EXAMPLE 1 (a) If x2 + y2- 25, find dy dx (b) Find an equation of the tangent to the circle x2 + y2 – 25 at the point (3, 4). SOLUTION 1 (a) Differentiating both sides of the equation x2+y25 )-) (25) + dx dx d 2x+2y X + dx 0. dx Remembering that y is a function of x and using the Chain Rule, we have 2y dy dx -2x X dx dx Thus 2y 2x + = 0. dx Now we solve this equation for dy\/dx dy dx X y (b) At the point (3, 4) we have x = 3 and y= 4, so dy -3\/4 dx An equation of the tangent to the circle at (3, 4) is therefore 4 -3\/4 (x y- or 3 25 y X 4 SOLUTION 2 (b) Solving the equation x2y-25, we get y t25-x2. The point (3, 4) lies on the upper semicircle y 25-x and so we consider the function fx)- 25-Differentiating f using the Chain Rule, + we have 125–\/2 V25- f(x) dx Your answer is incorrect. as- (25-2x) x ) So f(3) and, as in Solution 1, an equation of the tangent is<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n We start with the given equation of the circle:<\/p>\n\n\n\n [ Differentiating both sides with respect to ( x ), using implicit differentiation:<\/p>\n\n\n\n [ [ Solving for ( \\frac{dy}{dx} ):<\/p>\n\n\n\n [ [ At the given point ( (3,4) ), we substitute ( x = 3 ) and ( y = 4 ):<\/p>\n\n\n\n [ The equation of a tangent line at ( (x_1, y_1) ) with slope ( m ) is:<\/p>\n\n\n\n [ Substituting ( (3,4) ) and ( m = -\\frac{3}{4} ):<\/p>\n\n\n\n [ Expanding:<\/p>\n\n\n\n [ [ [ Thus, the equation of the tangent line is:<\/p>\n\n\n\n [ A circle<\/strong> is defined as the set of all points equidistant from a fixed center. The given equation, ( x^2 + y^2 = 25 ), represents a circle centered at the origin ( (0,0) ) with a radius of ( 5 ) since ( 25 = 5^2 ).<\/p>\n\n\n\n To find the tangent line<\/strong>, we first compute the derivative ( \\frac{dy}{dx} ) using implicit differentiation<\/strong>, which is useful when ( y ) is a function of ( x ) but cannot be explicitly solved for easily. Differentiating both sides, we apply the chain rule<\/strong> to ( y^2 ), which gives ( 2y \\frac{dy}{dx} ). Solving for ( \\frac{dy}{dx} ), we obtain ( -\\frac{x}{y} ).<\/p>\n\n\n\n At the specific point ( (3,4) ), substituting ( x = 3 ) and ( y = 4 ) into the derivative formula yields ( \\frac{dy}{dx} = -\\frac{3}{4} ).<\/p>\n\n\n\n The tangent line<\/strong> follows the equation ( y – y_1 = m(x – x_1) ), where ( (x_1, y_1) = (3,4) ) and ( m = -\\frac{3}{4} ). Substituting these values and simplifying, we obtain the final equation:<\/p>\n\n\n\n [ This line represents the tangent to the circle<\/strong> at ( (3,4) ), meaning it touches the circle at exactly that point without crossing it. This solution confirms that the previous answer was incorrect due to differentiation mistakes.<\/p>\n","protected":false},"excerpt":{"rendered":" EXAMPLE 1 (a) If x2 + y2- 25, find dy dx (b) Find an equation of the tangent to the circle x2 + y2 – 25 at the point (3, 4). SOLUTION 1 (a) Differentiating both sides of the equation x2+y25 )-) (25) + dx dx d 2x+2y X + dx 0. dx Remembering that […]<\/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-187140","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187140","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=187140"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187140\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187140"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187140"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187140"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Solution to the Given Problem<\/h3>\n\n\n\n
(a) Finding ( \\frac{dy}{dx} )<\/strong><\/h4>\n\n\n\n
x^2 + y^2 = 25
]<\/p>\n\n\n\n
\\frac{d}{dx} (x^2) + \\frac{d}{dx} (y^2) = \\frac{d}{dx} (25)
]<\/p>\n\n\n\n
2x + 2y \\frac{dy}{dx} = 0
]<\/p>\n\n\n\n
2y \\frac{dy}{dx} = -2x
]<\/p>\n\n\n\n
\\frac{dy}{dx} = \\frac{-x}{y}
]<\/p>\n\n\n\n(b) Finding the Equation of the Tangent Line<\/strong><\/h4>\n\n\n\n
\\frac{dy}{dx} = \\frac{-3}{4}
]<\/p>\n\n\n\n
y – y_1 = m(x – x_1)
]<\/p>\n\n\n\n
y – 4 = -\\frac{3}{4} (x – 3)
]<\/p>\n\n\n\n
y – 4 = -\\frac{3}{4} x + \\frac{9}{4}
]<\/p>\n\n\n\n
y = -\\frac{3}{4} x + \\frac{9}{4} + 4
]<\/p>\n\n\n\n
y = -\\frac{3}{4} x + \\frac{25}{4}
]<\/p>\n\n\n\n
y = -\\frac{3}{4}x + \\frac{25}{4}
]<\/p>\n\n\n\n
\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n
y = -\\frac{3}{4} x + \\frac{25}{4}
]<\/p>\n\n\n\n