A hipérbole \mathrm{z}=x^{2}-y^{2} e o cilindro x^{2}+y^{2}=1

13.2 Derivadas e Integrais de Funções Vetoriais

9-16 Determine a derivada da função vetorial.
9. \mathbf{r}(t)=\left\langle t \operatorname{sen} t, t^{2}, t \cos 2 t\right\rangle
10. \mathbf{r}(t)=\left\langle\operatorname{tg} t, \operatorname{sec} t, 1 / t^{2}\right\rangle
11. \mathbf{r}(t)=\mathbf{i}-\mathbf{j}+e^{4 t} \mathbf{k}
12. \mathbf{r}(t)=\frac{1}{1+t} \mathbf{i}+\frac{t}{1+t} \mathbf{j}+\frac{t^{2}}{1+t} \boldsymbol{k}
13. \mathbf{r}(t)=e^{t^{2}} \mathbf{i}-\mathbf{j}+\ln (1+3 t) \mathbf{k}
14. \mathbf{r}(t)=a t \cos 3 t \mathbf{i}+b \operatorname{sen}^{3} t \mathbf{j}+c \cos ^{3} t \mathbf{k}
15. \mathbf{r}(t)=\mathbf{a}+t \mathbf{b}+t^{2} \mathbf{c}
16. \mathbf{r}(t)=t \mathbf{a} \times(\mathbf{b}+t \mathbf{c})

Passo 1

O enunciado pede para determinar a derivada \mathbf{r}'(t) de funções vetoriais dadas. Vamos aplicar a derivação componente a componente, usando as regras de derivada: produto, quociente, cadeia e derivadas usuais de funções trigonométricas e exponenciais.

Passo 2

Para \mathbf{r}(t)=\langle t\operatorname{sen} t,\;t^{2},\;t\cos(2t)\rangle:
– Na primeira componente, use produto:
\frac{d}{dt}[t\operatorname{sen} t]=1\cdot\operatorname{sen} t + t\cdot\cos t
– Na segunda, \frac{d}{dt}[t^{2}]=2t.
– Na terceira, produto e cadeia:
\frac{d}{dt}[t\cos(2t)]=1\cdot\cos(2t)+t\cdot(-\sin(2t)\cdot2) =\cos(2t)-2t\sin(2t).

Passo 3

Para \mathbf{r}(t)=\langle \operatorname{tg} t,\;\operatorname{sec} t,\;t^{-2}\rangle:
\frac{d}{dt}[\operatorname{tg} t]=\operatorname{sec}^{2}t.
\frac{d}{dt}[\operatorname{sec} t]=\operatorname{sec} t\tan t.
\frac{d}{dt}[t^{-2}]=-2t^{-3}=-\frac{2}{t^{3}}.

Passo 4

Para \mathbf{r}(t)=\mathbf{i}-\mathbf{j}+e^{4t}\mathbf{k}:
\mathbf{i} e \mathbf{j} são constantes, derivada zero.
\frac{d}{dt}[e^{4t}]=4e^{4t}.

Passo 5

Para \mathbf{r}(t)=\frac{1}{1+t}\,\mathbf{i}+\frac{t}{1+t}\,\mathbf{j}+\frac{t^{2}}{1+t}\,\mathbf{k}:
\frac{d}{dt}\bigl[(1+t)^{-1}\bigr]=- (1+t)^{-2}=-\frac{1}{(1+t)^{2}}.
\frac{d}{dt}\bigl[\frac{t}{1+t}\bigr]=\frac{(1+t)-t}{(1+t)^{2}}=\frac{1}{(1+t)^{2}}.
\frac{d}{dt}\bigl[\frac{t^{2}}{1+t}\bigr] =\frac{2t(1+t)-t^{2}}{(1+t)^{2}} =\frac{2t+t^{2}}{(1+t)^{2}} =\frac{t(2+t)}{(1+t)^{2}}.

Passo 6

Para \mathbf{r}(t)=e^{t^{2}}\mathbf{i}-\mathbf{j}+\ln(1+3t)\mathbf{k}:
\frac{d}{dt}[e^{t^{2}}]=2t\,e^{t^{2}}.
\frac{d}{dt}[-\mathbf{j}]=\mathbf{0}.
\frac{d}{dt}[\ln(1+3t)]=\frac{3}{1+3t}.

Passo 7

Para \mathbf{r}(t)=a\,t\cos(3t)\,\mathbf{i}+b\,\operatorname{sen}^{3}t\,\mathbf{j}+c\,\cos^{3}t\,\mathbf{k}:
\frac{d}{dt}[t\cos(3t)] =\cos(3t)-3t\sin(3t), multiplicado por a.
\frac{d}{dt}[\operatorname{sen}^{3}t]=3\operatorname{sen}^{2}t\cos t, multiplicado por b.
\frac{d}{dt}[\cos^{3}t]=3\cos^{2}t(-\sin t)=-3\cos^{2}t\sin t, multiplicado por c.

Passo 8

Para \mathbf{r}(t)=\mathbf{a}+t\,\mathbf{b}+t^{2}\,\mathbf{c} com \mathbf{a},\mathbf{b},\mathbf{c} constantes:
\frac{d}{dt}[\mathbf{a}]=\mathbf{0},
\frac{d}{dt}[t\,\mathbf{b}]=\mathbf{b},
\frac{d}{dt}[t^{2}\,\mathbf{c}]=2t\,\mathbf{c}.

Passo 9

Para \mathbf{r}(t)=t\,\mathbf{a}\times\bigl(\mathbf{b}+t\,\mathbf{c}\bigr):
– Escreva t\,\mathbf{a}\times(\mathbf{b}+t\mathbf{c}) =t\,\mathbf{a}\times\mathbf{b}+t^{2}\,\mathbf{a}\times\mathbf{c}.
– Derive cada termo: \frac{d}{dt}[t\,\mathbf{a}\times\mathbf{b}] =\mathbf{a}\times\mathbf{b}, \frac{d}{dt}[t^{2}\,\mathbf{a}\times\mathbf{c}] =2t\,\mathbf{a}\times\mathbf{c}.

Resposta

  1. \mathbf{r}'(t)=\langle \operatorname{sen} t + t\cos t,\;2t,\;\cos(2t)-2t\sin(2t)\rangle
  2. \mathbf{r}'(t)=\langle \operatorname{sec}^{2}t,\;\operatorname{sec} t\tan t,\;-2/t^{3}\rangle
  3. \mathbf{r}'(t)=4e^{4t}\,\mathbf{k}
  4. \mathbf{r}'(t)=\bigl\langle -\tfrac{1}{(1+t)^{2}},\;\tfrac{1}{(1+t)^{2}},\;\tfrac{t(2+t)}{(1+t)^{2}}\bigr\rangle
  5. \mathbf{r}'(t)=\langle 2t\,e^{t^{2}},\;0,\;3/(1+3t)\rangle
  6. \mathbf{r}'(t)=\bigl\langle a\bigl(\cos(3t)-3t\sin(3t)\bigr),\;3b\,\operatorname{sen}^{2}t\cos t,\;-3c\,\cos^{2}t\sin t\bigr\rangle
  7. \mathbf{r}'(t)=\mathbf{b}+2t\,\mathbf{c}
  8. \mathbf{r}'(t)=\mathbf{a}\times\mathbf{b}+2t\,\mathbf{a}\times\mathbf{c}

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