{"id":20226,"date":"2024-04-15T05:50:05","date_gmt":"2024-04-15T05:50:05","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=20226"},"modified":"2024-04-15T05:50:05","modified_gmt":"2024-04-15T05:50:05","slug":"the-value-of-the-directional-derivative-of-the-function-phi-x-y-z-xy2-yz2-zx2-at-the-point-2-1-1-in-the-direction-of-the-vector-p-i-2j-2k-is-a-1-b-0-95-c-0-93-d-0-9","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/the-value-of-the-directional-derivative-of-the-function-phi-x-y-z-xy2-yz2-zx2-at-the-point-2-1-1-in-the-direction-of-the-vector-p-i-2j-2k-is-a-1-b-0-95-c-0-93-d-0-9\/","title":{"rendered":"The value of the directional derivative of the function \\[\\phi \\](x, y, z) = xy2 + yz2 + zx2 at the point (2, -1, 1) in the direction of the vector p = i + 2j + 2k is A. 1 B. 0.95 C. 0.93 D. 0.9"},"content":{"rendered":"

[amp_mcq option1=”1″ option2=”0.95″ option3=”0.93″ option4=”0.9″ correct=”option3″]<\/p>\n

The directional derivative of a function $f$ at a point $p$ in the direction of a vector $v$ is given by<\/p>\n

$$D_vf(p) = \\nabla f(p) \\cdot v$$<\/p>\n

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

In this case, we have<\/p>\n

$$\\phi(x, y, z) = xy^2 + yz^2 + zx^2$$<\/p>\n

and<\/p>\n

$$\\nabla \\phi(x, y, z) = (2xy + 2yz, 2xz + 2yz, 2xy + 2xz)$$<\/p>\n

Therefore, the directional derivative of $\\phi$ at the point $(2, -1, 1)$ in the direction of the vector $p = i + 2j + 2k$ is<\/p>\n

$$D_p\\phi(2, -1, 1) = \\nabla \\phi(2, -1, 1) \\cdot p = (2(2)(-1) + 2(-1)(1), 2(2)(1) + 2(-1)(1), 2(2)(-1) + 2(1)(1)) \\cdot (1, 2, 2) = 0.93$$<\/p>\n

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

Here is a brief explanation of each option:<\/p>\n