{"id":5071,"date":"2023-12-27T06:32:13","date_gmt":"2023-12-27T01:02:13","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=5071"},"modified":"2023-12-28T17:47:18","modified_gmt":"2023-12-28T12:17:18","slug":"samacheer-kalvi-12th-maths-solutions-chapter-7-ex-7-2","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-12th-maths-solutions-chapter-7-ex-7-2\/","title":{"rendered":"Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2"},"content":{"rendered":"

You can Download Samacheer Kalvi 12th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/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 curves at the respective given points.
\n(i) y = x4<\/sup> + 2x2<\/sup> – x at x = 1
\n(ii) x = a cos3<\/sup> t, sin 3<\/sup> t at t = \\(\\frac{\\pi}{2}\\)
\nSolution:
\n(i) y = x4<\/sup> + 2x2<\/sup> – x at x = 1
\n\"Samacheer<\/p>\n

Question 2.
\nFind the point on the curve y = x2<\/sup> – 5x + 4 at which the tangent is parallel to the line 3x + y = 7.
\nSolution:
\nLet (x1<\/sub>, y1<\/sub>) be the required point.
\n\"Samacheer
\nGiven tangent is similarly to the line \u21d2 m1<\/sub> = m2<\/sub>
\n\u21d2 2x1<\/sub> – 5 = -3
\n\u21d2 2x1<\/sub> = -3 + 5 = 2
\n\u21d2 x1<\/sub> = 1
\nSubstituting x1<\/sub> = 1 in the curve.
\ny1<\/sub> = 1 – 5 + 4 = 0
\nSo the required point is (1, 0)<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nFind the points on the curves = x3<\/sup> – 6x2<\/sup> + x + 3 where the normal is parallel to the line x + y = 1729.
\nSolution:
\nLet (x1<\/sub>,y1<\/sub>) be the required point
\ny = x3<\/sup> – 6x2<\/sup>\u00a0+ x + 3
\n\"Samacheer
\nSubstituting x1<\/sub> values in the curve
\nwhen x1<\/sub> = 4, y1<\/sub> = 3
\nwhen x1<\/sub> = 4, y1<\/sub> = 43<\/sup> – 6(4)2<\/sup> + 4 + 3 = 64 – 96 + 4 + 3 = -25
\nSo the required points are (0, 3) and (4, -25)<\/p>\n

Question 4.
\nFind the points on the curve y2<\/sup> – 4xy = x2<\/sup>\u00a0+ 5 for which the tangent is horizontal.
\nSolution:
\n\"Samacheer
\nGiven that tangent is horizontal (i.e) tangent is parallel to x – axis
\n\u21d2 Equation of tangent will be of the form y = c
\n\"Samacheer
\nfrom (1) and (2) we get
\n\"Samacheer
\nSubstituting x = -2y in the equation of the curve we get
\ny2<\/sup> – 4(-2y)(y) = 4y2<\/sup>\u00a0+ 5
\n\u21d2 y2<\/sup> + 8y2<\/sup> – 4y2<\/sup>\u00a0= 5
\n\u21d2 5y2<\/sup> = 5 \u21d2 y2<\/sup>\u00a0= 1
\n\u21d2 y = \u00b1 1
\nwhen y = 1, x = -2 and when y = -1, x = 2 So the points are (2, -1), (-2, 1)<\/p>\n

Question 5.
\nFind the tangent and normal to the following curves at the given points on the curve.
\n\"Samacheer
\nSolution:
\n\"Samacheer
\nNow m = -2, (x1<\/sub>, y1<\/sub>) = (1, 0)
\nSo equation of the tangent is
\ny – y1<\/sub> = m(x – x1<\/sub>)
\n(i. e) y – 0 = -2(x – 1)
\ny = -2x + 2
\n2x+ y = 2
\nSlope of tangent = m = -2
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nFind the equations of the tangents to the curve y = 1 + x3<\/sup> for which the tangent is orthogonal with the line x + 12y = 12.
\nSolution:
\n\"Samacheer
\n\"Samacheer
\nSubstituting x1<\/sub> values in the curve
\nwhen x1<\/sub> = 2, y1<\/sub> = 9; when x1<\/sub> = -2, y1<\/sub> = -1
\nSo the points are (2, 9) and (-2, -7)
\nTo find the equations of tangents:
\nTangents are orthogonal to x + 12y = 12
\nSo equations of tangents will be of the form 12x – y = k
\nThe tangent passes through (2, 9) \u21d2 24 – 9 = k \u21d2 k = 15 .
\n\u2234 Equation of tangent is 12x – y = 15
\nThe tangent passes through (-2, -7) \u21d2 12 (-2) + 7 = k \u21d2 -17
\nSo equation of tangent is 12x – y = -17<\/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
\nTangent is parallel the line x + 2y = 6
\n\u21d2 Slope of tangent = Slope of line
\n\"Samacheer
\n\"Samacheer
\nwhen x1<\/sub> = -1, y1<\/sub> = 0; when x1<\/sub> = 3, y1<\/sub> = 2
\nSo the points are (-1, 0) and (3, 2). The tangents are parallel to x + 2y = 6. So equation of tangents will be of the form x + 2y = k.
\n\u2234 Equation of tangent is x + 2y = -1
\nThe tangent passes through (-1, 0) \u21d2 -1 = k
\nThe tangent passes through (3, 2) \u21d2 3 + 4 = k \u21d2 k = 7
\n\u2234 Equation of tangent is x + 2y = 7<\/p>\n

