{"id":245130,"date":"2025-07-05T17:25:42","date_gmt":"2025-07-05T17:25:42","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245130"},"modified":"2025-07-05T17:25:44","modified_gmt":"2025-07-05T17:25:44","slug":"a-jeweler-resizes-a-ring-so-that-its-inner-circumference-is-8-centimeters","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/05\/a-jeweler-resizes-a-ring-so-that-its-inner-circumference-is-8-centimeters\/","title":{"rendered":"A jeweler resizes a ring so that its inner circumference is 8 centimeters."},"content":{"rendered":"\n<p>A jeweler resizes a ring so that its inner circumference is 8 centimeters. (a) What is the radius r (in cm) of the ring? r = cm (b) The inner circumference of the ring varies between 7.5 centimeters and 8.5 centimeters. How does the radius vary? &lt; r &lt; (c) Use the \ud835\udf00\u2013\ud835\udeff definition of limit to describe this situation if \ud835\udf00 = 0.5. (Round your value of \ud835\udeff to four decimal places.) lim r\u2192 = 8; \ud835\udf00 = 0.5; \ud835\udeff =<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (a): Finding the Radius of the Ring<\/h3>\n\n\n\n<p>To find the radius, we use the formula for the circumference of a circle:C=2\u03c0rC = 2 \\pi rC=2\u03c0r<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>CCC is the circumference,<\/li>\n\n\n\n<li>rrr is the radius,<\/li>\n\n\n\n<li>\u03c0\u22483.1416\\pi \\approx 3.1416\u03c0\u22483.1416.<\/li>\n<\/ul>\n\n\n\n<p>The given inner circumference of the ring is 8 cm, so we can solve for rrr:8=2\u03c0r8 = 2 \\pi r8=2\u03c0r<\/p>\n\n\n\n<p>Divide both sides of the equation by 2\u03c02\\pi2\u03c0:r=82\u03c0=86.2832\u22481.2732\u2009cmr = \\frac{8}{2\\pi} = \\frac{8}{6.2832} \\approx 1.2732 \\, \\text{cm}r=2\u03c08\u200b=6.28328\u200b\u22481.2732cm<\/p>\n\n\n\n<p>Thus, the radius rrr is approximately <strong>1.2732 cm<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (b): Variation in the Radius<\/h3>\n\n\n\n<p>The problem states that the inner circumference varies between 7.5 cm and 8.5 cm. To determine how the radius varies, we apply the formula for the radius based on the circumference:r=C2\u03c0r = \\frac{C}{2\\pi}r=2\u03c0C\u200b<\/p>\n\n\n\n<p>Let\u2019s calculate the radius for the two extreme values of the circumference:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>For C=7.5\u2009cmC = 7.5 \\, \\text{cm}C=7.5cm:<\/li>\n<\/ol>\n\n\n\n<p>r=7.52\u03c0=7.56.2832\u22481.1949\u2009cmr = \\frac{7.5}{2\\pi} = \\frac{7.5}{6.2832} \\approx 1.1949 \\, \\text{cm}r=2\u03c07.5\u200b=6.28327.5\u200b\u22481.1949cm<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>For C=8.5\u2009cmC = 8.5 \\, \\text{cm}C=8.5cm:<\/li>\n<\/ol>\n\n\n\n<p>r=8.52\u03c0=8.56.2832\u22481.3526\u2009cmr = \\frac{8.5}{2\\pi} = \\frac{8.5}{6.2832} \\approx 1.3526 \\, \\text{cm}r=2\u03c08.5\u200b=6.28328.5\u200b\u22481.3526cm<\/p>\n\n\n\n<p>Thus, the radius varies between approximately <strong>1.1949 cm<\/strong> and <strong>1.3526 cm<\/strong>. This gives the range:1.1949&lt;r&lt;1.35261.1949 &lt; r &lt; 1.35261.1949&lt;r&lt;1.3526<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Part (c): Using the \u03f5\\epsilon\u03f5-\u03b4\\delta\u03b4 Definition of Limit<\/h3>\n\n\n\n<p>The \u03f5\\epsilon\u03f5-\u03b4\\delta\u03b4 definition of the limit is a formal way of describing the behavior of a function as it approaches a particular value. Here, we are asked to use the definition to describe the situation where the radius approaches 1.2732 cm as the circumference approaches 8 cm.