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Category: Differentiation

Question-126859

Question Number 126859 by sdfg last updated on 24/Dec/20 Answered by mr W last updated on 24/Dec/20 G=10×sin60°×cos45°i+10×sin60°×cos45°j+10×cos60°kG=562i+562j+5k$${H}=−\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{sin}\:\mathrm{30}°\:{i}+\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{cos}\:\mathrm{30}°\:{j}+\mathrm{15}×\mathrm{cos}\:\mathrm{45}°\:{k} \

Question-126854

Question Number 126854 by sdfg last updated on 24/Dec/20 Answered by mathmax by abdo last updated on 24/Dec/20 $$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{r}^{\mathrm{2}} =−\mathrm{k}^{\mathrm{2}} \:\Rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{ik}\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\mathrm{ae}^{\mathrm{ikx}}…