{"id":8159,"date":"2021-11-15T00:08:22","date_gmt":"2021-11-14T18:38:22","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8159"},"modified":"2021-11-15T16:01:29","modified_gmt":"2021-11-15T10:31:29","slug":"if-the-radius-of-a-sphere-is-measured-as-9-cm-with-an-error-of-0-03-cm-then-find-the-approximating-error-in-calculating-its-volume","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/if-the-radius-of-a-sphere-is-measured-as-9-cm-with-an-error-of-0-03-cm-then-find-the-approximating-error-in-calculating-its-volume\/","title":{"rendered":"If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume."},"content":{"rendered":"

Solution :<\/h2>\n

Let r be the radius of a sphere and \\(\\delta\\)r be the error in measuring the radius. Then, r = 9 cm and \\(\\delta\\)r = 0.03 cm.<\/p>\n

Let V be the volume of the sphere. Then,<\/p>\n

V = \\({4\\over 3}\\pi r^3\\)\u00a0 \\(\\implies\\) \\(dV\\over dr\\) = \\(4\\pi r^2\\)<\/p>\n

\\(\\implies\\) \\(({dV\\over dr})_{r = 9}\\) = \\(4\\pi \\times 9^2\\) = \\(324 \\pi\\)<\/p>\n

Let \\(\\delta\\)V be the error<\/a> in V due to error in V due to error \\(\\delta\\)r in r. Then,<\/p>\n

\\(\\delta\\)V = \\(dV\\over dr\\) \\(\\delta\\)r\u00a0 \\(\\implies\\)\u00a0 \\(\\delta\\)V =\u00a0 \\(324 \\pi\\times 0.03\\) = \\(9.72 \\pi cm^3\\)<\/p>\n

\n

Similar Questions<\/h3>\n

Find the approximate value of f(3.02), where f(x) = \\(3x^2 + 5x + 3\\).<\/a><\/p>\n

Verify Rolle\u2019s theorem for the function f(x) = \\(x^2\\) \u2013 5x + 6 on the interval [2, 3].<\/a><\/p>\n

It is given that for the function f(x) = \\(x^3 \u2013 6x^2 + ax + b\\) on [1, 3], Rolles\u2019s theorem holds with c = \\(2 +{1\\over \\sqrt{3}}\\). Find the values of a and b, if f(1) = f(3) = 0.<\/a><\/p>\n

Find the point on the curve y = cos x \u2013 1, x \\(\\in\\) \\([{\\pi\\over 2}, {3\\pi\\over 2}]\\) at which tangent is parallel to the x-axis.<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Let r be the radius of a sphere and \\(\\delta\\)r be the error in measuring the radius. Then, r = 9 cm and \\(\\delta\\)r = 0.03 cm. Let V be the volume of the sphere. Then, V = \\({4\\over 3}\\pi r^3\\)\u00a0 \\(\\implies\\) \\(dV\\over dr\\) = \\(4\\pi r^2\\) \\(\\implies\\) \\(({dV\\over dr})_{r = 9}\\) = …<\/p>\n

If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[66,43],"tags":[],"yoast_head":"\nIf the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/if-the-radius-of-a-sphere-is-measured-as-9-cm-with-an-error-of-0-03-cm-then-find-the-approximating-error-in-calculating-its-volume\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximating error in calculating its volume.\" \/>\n<meta property=\"og:description\" content=\"Solution : Let r be the radius of a sphere and (delta)r be the error in measuring the radius. 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