<\/p>\n\n\n\n<p>The general form of the \u03f5\\epsilon\u03f5-\u03b4\\delta\u03b4 definition of a limit is:For&nbsp;every&nbsp;\u03f5&gt;0,&nbsp;there&nbsp;exists&nbsp;a&nbsp;\u03b4&gt;0&nbsp;such&nbsp;that&nbsp;if&nbsp;\u2223C\u22128\u2223&lt;\u03b4,&nbsp;then&nbsp;\u2223r\u22121.2732\u2223&lt;\u03f5.\\text{For every } \\epsilon &gt; 0, \\text{ there exists a } \\delta &gt; 0 \\text{ such that if } |C &#8211; 8| &lt; \\delta, \\text{ then } |r &#8211; 1.2732| &lt; \\epsilon.For&nbsp;every&nbsp;\u03f5&gt;0,&nbsp;there&nbsp;exists&nbsp;a&nbsp;\u03b4&gt;0&nbsp;such&nbsp;that&nbsp;if&nbsp;\u2223C\u22128\u2223&lt;\u03b4,&nbsp;then&nbsp;\u2223r\u22121.2732\u2223&lt;\u03f5.<\/p>\n\n\n\n<p>Given \u03f5=0.5\\epsilon = 0.5\u03f5=0.5, we are tasked with finding the corresponding \u03b4\\delta\u03b4.<\/p>\n\n\n\n<p>Start by writing the relationship between the radius rrr and the circumference CCC:r=C2\u03c0r = \\frac{C}{2\\pi}r=2\u03c0C\u200b<\/p>\n\n\n\n<p>Now, we want to find \u03b4\\delta\u03b4 such that:\u2223r\u22121.2732\u2223&lt;0.5|r &#8211; 1.2732| &lt; 0.5\u2223r\u22121.2732\u2223&lt;0.5<\/p>\n\n\n\n<p>Since r=C2\u03c0r = \\frac{C}{2\\pi}r=2\u03c0C\u200b, this becomes:\u2223C2\u03c0\u22121.2732\u2223&lt;0.5\\left|\\frac{C}{2\\pi} &#8211; 1.2732\\right| &lt; 0.5\u200b2\u03c0C\u200b\u22121.2732\u200b&lt;0.5<\/p>\n\n\n\n<p>Simplifying:\u2223C\u221282\u03c0\u2223&lt;0.5\\left|\\frac{C &#8211; 8}{2\\pi}\\right| &lt; 0.5\u200b2\u03c0C\u22128\u200b\u200b&lt;0.5<\/p>\n\n\n\n<p>Multiply both sides by 2\u03c02\\pi2\u03c0:\u2223C\u22128\u2223&lt;0.5\u00d72\u03c0\u22483.1416|C &#8211; 8| &lt; 0.5 \\times 2\\pi \\approx 3.1416\u2223C\u22128\u2223&lt;0.5\u00d72\u03c0\u22483.1416<\/p>\n\n\n\n<p>Thus, the value of \u03b4\\delta\u03b4 is approximately <strong>3.1416<\/strong>. So, when \u03f5=0.5\\epsilon = 0.5\u03f5=0.5, we can set \u03b4=3.1416\\delta = 3.1416\u03b4=3.1416 to ensure that the radius rrr is within 0.5 cm of 1.2732 cm.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary of Answers:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>(a) The radius is approximately <strong>1.2732 cm<\/strong>.<\/li>\n\n\n\n<li>(b) The radius varies between <strong>1.1949 cm<\/strong> and <strong>1.3526 cm<\/strong>.<\/li>\n\n\n\n<li>(c) Using the \u03f5\\epsilon\u03f5-\u03b4\\delta\u03b4 definition, when \u03f5=0.5\\epsilon = 0.5\u03f5=0.5, the corresponding \u03b4\\delta\u03b4 is <strong>3.1416<\/strong>.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-594.jpeg\" alt=\"\" class=\"wp-image-245131\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A jeweler resizes a ring so that its inner circumference is 8 centimeters. (a) What is the radius r (in cm) of the ring? r = cm (b) The inner circumference of the ring varies between 7.5 centimeters and 8.5 centimeters. How does the radius vary? &lt; r &lt; (c) Use the \ud835\udf00\u2013\ud835\udeff definition of [&hellip;]<\/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-245130","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245130","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=245130"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245130\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245130"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245130"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245130"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}