{"id":245722,"date":"2025-07-06T11:37:21","date_gmt":"2025-07-06T11:37:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245722"},"modified":"2025-07-06T11:37:24","modified_gmt":"2025-07-06T11:37:24","slug":"the-gradient-in-spherical-coordinate-system-for-a-scalar-function-f-is","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/the-gradient-in-spherical-coordinate-system-for-a-scalar-function-f-is\/","title":{"rendered":"The gradient in spherical coordinate system for a scalar function\u00a0f\u00a0is"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s break down the problem step by step.<\/p>\n\n\n\n

The given function in Cartesian coordinates is:<\/p>\n\n\n\n

f(x,y,z)=x2z2+xy2f(x, y, z) = x^2 z^2 + x y^2f(x,y,z)=x2z2+xy2<\/p>\n\n\n\n

We are asked to express this function in spherical coordinates and then calculate the gradient and evaluate it at a specific point.<\/p>\n\n\n\n

Step A: Expressing f(x,y,z)f(x, y, z)f(x,y,z) in spherical coordinates<\/h3>\n\n\n\n

To convert f(x,y,z)f(x, y, z)f(x,y,z) to spherical coordinates, recall the relations between Cartesian and spherical coordinates:x=rsin\u2061\u03b8cos\u2061\u03d5x = r \\sin\\theta \\cos\\phix=rsin\u03b8cos\u03d5y=rsin\u2061\u03b8sin\u2061\u03d5y = r \\sin\\theta \\sin\\phiy=rsin\u03b8sin\u03d5z=rcos\u2061\u03b8z = r \\cos\\thetaz=rcos\u03b8<\/p>\n\n\n\n

Substitute these into the function:f(x,y,z)=(rsin\u2061\u03b8cos\u2061\u03d5)2(rcos\u2061\u03b8)2+(rsin\u2061\u03b8cos\u2061\u03d5)(rsin\u2061\u03b8sin\u2061\u03d5)2f(x, y, z) = (r \\sin\\theta \\cos\\phi)^2 (r \\cos\\theta)^2 + (r \\sin\\theta \\cos\\phi) (r \\sin\\theta \\sin\\phi)^2f(x,y,z)=(rsin\u03b8cos\u03d5)2(rcos\u03b8)2+(rsin\u03b8cos\u03d5)(rsin\u03b8sin\u03d5)2<\/p>\n\n\n\n

Simplifying each term:<\/p>\n\n\n\n

    \n
  1. (rsin\u2061\u03b8cos\u2061\u03d5)2(rcos\u2061\u03b8)2=r4sin\u20612\u03b8cos\u20612\u03d5cos\u20612\u03b8(r \\sin\\theta \\cos\\phi)^2 (r \\cos\\theta)^2 = r^4 \\sin^2\\theta \\cos^2\\phi \\cos^2\\theta(rsin\u03b8cos\u03d5)2(rcos\u03b8)2=r4sin2\u03b8cos2\u03d5cos2\u03b8<\/li>\n\n\n\n
  2. (rsin\u2061\u03b8cos\u2061\u03d5)(rsin\u2061\u03b8sin\u2061\u03d5)2=r3sin\u20613\u03b8cos\u2061\u03d5sin\u20612\u03d5(r \\sin\\theta \\cos\\phi) (r \\sin\\theta \\sin\\phi)^2 = r^3 \\sin^3\\theta \\cos\\phi \\sin^2\\phi(rsin\u03b8cos\u03d5)(rsin\u03b8sin\u03d5)2=r3sin3\u03b8cos\u03d5sin2\u03d5<\/li>\n<\/ol>\n\n\n\n

    Thus, the function in spherical coordinates becomes:f(r,\u03b8,\u03d5)=r4sin\u20612\u03b8cos\u20612\u03d5cos\u20612\u03b8+r3sin\u20613\u03b8cos\u2061\u03d5sin\u20612\u03d5f(r, \\theta, \\phi) = r^4 \\sin^2\\theta \\cos^2\\phi \\cos^2\\theta + r^3 \\sin^3\\theta \\cos\\phi \\sin^2\\phif(r,\u03b8,\u03d5)=r4sin2\u03b8cos2\u03d5cos2\u03b8+r3sin3\u03b8cos\u03d5sin2\u03d5<\/p>\n\n\n\n

