{"id":20262,"date":"2024-04-15T05:50:33","date_gmt":"2024-04-15T05:50:33","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20262"},"modified":"2024-04-15T05:50:33","modified_gmt":"2024-04-15T05:50:33","slug":"the-directional-derivative-of-the-scalar-function-fx-y-z-x2-2y2-z-at-the-point-p-1-1-2-in-the-direction-of-the-vector-overrightarrow-rma-3rmhat-i-4rmhat-j","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-directional-derivative-of-the-scalar-function-fx-y-z-x2-2y2-z-at-the-point-p-1-1-2-in-the-direction-of-the-vector-overrightarrow-rma-3rmhat-i-4rmhat-j\/","title":{"rendered":"The directional derivative of the scalar function f(x, y, z) = x2 + 2y2 + z at the point P = (1, 1, 2) in the direction of the vector \\[\\overrightarrow {\\rm{a}} = 3{\\rm{\\hat i}} – 4{\\rm{\\hat j}}\\] is A. -4 B. -2 C. -1 D. 1"},"content":{"rendered":"

[amp_mcq option1=”-4″ option2=”-2″ option3=”-1″ option4=”1″ correct=”option3″]<\/p>\n

The directional derivative of a scalar function $f$ at a point $P$ in the direction of a vector $\\mathbf{a}$ is given by:<\/p>\n

$$D_{\\mathbf{a}}f(P) = \\nabla f(P) \\cdot \\mathbf{a}$$<\/p>\n

where $\\nabla f$ is the gradient of $f$.<\/p>\n

In this case, we have $f(x, y, z) = x^2 + 2y^2 + z$ and $\\mathbf{a} = 3\\hat{\\imath} – 4\\hat{\\jmath}$. Therefore, the directional derivative is:<\/p>\n

$$D_{\\mathbf{a}}f(P) = \\nabla f(P) \\cdot \\mathbf{a} = (2x + 0y + 1z)(3\\hat{\\imath} – 4\\hat{\\jmath}) = 3 – 4 = -1$$<\/p>\n

Therefore, the correct answer is $\\boxed{-1}$.<\/p>\n

To explain each option in brief:<\/p>\n