[amp_mcq option1=”\\[{\\rm{4\\hat i}} + 2{\\rm{\\hat j}} – 3{\\rm{\\hat k}}\\]” option2=”\\[{\\rm{4\\hat i}} – 6{\\rm{\\hat j}} + 2{\\rm{\\hat k}}\\]” option3=”\\[{\\rm{2\\hat i}} – 6{\\rm{\\hat j}} + 2{\\rm{\\hat k}}\\]” option4=”\\[{\\rm{4\\hat i}} – 6{\\rm{\\hat j}} – 2{\\rm{\\hat k}}\\]” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{(B)}}$.<\/p>\n
The directional derivative of a function $\\phi$ at a point $P$ in the direction of a unit vector $\\hat{u}$ is given by:<\/p>\n
$$D_{\\hat{u}} \\phi(P) = \\nabla \\phi(P) \\cdot \\hat{u}$$<\/p>\n
where $\\nabla \\phi(P)$ is the gradient of $\\phi$ at $P$.<\/p>\n
In this case, we have:<\/p>\n
$$\\phi(x, y, z) = 2xz – y^2$$<\/p>\n
$$\\nabla \\phi(x, y, z) = (2x + 2z) \\hat{i} + (2y) \\hat{j} – 2y \\hat{k}$$<\/p>\n
Substituting $x = 1$, $y = 3$, and $z = 2$, we get:<\/p>\n
$$\\nabla \\phi(1, 3, 2) = (4 + 4) \\hat{i} + (6) \\hat{j} – (4) \\hat{k} = 4 \\hat{i} – 6 \\hat{j} + 2 \\hat{k}$$<\/p>\n
The directional derivative of $\\phi$ at the point $(1, 3, 2)$ in the direction of $\\hat{u}$ is then:<\/p>\n
$$D_{\\hat{u}} \\phi(1, 3, 2) = (4 \\hat{i} – 6 \\hat{j} + 2 \\hat{k}) \\cdot \\hat{u}$$<\/p>\n
The maximum value of the directional derivative occurs when $\\hat{u}$ is in the same direction as $\\nabla \\phi(1, 3, 2)$. Therefore, the maximum value of the directional derivative is:<\/p>\n
$$\\max_{\\hat{u}} D_{\\hat{u}} \\phi(1, 3, 2) = (4 \\hat{i} – 6 \\hat{j} + 2 \\hat{k}) \\cdot (4 \\hat{i} – 6 \\hat{j} + 2 \\hat{k}) = 24$$<\/p>\n
Therefore, the directional derivative of $\\phi$ at the point $(1, 3, 2)$ becomes maximum in the direction of $\\boxed{\\text{(B)}}$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”\\[{\\rm{4\\hat i}} + 2{\\rm{\\hat j}} – 3{\\rm{\\hat k}}\\]” option2=”\\[{\\rm{4\\hat i}} – 6{\\rm{\\hat j}} + 2{\\rm{\\hat k}}\\]” option3=”\\[{\\rm{2\\hat i}} – 6{\\rm{\\hat j}} + 2{\\rm{\\hat k}}\\]” option4=”\\[{\\rm{4\\hat i}} – 6{\\rm{\\hat j}} – 2{\\rm{\\hat k}}\\]” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20185","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n