{"id":25330,"date":"2024-12-20T08:56:27","date_gmt":"2024-12-20T03:26:27","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=25330"},"modified":"2024-12-21T10:02:40","modified_gmt":"2024-12-21T04:32:40","slug":"samacheer-kalvi-12th-maths-guide-chapter-7-ex-7-2","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-12th-maths-guide-chapter-7-ex-7-2\/","title":{"rendered":"Samacheer Kalvi 12th Maths Guide Chapter 7 Applications of Differential Calculus Ex 7.2"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide<\/a> Pdf Chapter 7 Applications of Differential Calculus Ex 7.2 Textbook Questions and Answers, Notes. BANDHANBNK Pivot Point Calculator<\/a><\/p>\n

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2<\/h2>\n

Question 1.
\nFind the slope of the tangent to the following curves at the respective given points.
\n(i) y = x4<\/sup> + 2x\u00b2 – x at x = 1
\n(ii) x = a cos\u00b3 t, y = b sin\u00b3 t at t = \\(\\frac { \u03c0 }{ 2 }\\)
\nSolution:
\n(i) y = x4<\/sup> + 2x\u00b2 – x
\nDifferentiating w.r.t. ‘x’
\n\\(\\frac { dy }{ dx }\\) = 4x\u00b3 + 4x – 1
\nSlope of the tangent (\\(\\frac { dy }{ dx }\\))(x=1)<\/sub>
\n= 4(1)\u00b3 + 4(1) – 1
\n= 4 + 4 – 1 = 7
\n(ii) x = a cos\u00b3 t, y = b sin\u00b3 t
\nDifferenriating w.r.t. \u2018t’
\n\\(\\frac { dx }{ dt }\\) = – 3a cos\u00b2 t sin t
\n\\(\\frac { dy }{ dt }\\) = 3b sin\u00b2 t sin t
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nFind the point on the curve y = x\u00b2 – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.
\nSolution:
\ny = x\u00b2 – 5x + 4
\nDifferentiating w.r.t. ‘x’
\nSlope of the tangent \\(\\frac { dy }{ dx }\\) = 2x – 5
\nGiven line 3x + y = 7
\nSlope of the line = –\\(\\frac { 3 }{ 1 }\\) = -3
\nSince the tangent is parallel to the line, their slopes are equal.
\n\u2234 \\(\\frac { dy }{ dx }\\) = -3
\n\u21d2 2x – 5 = -3
\n2x = 2
\nx = 1
\nWhen x = 1, y = (1)\u00b2 – 5 (1) + 4 = 0
\n\u2234 Point on the curve is (1, 0).<\/p>\n

Question 3.
\nFind the points on curve y = x\u00b3 – 6x\u00b2 + x + 3 where the normal is parallel to the line x + y = 1729.
\nSolution:
\ny = x\u00b3 – 6x\u00b2 + x+ 3
\nDifferentiating w.r.t. \u2018x’
\nSlope of the tangent \\(\\frac { dy }{ dx }\\) = 3x\u00b2 – 12x + 1
\nSlope of the normal = \\(\\frac { 1 }{ 3x^2 – 12x + 1 }\\)
\nGiven line is x + y = 1729
\nSlope of the line is – 1
\nSince the normal is parallel to the line, their slopes are equal.
\n\\(\\frac { 1 }{ 3x^2 – 12x + 1 }\\) = -1
\n3x\u00b2 – 12x + 1 = 1
\n3x\u00b2 – 12x =0
\n3x(x – 4) = 0
\nx = 0, 4
\nWhen x = 0, y = (0)\u00b3 – 6(0)\u00b2 + 0 + 3 = 3
\nWhen x = 4, y = (4)\u00b3 – 6(4)\u00b2 + 4 + 3
\n= 64 – 96 + 4 + 3 = -25
\n\u2234 The points on the curve are (0, 3) and (4, -25).<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nFind the points on the curve y\u00b2 – 4xy = x\u00b2 + 5 for which the tangent is horizontal.
\nSolution:
\ny\u00b2 – 4xy = x\u00b2 + 5 ………… (1)
\nDifferentiating w.r.t. \u2018x’
\n\"Samacheer
\nWhen the tangent is horizontal(Parallel to X-axis) then slope of the tangent is zero.
\n\\(\\frac { dy }{ dx }\\) = 0 \u21d2 \\(\\frac { x+2y }{ y-2x }\\) = 0
\n\u21d2 x + 2y = 0
\nx = -2y
\nSubstituting in (1)
\ny\u00b2 – 4 (-2y) y = (-2y)\u00b2 + 5
\ny\u00b2 + 8y\u00b2 = 4y\u00b2 + 5
\n5y\u00b2 = 5 \u21d2 y\u00b2 = 1
\ny = \u00b11
\nWhen y = 1, x = -2
\nWhen y = – 1, x = 2
\n\u2234 The points on the curve are (- 2, 1) and (2, -1).<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nFind the tangent and normal to the following curves at the given points on the curve.
\n(i) y = x\u00b2 – x4<\/sup> at (1, 0)
\n(ii) y = x4<\/sup> + 2ex<\/sup> at (0, 2)
\n(iii) y = x sin x at (\\(\\frac { \u03c0 }{ 2 }\\), \\(\\frac { \u03c0 }{ 2 }\\))
\n(iv) x = cos t, y = 2 sin\u00b2 t at t = \\(\\frac { \u03c0 }{ 3 }\\)
\nSolution:
\n(i) y = x\u00b2 – x4<\/sup> at (1, 0)
\nDifferentiating w.r.t. \u2018x’
\n\\(\\frac { dx }{ dy }\\) = 2x – 4x\u00b3
\nSlope of the tangent \u2018m\u2019 = (\\(\\frac { dx }{ dy }\\))(1, 0)<\/sub>
\n= 2 (1) – 4 (1)\u00b3 = -2
\nSlope of the normal –\\(\\frac { 1 }{ m }\\) = \\(\\frac { -1 }{ -2 }\\) = \\(\\frac { 1 }{ 2 }\\)
\nEquation of tangent is
\ny – y1<\/sub> = m (x – x1<\/sub>)
\ny – 0 = – 2 (x – 1)
\ny = -2x + 2
\n2x + y – 2 = 0
\nEquation of Normal is
\ny – y1<\/sub> = –\\(\\frac { 1 }{ m }\\)(x – x1<\/sub>)
\ny – 0 = \\(\\frac { 1 }{ 2 }\\)(x – 1)
\n2y = x- 1
\nx – 2y – 1 = 0<\/p>\n

