The point P(2, ln 2) lies on the curve y = ln x.<\/p>\n\n\n\n
(a) If Q is the point (x, in x), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x:<\/p>\n\n\n\n
(i) 1.5<\/p>\n\n\n\n
(ii) 1.9<\/p>\n\n\n\n
(iii) 1.99<\/p>\n\n\n\n
(iv) 1.999<\/p>\n\n\n\n
(v) 2.5<\/p>\n\n\n\n
(vi) 2.1<\/p>\n\n\n\n
(vii) 2.01<\/p>\n\n\n\n
(viii) 2.001<\/p>\n\n\n\n
(b) Using the results of part (a), guess the value of the slope of the tangent line to the curve at P(2, ln 2).<\/p>\n\n\n\n
(c) Using the slope from part (b), find an equation of the tangent line to the curve at P(2, ln 2).<\/p>\n\n\n\n
(d) Sketch the curve, two of the secant lines, and the tangent line.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The slope of the secant line joining two points ( P(x_1, y_1) ) and ( Q(x_2, y_2) ) on the curve is given by the formula:<\/p>\n\n\n\n [ where ( P(2, \\ln 2) ) and ( Q(x, \\ln x) ). Here, ( y_1 = \\ln 2 ) and ( x_1 = 2 ).<\/p>\n\n\n\n So the slope of the secant line becomes:<\/p>\n\n\n\n [ Now we calculate for each value of ( x ):<\/p>\n\n\n\n (i) For ( x = 1.5 ):<\/p>\n\n\n\n [ (ii) For ( x = 1.9 ):<\/p>\n\n\n\n [ (iii) For ( x = 1.99 ):<\/p>\n\n\n\n [ (iv) For ( x = 1.999 ):<\/p>\n\n\n\n [ (v) For ( x = 2.5 ):<\/p>\n\n\n\n [ (vi) For ( x = 2.1 ):<\/p>\n\n\n\n [ (vii) For ( x = 2.01 ):<\/p>\n\n\n\n [ (viii) For ( x = 2.001 ):<\/p>\n\n\n\n [ By examining the slopes of the secant lines as ( x ) approaches 2, we see that the slopes increase rapidly as ( x ) approaches 2 from the left and stabilize around 0.5 as ( x ) approaches 2 from the right. The slopes near ( x = 2 ) are:<\/p>\n\n\n\n Therefore, we can reasonably guess that the slope of the tangent line at ( P(2, \\ln 2) ) is approximately ( 0.5 ).<\/p>\n\n\n\n The general equation of a line is:<\/p>\n\n\n\n [ where ( m ) is the slope, and ( (x_1, y_1) ) is a point on the line. Using ( P(2, \\ln 2) ) and the slope ( m = 0.5 ), we get:<\/p>\n\n\n\n [ Solving for ( y ):<\/p>\n\n\n\n [ [ Thus, the equation of the tangent line is:<\/p>\n\n\n\n [ For this step, we would plot the curve ( y = \\ln x ), the tangent line ( y = 0.5x – 1 + \\ln 2 ), and two secant lines for chosen values of ( x ) such as ( x = 2.01 ) and ( x = 1.99 ). The secant lines should approach the tangent line as ( x ) gets closer to 2.<\/p>\n\n\n\n The tangent line approximates the curve closely at ( x = 2 ), and as the points ( Q(x, \\ln x) ) approach ( P(2, \\ln 2) ), the secant lines become increasingly similar to the tangent line, indicating the concept of the derivative at a point.<\/p>\n\n\n\n The secant lines show how the slope between two points on the curve changes as the second point gets closer to the point of tangency. As we move closer to ( x = 2 ), the slope of the secant lines approaches the slope of the tangent line. The derivative of ( \\ln x ) at ( x = 2 ) provides the slope of the tangent line at that point, and we approximate this value by computing the secant slopes for values of ( x ) close to 2. The value ( 0.5 ) is a reasonable estimate for the derivative of ( \\ln x ) at ( x = 2 ), since the actual derivative of ( \\ln x ) is ( \\frac{1}{x} ), and at ( x = 2 ), this is exactly ( \\frac{1}{2} ).<\/p>\n","protected":false},"excerpt":{"rendered":" The point P(2, ln 2) lies on the curve y = ln x. (a) If Q is the point (x, in x), use your calculator to find the slope of the secant line PQ (correct to six decimal places) for the following values of x: (i) 1.5 (ii) 1.9 (iii) 1.99 (iv) 1.999 (v) 2.5 […]<\/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-186968","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186968","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=186968"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186968\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186968"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186968"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186968"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Finding the slope of the secant line PQ for the given values of ( x ):<\/h3>\n\n\n\n
m_{\\text{secant}} = \\frac{y_2 – y_1}{x_2 – x_1}
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln x – \\ln 2}{x – 2}
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 1.5 – \\ln 2}{1.5 – 2} = \\frac{\\ln \\frac{1.5}{2}}{-0.5} = \\frac{\\ln 0.75}{-0.5} \\approx 0.287682
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 1.9 – \\ln 2}{1.9 – 2} = \\frac{\\ln \\frac{1.9}{2}}{-0.1} \\approx 1.156820
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 1.99 – \\ln 2}{1.99 – 2} = \\frac{\\ln \\frac{1.99}{2}}{-0.01} \\approx 6.273197
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 1.999 – \\ln 2}{1.999 – 2} = \\frac{\\ln \\frac{1.999}{2}}{-0.001} \\approx 69.314718
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 2.5 – \\ln 2}{2.5 – 2} = \\frac{\\ln \\frac{2.5}{2}}{0.5} \\approx 0.405465
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 2.1 – \\ln 2}{2.1 – 2} = \\frac{\\ln \\frac{2.1}{2}}{0.1} \\approx 0.487227
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 2.01 – \\ln 2}{2.01 – 2} = \\frac{\\ln \\frac{2.01}{2}}{0.01} \\approx 0.497588
]<\/p>\n\n\n\n
m_{\\text{secant}} = \\frac{\\ln 2.001 – \\ln 2}{2.001 – 2} = \\frac{\\ln \\frac{2.001}{2}}{0.001} \\approx 0.499751
]<\/p>\n\n\n\n(b) Guessing the slope of the tangent line at P(2, ln 2):<\/h3>\n\n\n\n
\n
(c) Finding the equation of the tangent line at P(2, ln 2):<\/h3>\n\n\n\n
y – y_1 = m(x – x_1)
]<\/p>\n\n\n\n
y – \\ln 2 = 0.5(x – 2)
]<\/p>\n\n\n\n
y = 0.5(x – 2) + \\ln 2
]<\/p>\n\n\n\n
y = 0.5x – 1 + \\ln 2
]<\/p>\n\n\n\n
y = 0.5x – 1 + \\ln 2
]<\/p>\n\n\n\n(d) Sketching the curve, two secant lines, and the tangent line:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n