1 . Derivada
d r ⃗ d x = r ⃗ ˙ ( t ) = lim h → 0 r ( t + h ) − r ( t ) h \frac{d\vec{r}}{dx} = {\dot{\vec{r}}(t)} = \lim_{h \rightarrow 0} \frac{\textbf{r}(t+h) - \textbf{r}(t)}{h} d x d r = r ˙ ( t ) = h → 0 lim h r ( t + h ) − r ( t ) O vetor r ⃗ ˙ ( t ) \dot{\vec{r}}(t) r ˙ ( t ) r ⃗ ˙ ( t ) \dot{\vec{r}}(t) r ˙ ( t ) r ⃗ ˙ ( t ) \dot{\vec{r}}(t) r ˙ ( t ) reta tangente \textbf{reta tangente} reta tangente r ⃗ ˙ ( t ) \dot{\vec{r}}(t) r ˙ ( t )
1.1 Teorema
Se
r ⃗ ( t ) = < f ( t ) , g ( t ) , h ( t ) > \vec{r}(t) = \left<f(t), g(t), h(t)\right> r ( t ) = ⟨ f ( t ) , g ( t ) , h ( t ) ⟩ então
r ⃗ ˙ ( t ) = < f ′ ( t ) , g ′ ( t ) , h ′ ( t ) > \dot{\vec{r}}(t) = \left<f^{'}(t), g^{'}(t), h^{'}(t)\right> r ˙ ( t ) = ⟨ f ′ ( t ) , g ′ ( t ) , h ′ ( t ) ⟩ 1.2 Exemplo 1
(a) Determine a derivada de r ( t ) = ( 1 + t 3 ) i ^ + t e − t j ^ + s e n 2 t k ^ r(t) = (1+t^3) \hat{i} + t e^{-t} \hat{j} + sen2t \hat{k} r ( t ) = ( 1 + t 3 ) i ^ + t e − t j ^ + s e n 2 t k ^
r ⃗ ˙ ( t ) = < 3 t 2 , ( 1 − t ) e − t , 2 c o s ( 2 t ) > \dot{\vec{r}}(t) = \left<3t^2, (1-t)e^{-t}, 2cos(2t)\right> r ˙ ( t ) = ⟨ 3 t 2 , ( 1 − t ) e − t , 2 c o s ( 2 t ) ⟩ (b) Encontre a derivada tangente no ponto onde t=0.
r ⃗ ˙ ( t ) = < 0 , 1 , 2 > \dot{\vec{r}}(t) = \left<0,1,2\right> r ˙ ( t ) = ⟨ 0 , 1 , 2 ⟩ 1.3 Exemplo 2
Determine as equações paramétricas para a reta tangente à hélice com equações paramétricas
{ x = 2 c o s t y = s e n t z = t \begin{cases} x = 2cost\\ y = sen t\\ z=t \end{cases} ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x = 2 c o s t y = s e n t z = t no ponto (0,1,π \pi π
r ( t ) = < 2 c o s t , s e n t , t > r ′ ( t ) = < − 2 s e n t , c o s t , 1 > t = π / 2 : r ′ ( π / 2 ) = < − 2 , 0 , 1 > r(t) = \left<2cos t, sen t, t\right>\\ r'(t) = \left<-2sen t, cos t, 1\right>\\ t = \pi / 2: r'(\pi/2) = \left<-2,0,1\right> r ( t ) = ⟨ 2 c o s t , s e n t , t ⟩ r ′ ( t ) = ⟨ − 2 s e n t , c o s t , 1 ⟩ t = π / 2 : r ′ ( π / 2 ) = ⟨ − 2 , 0 , 1 ⟩ As equações paramétricas da reta são:
( x , y , z ) = ( x 0 , y 0 , z 0 ) + t ( a , b , c ) ( x , y , z ) = ( 0 , 1 , π / 2 ) + t ( − 2 , 0 , 1 ) ( x , y , z ) = ( − 2 t , 1 , π 2 + t ) (x,y,z) = (x_0, y_0, z_0) + t(a,b,c)\\ \\ (x,y,z) = (0,1,\pi/2) + t(-2,0,1)\\ \\ (x,y,z) = (-2t, 1, \frac{\pi}{2} + t) ( x , y , z ) = ( x 0 , y 0 , z 0 ) + t ( a , b , c ) ( x , y , z ) = ( 0 , 1 , π / 2 ) + t ( − 2 , 0 , 1 ) ( x , y , z ) = ( − 2 t , 1 , 2 π + t ) 2. Regras de derivação