(ii) y = x4<\/sup> + 2ex<\/sup> at (0, 2)
\nDifferentiating w.r.t. \u2018x’
\n\\(\\frac { dy }{ dx }\\) = 4x3<\/sup> + 2ex<\/sup>
\nSlope of the tangent \u2018m\u2019
\n(\\(\\frac { dy }{ dx }\\))(0, 2)<\/sub> = 4(0)\u00b3 + 2e0<\/sup> = 2
\nSlope of the Normal –\\(\\frac { 1 }{ m }\\) =-\\(\\frac { 1 }{ 2 }\\)
\nEquation of tangent is
\ny – y1<\/sub> = m(x – x1<\/sub>)
\n\u21d2 y – 2 = 2(x – 0)
\n\u21d2 y – 2 = 2x
\n\u21d2 2x – y + 2 = 0
\nEquation of Normal is
\ny – y1<\/sub> = –\\(\\frac { 1 }{ m }\\) (x – x1<\/sub>)
\ny – 2 = –\\(\\frac { 1 }{ 2 }\\)(x – 0)
\n2y – 4 = -x
\nx + 2y – 4 = 0<\/p>\n

(iii) y = x sin x at (\\(\\frac { \u03c0 }{ 2 }\\), \\(\\frac { \u03c0 }{ 2 }\\))
\nDifferentiating w.r.t. \u2018x’
\n\\(\\frac { dy }{ dx }\\) = x cos x + sin x
\nSlope of the tangent \u2018m\u2019 = (\\(\\frac { dy }{ dx }\\))(\u03c0\/2, \u03c0\/2)<\/sub>
\n= \\(\\frac { \u03c0 }{ 2 }\\) cos \\(\\frac { \u03c0 }{ 2 }\\) + sin \\(\\frac { \u03c0 }{ 2 }\\) = 1
\nSlope of the Normal –\\(\\frac { 1 }{ m }\\) = -1
\nEquation of tangent is
\ny – y1<\/sub> = m(x – x1<\/sub>)
\n\u21d2 y – \\(\\frac { \u03c0 }{ 2 }\\) = 1 (x – \\(\\frac { \u03c0 }{ 2 }\\))
\n\u21d2 x – y = 0
\nEquation of Normal is
\ny – y1<\/sub> = –\\(\\frac { 1 }{ m }\\)(x – x1<\/sub>)
\n\u21d2 y – \\(\\frac { \u03c0 }{ 2 }\\) = -1(x – \\(\\frac { \u03c0 }{ 2 }\\))
\n\u21d2 y – \\(\\frac { \u03c0 }{ 2 }\\) = -x + \\(\\frac { \u03c0 }{ 2 }\\)
\n\u21d2 x + y – \u03c0 = 0<\/p>\n

