3x 5. GiveX: Z= f(x,y) = e parlay21) PETERMINE THE RATE OF CHANGE OF THE FUNCTION AT THE POINT P(0,0) WHEN MOVING IN THE DIRECTION OF THE POINT Q(2,2). DETERMINE THE DIRECTION TO MOVE FROM P(1, 1) FOR THE MAXIMUM RATE OF DECREASE IN THE FUNCTION 6. DETERMINE THE EQUATION OF THE TANGENT PLANE TO THE SURFACE: 7?- ex*y? + 2x + y \u00b2 AT THE POINT P(1,1,2)?<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n Let’s break this into three parts<\/strong> based on the prompt:<\/p>\n\n\n\n Given: Step 1: Compute the Gradient<\/strong><\/p>\n\n\n\n $$ At $P(0,0)$:<\/p>\n\n\n\n $$ Step 2: Direction Vector from P to Q<\/strong><\/p>\n\n\n\n $$ Step 3: Directional Derivative<\/strong><\/p>\n\n\n\n $$ Answer:<\/strong> Use the same function:<\/strong><\/p>\n\n\n\n $$ Gradient at (1,1):<\/strong><\/p>\n\n\n\n $$ Direction of Maximum Decrease:<\/strong> $$ Unit Vector Direction:<\/strong><\/p>\n\n\n\n $$ Answer:<\/strong> Given:<\/p>\n\n\n\n $$ Step 1: Compute Partial Derivatives<\/strong><\/p>\n\n\n\n $$ At $(1,1)$:<\/p>\n\n\n\n $$ Step 2: Use Point-Slope Form for Tangent Plane<\/strong><\/p>\n\n\n\n $$ First find $f(1,1)$:<\/p>\n\n\n\n $$ Final equation:<\/p>\n\n\n\n $$ This set of problems tests your understanding of gradients<\/strong>, directional derivatives<\/strong>, and tangent planes<\/strong>, which are key concepts in multivariable calculus.<\/p>\n\n\n\n In the first part<\/strong>, the rate of change of a scalar field in a specific direction is computed using the directional derivative<\/strong>, which is the dot product of the gradient<\/strong> and a unit vector<\/strong> in the given direction. The gradient at $(0,0)$ turns out to be zero for this function, meaning the function does not change at all<\/strong> in any direction at that point\u2014indicating a critical point<\/strong>, potentially a local minimum or maximum.<\/p>\n\n\n\n In the second part<\/strong>, you determine the direction of maximum decrease<\/strong>. The gradient vector points in the direction of the steepest ascent<\/strong>, and its opposite<\/strong> points to the steepest descent<\/strong>. Hence, moving in the direction opposite to the gradient at point $(1,1)$, specifically in the direction $(-1, -1)$, will cause the function to decrease as rapidly as possible.<\/p>\n\n\n\n In the third part<\/strong>, you’re asked to find the tangent plane<\/strong> to a surface. This is the 3D analog of the tangent line. The partial derivatives<\/strong> with respect to $x$ and $y$ provide the slope of the surface in each direction, and by plugging these into the point-slope formula<\/strong>, you can build the equation for the tangent plane. This plane approximates the surface near the point $(1,1,2)$ and is important in optimization, physics, and differential geometry.<\/p>\n\n\n\n Understanding these concepts builds your intuition for how multivariable functions behave and change in space.<\/p>\n","protected":false},"excerpt":{"rendered":" 3x 5. GiveX: Z= f(x,y) = e parlay21) PETERMINE THE RATE OF CHANGE OF THE FUNCTION AT THE POINT P(0,0) WHEN MOVING IN THE DIRECTION OF THE POINT Q(2,2). DETERMINE THE DIRECTION TO MOVE FROM P(1, 1) FOR THE MAXIMUM RATE OF DECREASE IN THE FUNCTION 6. DETERMINE THE EQUATION OF THE TANGENT PLANE TO […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-216267","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216267","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=216267"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/216267\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=216267"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=216267"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=216267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nProblem 1: Rate of Change at P(0, 0) in the Direction of Q(2, 2)<\/strong><\/h3>\n\n\n\n
$f(x, y) = e^{x^2 + y^2}$
Find the rate of change at $P(0,0)$ in the direction of $Q(2,2)$.<\/p>\n\n\n\n
\\nabla f = \\left( \\frac{\\partial f}{\\partial x}, \\frac{\\partial f}{\\partial y} \\right)
= \\left( 2xe^{x^2 + y^2}, 2ye^{x^2 + y^2} \\right)
$$<\/p>\n\n\n\n
\\nabla f(0, 0) = (0, 0)
$$<\/p>\n\n\n\n
\\vec{v} = Q – P = (2,2)
\\Rightarrow \\hat{v} = \\left( \\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}} \\right)
$$<\/p>\n\n\n\n
D_{\\hat{v}} f = \\nabla f \\cdot \\hat{v} = (0, 0) \\cdot \\left( \\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}} \\right) = 0
$$<\/p>\n\n\n\n
The rate of change is 0.<\/strong><\/p>\n\n\n\n
\n\n\n\nProblem 2: Direction of Maximum Decrease at P(1, 1)<\/strong><\/h3>\n\n\n\n
f(x, y) = e^{x^2 + y^2}
$$<\/p>\n\n\n\n
\\nabla f(1,1) = \\left( 2(1)e^{2}, 2(1)e^{2} \\right) = (2e^2, 2e^2)
$$<\/p>\n\n\n\n
Opposite of the gradient vector:<\/p>\n\n\n\n
\\Rightarrow \\boxed{-\\nabla f(1,1) = (-2e^2, -2e^2)}
$$<\/p>\n\n\n\n
\\left( -\\frac{1}{\\sqrt{2}}, -\\frac{1}{\\sqrt{2}} \\right)
$$<\/p>\n\n\n\n
Move in the direction $(-1, -1)$ for the greatest decrease.<\/strong><\/p>\n\n\n\n
\n\n\n\nProblem 3: Tangent Plane Equation at P(1,1,2)<\/strong><\/h3>\n\n\n\n
z = f(x, y) = e^{-xy^2} + 2x + y^2
$$<\/p>\n\n\n\n
\\frac{\\partial f}{\\partial x} = -y^2 e^{-xy^2} + 2
\\frac{\\partial f}{\\partial y} = -2xy e^{-xy^2} + 2y
$$<\/p>\n\n\n\n
f_x = -1 \\cdot e^{-1} + 2 = 2 – \\frac{1}{e}
f_y = -2 \\cdot 1 \\cdot 1 \\cdot e^{-1} + 2(1) = 2 – \\frac{2}{e}
$$<\/p>\n\n\n\n
z – f(1,1) = f_x(1,1)(x – 1) + f_y(1,1)(y – 1)
$$<\/p>\n\n\n\n
f(1,1) = e^{-1} + 2 + 1 = \\frac{1}{e} + 3
$$<\/p>\n\n\n\n
z – \\left(\\frac{1}{e} + 3\\right) = \\left(2 – \\frac{1}{e}\\right)(x – 1) + \\left(2 – \\frac{2}{e}\\right)(y – 1)
$$<\/p>\n\n\n\n
\n\n\n\nSummary and Explanation (300+ words):<\/strong><\/h3>\n\n\n\n