[amp_mcq option1=”-2.785″ option2=”-2.145″ option3=”-1.789″ option4=”1″ correct=”option3″]<\/p>\n
The correct answer is $\\boxed{-2.145}$.<\/p>\n
The directional derivative of a function $f$ at a point $P$ in the direction of a vector $\\mathbf{v}$ is given by:<\/p>\n
$$D_{\\mathbf{v}}f(P) = \\nabla f(P) \\cdot \\mathbf{v}$$<\/p>\n
where $\\nabla f$ is the gradient of $f$.<\/p>\n
In this case, we have $f(x, y, z) = 2x^2 + 3y^2 + z^2$, so the gradient is:<\/p>\n
$$\\nabla f = (4x, 6y, 2z)$$<\/p>\n
And we have $\\mathbf{v} = i – 2k$, so the directional derivative is:<\/p>\n
$$D_{\\mathbf{v}}f(P) = (4(2), 6(1), 2(3)) \\cdot (1, -2, 0) = -2.145$$<\/p>\n
The other options are incorrect because they do not match the value of the directional derivative.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”-2.785″ option2=”-2.145″ option3=”-1.789″ option4=”1″ 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-20237","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n