    Step B: Finding the gradient \u2207f(r,\u03b8,\u03d5)\\nabla f(r, \\theta, \\phi)\u2207f(r,\u03b8,\u03d5)<\/h3>\n\n\n\n

    The gradient in spherical coordinates is given by:\u2207f(r,\u03b8,\u03d5)=r^\u2202f\u2202r+\u03b8^1r\u2202f\u2202\u03b8+\u03d5^1rsin\u2061\u03b8\u2202f\u2202\u03d5\\nabla f(r, \\theta, \\phi) = \\hat{r} \\frac{\\partial f}{\\partial r} + \\hat{\\theta} \\frac{1}{r} \\frac{\\partial f}{\\partial \\theta} + \\hat{\\phi} \\frac{1}{r \\sin\\theta} \\frac{\\partial f}{\\partial \\phi}\u2207f(r,\u03b8,\u03d5)=r^\u2202r\u2202f\u200b+\u03b8^r1\u200b\u2202\u03b8\u2202f\u200b+\u03d5^\u200brsin\u03b81\u200b\u2202\u03d5\u2202f\u200b<\/p>\n\n\n\n

    Now, let’s compute the partial derivatives.<\/p>\n\n\n\n

    1. \u2202f\u2202r\\frac{\\partial f}{\\partial r}\u2202r\u2202f\u200b<\/h4>\n\n\n\n

    This will be:\u2202\u2202r(r4sin\u20612\u03b8cos\u20612\u03d5cos\u20612\u03b8+r3sin\u20613\u03b8cos\u2061\u03d5sin\u20612\u03d5)\\frac{\\partial}{\\partial r} \\left( r^4 \\sin^2\\theta \\cos^2\\phi \\cos^2\\theta + r^3 \\sin^3\\theta \\cos\\phi \\sin^2\\phi \\right)\u2202r\u2202\u200b(r4sin2\u03b8cos2\u03d5cos2\u03b8+r3sin3\u03b8cos\u03d5sin2\u03d5)<\/p>\n\n\n\n

    2. \u2202f\u2202\u03b8\\frac{\\partial f}{\\partial \\theta}\u2202\u03b8\u2202f\u200b<\/h4>\n\n\n\n

    For this derivative, we will differentiate each term with respect to \u03b8\\theta\u03b8.<\/p>\n\n\n\n

    3. \u2202f\u2202\u03d5\\frac{\\partial f}{\\partial \\phi}\u2202\u03d5\u2202f\u200b<\/h4>\n\n\n\n

    Here, we will differentiate each term with respect to \u03d5\\phi\u03d5.<\/p>\n\n\n\n

    Step C: Calculating \u2207f(2,\u03c0\/4,\u03c0\/3)\\nabla f(2, \\pi\/4, \\pi\/3)\u2207f(2,\u03c0\/4,\u03c0\/3)<\/h3>\n\n\n\n

    Once we have the gradient expression, we substitute r=2r = 2r=2, \u03b8=\u03c0\/4\\theta = \\pi\/4\u03b8=\u03c0\/4, and \u03d5=\u03c0\/3\\phi = \\pi\/3\u03d5=\u03c0\/3 into the formula for \u2207f\\nabla f\u2207f.<\/p>\n\n\n\n

    \n\n\n\n

    Given the complexity of the derivatives, it’s ideal to first work through the detailed computations of the partial derivatives for rrr, \u03b8\\theta\u03b8, and \u03d5\\phi\u03d5. <\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The Correct Answer and Explanation is: Let’s break down the problem step by step. The given function in Cartesian coordinates is: f(x,y,z)=x2z2+xy2f(x, y, z) = x^2 z^2 + x y^2f(x,y,z)=x2z2+xy2 We are asked to express this function in spherical coordinates and then calculate the gradient and evaluate it at a specific point. Step A: Expressing […]<\/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-245722","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245722","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=245722"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245722\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245722"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245722"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}