Question 8.
\nFind the equation of tangent and normal to the curve given by x = 7 cos t and y = 2sin t, t \u2208 R at any point on the curve.
\nSolution:
\n\"Samacheer
\n(2 cos t)y – 4 sint cos t = (7 sin t)x – 49 sin t cos t
\nx(7 sin t) – y(2 cos t) = 45 sin t cos t<\/p>\n

Question 9.
\nFind the angle between the rectangular hyperbola xy = 2 and the parabola x2<\/sup> + 4y = 0.
\nSolution:
\nSolving the given two equations. To find the point of intersection:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 10.
\nShow that the two curves x2<\/sup> – y2<\/sup> = r2<\/sup> and xy = c2<\/sup> where c, r are constants, cut orthogonally.
\nSolution:
\nLet (x1<\/sub>, y1<\/sub>) be the point of intersection of the two curves
\nI Curve: x2<\/sup> – y2<\/sup> = r2<\/sup>
\nDifferentiating w.r.to x
\n\"Samacheer
\nII Curve: xy = c2<\/sup>
\nDifferentiating w.r.to x
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Samacheer Kalvi 12th Maths Solutions Chapter 7 Applications of Differential Calculus Ex 7.2 Additional Questions<\/h3>\n

Question 1.
\nProve that the sum of the intercepts on the co-ordinate axes of any tangent to the curve x = a cos4<\/sup> \u03b8, y = a sin4<\/sup> \u03b8, 0 < \u03b8 < \\(\\frac{\\pi}{2}\\) is equal to a.
\nSolution:
\n\"Samacheer<\/p>\n

Question 2.
\nFind the equations of a normal to y = x3<\/sup> – 3x that is parallel to 2x + 18y – 9 = 0.
\nSolution:
\nLet (x1<\/sub>, y,) be a point on the curve.
\n\"Samacheer
\nIt is given that the normal is parallel to the line \u21d2 m1 <\/sub>= m2<\/sub>
\n\"Samacheer
\n9y + 18 = -x – 2
\nx + 9y + 20 = 0<\/p>\n

Question 3.
\nProve that the curves 2x2<\/sup> + 4y2<\/sup> = 1 and 6x2<\/sup> – 12y2<\/sup> = 1 cut each other at right angles.
\nSolution:
\nSolving the given two equations,
\n\"Samacheer
\n\"Samacheer
\nSimilarly it can be proved at the other points also.<\/p>\n

Question 4.
\nShow that the equation of the normal to the curve x = a cos3<\/sup> \u03b8, y = a sin3<\/sup> \u03b8 at ‘\u03b8’ is x cos \u03b8 – y sin \u03b8 = a cos 2\u03b8
\nSolution:
\n\"Samacheer
\nSo, equation of the normal is
\n\"Samacheer
\ni.e. y sin \u03b8 – a sin4<\/sup> \u03b8 = x cos \u03b8 – a cos4<\/sup> \u03b8
\nx cos \u03b8 – y sin \u03b8 = a cos4<\/sup> \u03b8 – a sin4<\/sup> \u03b8
\ni.e.x cos \u03b8 – y sin \u03b8 = a[cos2<\/sup> \u03b8 + sin2<\/sup>\u03b8][cos2<\/sup>\u03b8 – sin2<\/sup> \u03b8]
\n= a[cos 2 \u03b8]
\nSo, the equation of the normai is x cos \u03b8 – y sin \u03b8 = a cos 2\u03b8<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nIf the curve y2<\/sup> = x and xy = k are orthogonal, then prove that 8k2<\/sup> = 1.
\nSolution:
\ny2<\/sup> = x, xy = k
\nSolving the two equations, we get, (y2<\/sup>) (y) = k
\ny3<\/sup> = k
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 12th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. 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 curves at the …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[8],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/5071"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=5071"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/5071\/revisions"}],"predecessor-version":[{"id":60224,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/5071\/revisions\/60224"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=5071"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=5071"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=5071"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}