\"Samacheer<\/p>\n

(iv) x = cos t, y = 2 sin\u00b2 t at t = \\(\\frac { \u03c0 }{ 2 }\\)
\nat t = \\(\\frac { \u03c0 }{ 3 }\\), x = cos \\(\\frac { \u03c0 }{ 3 }\\) = \\(\\frac { 1 }{ 2 }\\)
\nat t = \\(\\frac { \u03c0 }{ 3 }\\), y = 2 sin\u00b2 \\(\\frac { \u03c0 }{ 3 }\\) = 2(\\(\\frac { 3 }{ 4}\\)) = \\(\\frac { 3 }{ 2 }\\)
\nPoint is (\\(\\frac { 1 }{ 2 }\\), \\(\\frac { 3 }{ 2 }\\))
\nNow x = cos t y = 2 sin\u00b2 t
\nDifferentiating w.r.t. \u2018t\u2019,
\n\\(\\frac { dx }{ dt }\\) = -sin t; \\(\\frac { dy }{ dt }\\) = 4 sin t cos t
\nSlope of the tangent
\n\"Samacheer
\nSlope of the Normal –\\(\\frac { 1 }{ m }\\) = \\(\\frac { 1 }{ 2 }\\)
\nEquation of tangent is
\ny – y1<\/sub> = m(x – x1<\/sub>)
\n\u21d2 y – \\(\\frac { 3 }{ 2 }\\) = -2(x – \\(\\frac { 1 }{ 2 }\\))
\n\u21d2 2y – 3 = – 4x + 2
\n\u21d2 4x + 2y – 5 = 0
\nEquation of Normal is
\ny – y1<\/sub> = –\\(\\frac { 1 }{ m }\\)(x – x1<\/sub>)
\n\u21d2 y – \\(\\frac { 3 }{ 2 }\\) = \\(\\frac { 1 }{ 2 }\\)(x – \\(\\frac { 1 }{ 2 }\\))
\n\u21d2 2 (2y – 3) = 2x – 1
\n\u21d2 4y – 6 = 2x – 1
\n\u21d2 2x – 4y + 5 = 0<\/p>\n

Question 6.
\nFind the equations of the tangents to the curve y = 1 + x\u00b3 for which the tangent is orthogonal with the line x + 12y = 12.
\nSolution:
\nCurve is y = 1 + x\u00b3
\nDifferentiating w.r.t \u2018x\u2019,
\nSlope of the tangent \u2018m\u2019 = \\(\\frac { dy }{ dx }\\) = 3x\u00b2
\nGiven line is x + 12y = 12
\nSlope of the line is –\\(\\frac { 1 }{ 12 }\\)
\nSince the tangent is orthogonal with the line, the slope of the tangent is 12.
\n\u2234 \\(\\frac { dy }{ dx }\\) = 12
\ni.e 3x\u00b2 = 12
\nx\u00b2 = 4
\nx = \u00b12
\nWhen x = 2, y = 1 + 8 = 9 \u21d2 point is (2, 9)
\nWhen x = -2, y = 1 – 8 = -7 \u21d2 point is (-2, -7)
\nEquation of tangent with slope 12 and at the j point (2, 9) is
\ny – 9 = 12 (x – 2)
\ny – 9 = 12x – 24
\n12x – y – 15 = 0
\nEquation of tangent with slope 12 and at the point (-2, -7) is
\ny + 7 = 12 (x + 2)
\ny + 7 = 12x + 24
\n12x – y + 17 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nFind the equations of the tangents to the curve y = –\\(\\frac { x+1 }{ x-1 }\\) which are parallel to the line x + 2y = 6.
\nSolution:
\n\"Samacheer
\nGiven line is x + 2y = 6
\nSlope of the line = –\\(\\frac { 1 }{ 2 }\\)
\nSince the tangent is parallel to the line, then the slope of the tangent is –\\(\\frac { 1 }{ 2 }\\)
\n\u2234 \\(\\frac { dy }{ dx }\\) = \\(\\frac { 2 }{ (x-1)^2 }\\) = –\\(\\frac { 1 }{ 2 }\\)
\n(x – 1)\u00b2 = 4
\nx – 1 = \u00b12
\nx = -1, 3
\nWhen x = – 1, y = 0 \u21d2 point is (-1, 0)
\nWhen x = 3, y = 2 \u21d2 point is (3, 2)
\nEquation of tangent with slope –\\(\\frac { 1 }{ 2 }\\) and at the point (-1, 0) is
\ny – o = –\\(\\frac { 1 }{ 2 }\\)(x + 1)
\n2y = -x – 1 \u21d2 x + 2y + 1 = 0
\nEquation of tangent with slope –\\(\\frac { 1 }{ 2 }\\) and at the point (3, 2) is 2
\ny – 2 = –\\(\\frac { 1 }{ 2 }\\) (x – 3)
\n2y – 4 = -x + 3
\nx + 2y – 7 = 0.<\/p>\n

