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).


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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

	$\vec{A} . \vec{B} = 5 t^{2} \sin (t) - t \cos t$ $\Rightarrow \frac{d (\vec{A} . \vec{B})}{d t} = 5 t^{2} \cos t + 10 t \sin t + t \sin t - \cos t$ $= 5 t^{2} \cos t + 11 t \sin t - \cos t$ $A \land B = (\begin{matrix} 5 t^{2} \\ t \\ - t^{3} \end{matrix}) \times (\begin{matrix} \sin t \\ - \cos t \\ 0 \end{matrix}) = (\begin{matrix} t^{3} \cos t \\ - t^{3} \sin t \\ t^{3} \cos t \end{matrix})$ $\frac{d (A \times B)}{d t} = (- t^{3} \sin t + 3 t^{2} \cos t) i + . . .$
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