Question #190769

If š‘… š‘¢ = š‘¢ āˆ’ š‘¢ 2 š‘– + 2š‘¢ 3 š‘— āˆ’ 3š‘˜, find ļæ½š‘‘ļæ½ �� �� ׬ .�� š‘. ą¶± 1 2 š‘…(š‘¢)š‘‘ļæ½


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š‘…(š‘¢)=š‘¢āˆ’š‘¢2š‘–+2š‘¢3š‘—āˆ’3š‘˜š‘… (š‘¢ )= š‘¢ āˆ’ š‘¢ 2 š‘– + 2š‘¢ 3 š‘— āˆ’ 3š‘˜\\


dRdu=lim⁔u→0R(u+Ī”uāˆ’R(u))Ī”u\dfrac{dR}{du}=\lim_{u\to0}\dfrac{R(u+\Delta u-R(u))}{\Delta u}\\


=lim⁔u→0[x(u+Ī”u)2i+y(u+Ī”u)6j]+z(u+Ī”u)3k]āˆ’[x(u)i+y(u)j+z(u)k]Ī”u=\lim_{u\to0}\dfrac{[x(u+\Delta u)2i+y(u+\Delta u)6j]+z(u+\Delta u)3k]-[x(u)i+y(u)j+z(u)k]}{\Delta u}


=lim⁔u→0x(u+Ī”u)āˆ’x(u)Ī”u2i+y(u+Ī”u)āˆ’y(u)Ī”u6j+z(u+Ī”u)āˆ’z(u)Ī”u3k=\lim_{u\to0}\dfrac{ x(u+\Delta u)-x(u)}{\Delta u}2i+\dfrac{ y(u+\Delta u)-y(u)}{\Delta u}6j+\dfrac{ z(u+\Delta u)-z(u)}{\Delta u}3k


=dxdu2i+dydu6j+dzdu3k=\dfrac{dx}{du}2i+\dfrac{dy}{du}6j+\dfrac{dz}{du}3k




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