Question #190769

If 𝑅 đ‘ĸ = đ‘ĸ − đ‘ĸ 2 𝑖 + 2đ‘ĸ 3 𝑗 − 3𝑘, find īŋŊ𝑑īŋŊ īŋŊīŋŊ īŋŊīŋŊ ×Ŧ .īŋŊīŋŊ 𝑏. āļą 1 2 𝑅(đ‘ĸ)𝑑īŋŊ


1
Expert's answer
2021-05-11T11:25:22-0400

Ans:-

𝑅(đ‘ĸ)=đ‘ĸ−đ‘ĸ2𝑖+2đ‘ĸ3𝑗−3𝑘𝑅 (đ‘ĸ )= đ‘ĸ − đ‘ĸ 2 𝑖 + 2đ‘ĸ 3 𝑗 − 3𝑘\\


dRdu=lim⁡u→0R(u+Δu−R(u))Δu\dfrac{dR}{du}=\lim_{u\to0}\dfrac{R(u+\Delta u-R(u))}{\Delta u}\\


=lim⁡u→0[x(u+Δu)2i+y(u+Δu)6j]+z(u+Δu)3k]−[x(u)i+y(u)j+z(u)k]Δu=\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+Δu)−x(u)Δu2i+y(u+Δu)−y(u)Δu6j+z(u+Δu)−z(u)Δu3k=\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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