Question

Determinar os vetores velocidade e aceleração para qualquer instante t da função vetorial r (t)=(5cos(π t),8t-t^2,3sen(π t)+4)

Answer 1

$\displaystyle \vec{\text r}(\text t)=[\text{x(t) , y(t) , z(t) }] \\\\ \underline{\text{Sabemos que}}:\\\\ \vec{\text v}(\text t )=\frac{\text{d}{\vec{\text {r}}}}{\text{dt} } \ \ ; \ \ \vec{\text a}(\text t)=\frac{\text{d}\vec{\text v}}{\text {dt}}$

Temos :

$\displaystyke \vec{\text r}(\text t)=[\ \text{5.cos}(\pi.\text t) \ , \ 8\text t-\text t^2 \ , \ \text{3.sen}(\pi.\text t)+4\ ] \\\\\\ \underline{\text{Velocidade}}: \\\\ \vec{\text v}(\text t)= [\ [\text{5.cos}(\pi.\text t)]' \ , \ [8\text t-\text t^2]' \ , \ [\text{3.sen}(\pi.\text t)+4]' \ ] \\\\ \boxed{\vec{\text v}{\text t}=[\ -5.\pi.\text{sen}(\pi.\text t) \ , \ 8-2\text t \ , \ 3.\pi.\text{cos}(\pi.\text t) \ ]\ }\checkmark$

$\displaystyle \underline{\text{Acelera{\c c}{\~a}o}}: \\\\ \vec{\text a}(\text t)=[\ [-5\pi.\text{sen}(\pi.\text t)]' \ , \ [8-2\text t]'\ , \ [3\pi.\text{cos}(\pi.\text t)]' \ ] \\\\ \boxed{\vec{\text a}(\text t)=[-5.\pi^2.\text{cos}(\pi.\text t) \ , \ -2 \ ,\ -3.\pi^2.\text{sen}(\pi.\text t)] } \checkmark$