t > - 1 ^ t ≠ 0 ^ ( t + 2 ) ( t + 4 ) ≤ 0 ( t - 2)² > 0t ≠ 2
Prescindimos de este valor
Pero teniendo encuenta t≠2
→ Dom f = { ( x , y ) IR² / t IR t ( - 1, 1) ( 1 , 2) ( 2 , + ) }

1.calcule lim f (t
A. f (t) = ( ; t² ,sen t ) , t₀ = 2
limf (t) = (lim ; lim t² ; lim sen t )
t2 t2 t2
limf(t) = ( ; 4 ; sen 2)
B .f (t) = (ln t ; ; ) ; t₀ = 3
lim f (t) = ( lim ln ; lim lim )
= (lim(3) ; ; )
Lim f(t)= ( ln (3) ; , )
C. f(t) = ; ; 5t² ) , t₀ = 3
Lim f(t)= (lim ; lim ; 5(3)² ;lim 5t²)
= ( ; + 5(3)²)
Lim f(t)= ( ; ; 45 ) = ( ; 2; 45)
D. f(t) = ( ; ; ) ; t₀= 0
Lim f(t) = (lim ; lim ; lim )
Aplicando H´OSPITAL | lim f(t) = |
Lim f (t) = ( lim ; lim ; lim )
= 7 ; ; )
Lim f (t) = ( 7 ; ; )
DOMINIO =
D. 4f – 2g :
4 f (t) =(8 t¯¹ ; 4 )
2g(t) = ( 2 ln ( t+1) ; 2 )
4f – 2g = ( 8t¯¹ - 2 lm (t + 1) ; 4 - 2 )
= ( – 2 ln (t+1) ; ; 4 - 2 )
Dominio : t ≠ 0 ^ t + 1 > 0 ^ 4 - t² ≥ 0 ^ t² + 2t – 8 ≥ 0
t ≠ 0 ^ t – 1 ^ t² - 4 ≤ 0 ^ t² + 2t – 8 ≥ 0
t ≠ 0 ^ t – 1 ^ ( t – 2 ) ( t + 2...