Question 8.
\nFind the equation of tangent and normal to the curve given by x – 7 cos t andy = 2 sin t, t \u2208 R at any point on the curve.
\nSolution:
\nx = 7 cos t and y = 2 sin t, t \u2208 R
\nDifferentiating w.r.t. \u2018t\u2019,
\n\\(\\frac { dx }{ dt }\\) = -7 sin t and \\(\\frac { dy }{ dt }\\) = 2 cos t
\nSlope of the tangent \u2018m\u2019
\n\\(\\frac { dy }{ dx }\\) = \\(\\frac{\\frac { dy }{ dt }}{\\frac{ dx }{ dt }}\\) = \\(\\frac { 2 cot t }{ -7 sin t }\\)
\nAny point on the curve is (7 Cos t, 2 sin t)
\nEquation of tangent is y – y1<\/sub> = m (x – x1<\/sub>)
\ny – 2 sint = –\\(\\frac { 2 cot t }{ 7 sin t }\\) (x – 7 cos t)
\n7y sin t – 14 sin\u00b2 t = -2x cos t + 14 cos\u00b2 t
\n2x cos t + 7 y sin t – 14 (sin\u00b2 t + cos\u00b2 t) = 0
\n2x cos t + 7y sin t – 14 = 0
\nNow slope of normal is –\\(\\frac { 1 }{ 3 }\\) = \\(\\frac { 7 sin t }{ 2 cos t }\\)
\nEquation of normal is y – y1<\/sub> = –\\(\\frac { 1 }{ m }\\)(x – x1<\/sub>)
\ny – 2 sin t = \\(\\frac { 7 sin t }{ 2 cos t }\\) (x – 7 cos t)
\n2y cos t – 4 sin t cos t = 7x sin t – 49 sin t cos t 7x sin t – 2y cos t – 45 sin t cos t = 0<\/p>\n

\"Samacheer<\/p>\n

Question 9.
\nFind the angle between the rectangular hyperbola xy = 2 and the parabola x\u00b2 + 4y = 0
\nSolution:
\nGiven curves are xy = 2 ……… (1)
\nx\u00b2 + 4y = 0 ………. (2)
\nNow solving (1) and (2)
\nSubstituting (1) in (2)
\n\u21d2 x\u00b2 + 4(2\/x) = 0
\nx\u00b3 + 8 = 0
\nx\u00b3 = -8
\nx = -2
\nSubstituting in (1) \u21d2 y = \\(\\frac { 2 }{ -2 }\\) = -1
\n\u2234 Point of intersection of (1) and (2) is (-2, -1)
\nxy = 2 \u21d2 y = \\(\\frac { 2 }{ x }\\) ……….. (1)
\nDifferentiating w.r.t. \u2018x’
\n\"Samacheer
\nThe angle between the curves
\n\"Samacheer<\/p>\n

Question 10.
\nShow that the two curves x\u00b2 – y\u00b2 = r\u00b2 and xy = c\u00b2 where c, r are constants, cut orthogonally.
\nSolution:
\nGiven curves are x\u00b2 – y\u00b2 = r\u00b2 ……….. (1)
\nxy = c\u00b2 …….. (2)
\nLet (x1<\/sub>, y1<\/sub>) be the point of intersection of the given curves.
\n(1) \u21d2 x\u00b2 – y\u00b2 = r\u00b2
\nDifferentiating w.r.t \u2018x\u2019,
\n2x – 2y \\(\\frac { dx }{ dy }\\) = 0
\n\\(\\frac { dx }{ dy }\\) = \\(\\frac { x }{ y }\\)
\nnow (\\(\\frac { dx }{ dy }\\))(x1<\/sub>,y1<\/sub><\/sub>) = m1<\/sub> = \\(\\frac { x_1 }{ y_1 }\\)
\n(2) \u21d2 xy = c\u00b2
\nDifferentiating w.r.t \u2018x\u2019,
\n\"Samacheer
\nHence, the given curves cut orthogonally.<\/p>\n

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

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 12th Maths Guide Pdf Chapter 7 Applications of Differential Calculus Ex 7.2 Textbook Questions and Answers, Notes. BANDHANBNK Pivot Point Calculator Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2 Question 1. Find the slope of the tangent to the following curves at the …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[5],"tags":[],"class_list":["post-25330","post","type-post","status-publish","format-standard","hentry","category-class-12"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/25330"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=25330"}],"version-history":[{"count":3,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/25330\/revisions"}],"predecessor-version":[{"id":41723,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/25330\/revisions\/41723"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=25330"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=25330"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=25330"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}