Answer to Question #285330 in Electricity and Magnetism for Shitan oinam

Question #285330

calculate the directional derivative of scalar field f(x,y,z)=(x ²+yz+z²)at (1,2,1) along n^=1/3(2i-j+k^)

1
Expert's answer
2022-01-07T09:29:46-0500

The directional derivative in the direction of "\\mathbf{n} = \\dfrac23\\mathbf{i}-\\dfrac13\\mathbf{j} + \\dfrac13\\mathbf{k}" is given as follows:


"\\nabla_{\\mathbf{n}}f = \\nabla f\\cdot \\dfrac{\\mathbf{n}}{|\\mathbf{n}|}"

where "\\nabla f" is the gradient of "f". Thus, obtain:


"\\nabla f = 2x\\mathbf{i} + z\\mathbf{j}+(y+2z)\\mathbf{k}\\\\\n|\\mathbf{n}| = \\sqrt{\\dfrac49 + \\dfrac19 + \\dfrac19} = \\dfrac{\\sqrt{6}}{3}\\\\\n\\dfrac{\\mathbf{n}}{|\\mathbf{n}|} = \\dfrac{1}{\\sqrt{6}}(2\\mathbf{i}-\\mathbf{j} + \\mathbf{k})"

Finally,


"\\nabla_{\\mathbf{n}}f = (2x\\mathbf{i} + z\\mathbf{j}+(y+2z)\\mathbf{k})\\cdot \\dfrac{1}{\\sqrt{6}}(2\\mathbf{i}-\\mathbf{j} + \\mathbf{k})=\\\\\n=\\dfrac{1}{\\sqrt{6}}(4x-z+y+2z) = \\dfrac{1}{\\sqrt{6}}(4x+y+z)"

Substituting point (1,2,1), obtain:


"\\dfrac{1}{\\sqrt{6}}(4\\cdot 1+2+1) = \\dfrac{7}{\\sqrt{6}}"

Answer. "\\dfrac{7}{\\sqrt{6}}".


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