Given-A-5t-2-i-tj-t-3-k-and-B-sin-t-i-cos-t-j-Calculate-d-A-B-dx-d-A-B-dx-and-d-A-A-dx-

Question Number 159111 by mathocean1 last updated on 13/Nov/21

G i v e n \vec{A} = 5 t^{2} \vec{i} + t \vec{j} - t^{3} \vec{k} a n d

\vec{B} = s i n (t) \vec{i} - c o s (t) \vec{j} .

C a l c u l a t e \frac{d (\vec{A} . \vec{B})}{d x}; \frac{d (\vec{A} \land \vec{B})}{d x} a n d

\frac{d (\vec{A} . \vec{A})}{d x} .

Answered by physicstutes last updated on 13/Nov/21

\vec{A} . \vec{B} = 5 t^{2} \sin (t) - t \cos t

\Rightarrow \frac{d (\vec{A} . \vec{B})}{d t} = 5 t^{2} \cos t + 10 t \sin t + t \sin t - \cos t

= 5 t^{2} \cos t + 11 t \sin t - \cos t

A \land B = (\begin{matrix} 5 t^{2} \\ t \\ - t^{3} \end{matrix}) \times (\begin{matrix} \sin t \\ - \cos t \\ 0 \end{matrix}) = (\begin{matrix} t^{3} \cos t \\ - t^{3} \sin t \\ t^{3} \cos t \end{matrix})

\frac{d (A \times B)}{d t} = (- t^{3} \sin t + 3 t^{2} \cos t) i + \dots