d d t [ u ( t ) + v ( t ) ] = u ′ ( t ) + v ′ ( t ) \frac{d}{dt}[u(t)+v(t)] = u'(t)+v'(t) d t d [ u ( t ) + v ( t ) ] = u ′ ( t ) + v ′ ( t ) d d t [ c v ( t ) ] = c v ′ ( t ) \frac{d}{dt}[cv(t)] =cv'(t) d t d [ c v ( t ) ] = c v ′ ( t ) d d t [ f ( t ) u ( t ) ] = f ′ ( t ) u ( t ) + f ( t ) u ′ ( t ) \frac{d}{dt}[f(t)u(t)] =f'(t)u(t)+f(t)u'(t) d t d [ f ( t ) u ( t ) ] = f ′ ( t ) u ( t ) + f ( t ) u ′ ( t ) d d t [ u ( t ) ⋅ v ( t ) ] = u ′ ( t ) ⋅ v ( t ) + u ( t ) ⋅ v ′ ( t ) \frac{d}{dt}[u(t) \cdot v(t)] =u'(t) \cdot v(t)+u(t) \cdot v'(t) d t d [ u ( t ) ⋅ v ( t ) ] = u ′ ( t ) ⋅ v ( t ) + u ( t ) ⋅ v ′ ( t ) d d t [ u ( f ( t ) ) ] = f ′ ( t ) u ′ ( f ( t ) ) \frac{d}{dt}[u(f(t))]=f'(t)u'(f(t)) d t d [ u ( f ( t ) ) ] = f ′ ( t ) u ′ ( f ( t ) ) d d t [ u ( t ) ⋅ v ( t ) ] = u ′ ( t ) ⋅ v ( t ) + u ( t ) ⋅ v ′ ( t ) \frac{d}{dt}[u(t) \cdot v(t)] =u'(t) \cdot v(t)+u(t) \cdot v'(t) d t d [ u ( t ) ⋅ v ( t ) ] = u ′ ( t ) ⋅ v ( t ) + u ( t ) ⋅ v ′ ( t ) 3. Integrais
Podemos definir a integral definida de uma função vetorial como é feito para funções reais, mas, nesse caso, a integral definida é um vetor. Mas precisamente,a integral definida no intervalo [a,b] da função vetorial r ( t ) i ^ + g ( t ) j ^ + h ( t ) k ^ r(t)\hat{i}+g(t)\hat{j}+h(t)\hat{k} r ( t ) i ^ + g ( t ) j ^ + h ( t ) k ^
∫ a b r ( t ) d t = lim n → ∞ ∑ i = 1 n r ⃗ ( t i ) Δ t \int_{a}^b r(t) dt= \lim_{n \rightarrow \infty} \sum_{i=1}^n \vec{r}(t_i) \Delta t ∫ a b r ( t ) d t = n → ∞ lim i = 1 ∑ n r ( t i ) Δ t lim n → ∞ [ ( ∑ i = 1 n f ( t i ) Δ t ) i ^ + ( ∑ i = 1 n g ( t i ) Δ t ) j ^ + ( ∑ i = 1 n h ( t i ) Δ t ) k ^ ] \lim_{n \rightarrow \infty} \left[\left( \sum_{i=1}^n f(t_i) \Delta t\right)\hat{\mathbf{i}} + \left( \sum_{i=1}^n g(t_i) \Delta t\right)\hat{\mathbf{j}} +\left( \sum_{i=1}^n h(t_i) \Delta t\right)\hat{\mathbf{k}} \right] n → ∞ lim [ ( i = 1 ∑ n f ( t i ) Δ t ) i ^ + ( i = 1 ∑ n g ( t i ) Δ t ) j ^ + ( i = 1 ∑ n h ( t i ) Δ t ) k ^ ] Assim, a integral de uma função vetorial é o vetor cujas componentes são as integrais definidas das funções componentes.
∫ a b r ( t ) d t = ( ∫ a b f ( t ) d t ) i ^ + ( ∫ a b g ( t ) d t ) j ^ + ( ∫ a b h ( t ) d t ) k ^ \int_{a}^b r(t)dt = \left(\int_{a}^b f(t)dt\right)\hat{\mathbf{i}} + \left(\int_{a}^b g(t)dt\right)\hat{\mathbf{j}} +\left(\int_{a}^b h(t)dt\right)\hat{\mathbf{k}} ∫ a b r ( t ) d t = ( ∫ a b f ( t ) d t ) i ^ + ( ∫ a b g ( t ) d t ) j ^ + ( ∫ a b h ( t ) d t ) k ^ também podemos estender o torema fundamental do cálculo:
∫ a b r ( t ) d t = R ( t ) ∣ a b = R ( b ) − R ( a ) \int_{a}^b r(t)dt =R(t)|_{a}^b =R(b)-R(a) ∫ a b r ( t ) d t = R ( t ) ∣ a b = R ( b ) − R ( a ) Em que R é uma primitiva de r, ou seja, R ⃗ ˙ ( t ) = r ( t ) . \dot{\vec{R}}(t)= r(t). R ˙ ( t ) = r ( t ) .
Escrito por Thays Simeia Rocha Barros
17 de Dezembro de 2020
