{"id":22154,"date":"2023-10-20T12:50:31","date_gmt":"2023-10-20T07:20:31","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22154"},"modified":"2023-10-20T17:08:56","modified_gmt":"2023-10-20T11:38:56","slug":"inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-1st-year-maths-1b-applications-of-derivatives-solutions-ex-10c\/","title":{"rendered":"Inter 1st Year Maths 1B Applications of Derivatives Solutions Ex 10(c)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the lengths of subtangent and sub-normal at a point of the curve
\nSolution:
\nEquation of the curve is y = b. sin \\(\\frac{x}{a}\\)
\n\\(\\frac{dy}{dx}\\) = b. cos \\(\\frac{x}{a}.\\frac{1}{a}=\\frac{b}{a}\\). cos \\(\\frac{x}{a}\\)
\nLength of the sub-tangent = |\\(\\frac{y_{1}}{f^{\\prime}\\left(x_{1}\\right)}\\)|
\n= \\(\\frac{b \\cdot \\sin \\frac{x}{a}}{\\frac{b}{a} \\cdot \\cos \\frac{x}{a}}=\\left|a \\cdot \\tan \\frac{x}{a}\\right|\\)
\nLength of the sub-normal = |y1<\/sub>, f'(x1<\/sub>)|
\n\"Inter<\/p>\n

Question 2.
\nShow that the length of sub normal at any point on the curve xy = a2 varies as the cube of the ordinate of the point.
\nSolution:
\nEquation of the curve is xy = a\u00b2
\ny = \\(\\frac{a^{2}}{x}\\)
\n\\(\\frac{dy}{dx}=\\frac{-a^{2}}{x^{2}}\\)
\nLength of the sub-normal = |y1<\/sub>, f’ (x1<\/sub>)|
\n\"Inter
\n\\(\\frac{y_{1}^{3}}{a^{2}}\\) \u221d y3<\/sup>1<\/sub> = cube of the ordinate.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nShow that at any point (x, y) on the curve y = bex\/a<\/sup>, the length of the sub\u00actangent is a constant and the length of the sub normal is \\(\\frac{y^{2}}{a}\\).
\nSolution:
\nEquation of the curve is y = b.ex\/a<\/sup>
\n\\(\\frac{a^{2}}{x}\\) = b.ex\/a<\/sup>.\\(\\frac{1}{a}=\\frac{y}{a}\\)
\nLength of the sub tangent
\n= |\\(\\frac{y_{1}}{f^{\\prime}\\left(x_{1}\\right)}\\)| = \\(\\frac{y_{1}}{\\left(\\frac{y_{1}}{a}\\right)}\\) = a = constant
\nLength of the sub-normal
\n= |y1<\/sub>f'(x1<\/sub>)| = |y1<\/sub>\\(\\frac{y_{1}}{a}\\)| = \\(\\frac{y_{1}^{2}}{a}\\).<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the value of k so that the length of the sub normal at any point on the curve xyk<\/sup> = ak + 1<\/sup> is a constant.
\nSolution:
\nEquation of the curve is x.yk<\/sup> = ak + 1<\/sup>
\nDifferentiating w.r.to x
\nx.k.yk-1<\/sup> \\(\\frac{dy}{dx}\\) + yk<\/sup>.1 = 0
\nx.k.yk-1<\/sup> \\(\\frac{dy}{dx}\\) = -yk<\/sup>
\n\\(\\frac{dy}{dx}=\\frac{-y^{k}}{k \\cdot x \\cdot y^{k-1}}=-\\frac{y}{kx}\\)
\nLength of the sub-normal = |y1<\/sub>f'(x1<\/sub>|
\n\"Inter
\nLength of the sub-normal is constant at any point on the curve is independent of x1<\/sub> and y1<\/sub>
\n\u2234 \\(\\frac{y_{1}^{k+2}}{k \\cdot a^{k+1}}\\) is independent of x1<\/sub>, y1<\/sub>
\n\u21d2 k + 2 = 0 \u21d2 k = -2<\/p>\n

Question 2.
\nAt any point t on the curve x = a (t + sin t), y = a (1 – cos t), find the lengths of tangent, normal, sub tangent and sub
\nnormal.
\nSolution:
\nEquation of the curve is x = a (t + sin t), y = a (1 – cost)
\n\\(\\frac{dx}{dt}\\) = a (1 + cos t), \\(\\frac{dy}{dt}\\) = a. sin t
\n\"Inter
\nLength of the tangent = |y1<\/sub>|\\(\\sqrt{1+\\frac{1}{\\left[f^{\\prime}\\left(x_{1}\\right)\\right]^{2}}}\\)
\n\"Inter
\nLength of the normal = |y1<\/sub>|\\(\\sqrt{1+\\left[f\\left(x_{1}\\right)\\right]^{2}}\\)
\n\"Inter
\nLength of the sub tangent.
\n\"Inter
\n= |a (2 sin t\/2. cos t\/2|
\n= |a. sin t|
\nLength of the sub normal
\n= |y1<\/sub>.f'(x1<\/sub>)| = |a(1 – cos t).\\(\\frac{\\sin t \/ 2}{\\cos t \/ 2}\\)|
\n= |2a sin\u00b2 t\/2. tan t\/2|
\n= |2a sin\u00b2 t\/2. tan t\/2|<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the lengths of normal and sub normal at a point on the curve y = \\(\\frac{a}{2}\\) (ex\/a<\/sup> + e-x\/a<\/sup>).
\nSolution:
\nEquation of the curve is y = \\(\\frac{a}{2}\\) (ex\/a<\/sup> + e-x\/a<\/sup>)
\n= a. cosh (\\(\\frac{x}{a}\\))
\nLength of the normal = |y1<\/sub>|\\(\\sqrt{1+\\left[f\\left(x_{1}\\right)\\right]^{2}}\\)
\n= |a.cosh \\(\\frac{x}{a}\\)|\\(\\sqrt{1+\\sinh ^{2} \\frac{x}{a}}\\)
\n= a. cosh \\(\\frac{x}{a}\\). cosh \\(\\frac{x}{a}\\) = a. cosh\u00b2 \\(\\frac{x}{a}\\)
\nLength of the sub normal = |y1<\/sub> f'(x1<\/sub>)|
\n\"Inter<\/p>\n

Question 4.
\nFind the lengths of subtangent, sub normal at a point t on the curve x = a (cos t + t sin t), y=a (sin t – t cos t)
\nSolution:
\nEquations of the curve are x = a (cos t + t sin t)
\n\\(\\frac{dx}{dt}\\) = a (- sin t + sin t +1 cost) = at cos t
\ny = a (sin t – t cos t)
\n\\(\\frac{dy}{dt}\\) = a (cos t – cos t +1 sin t) = a t sin t
\n\"Inter
\nLength of the sub tangent
\n\"Inter
\n= |a cot t (sin t – t cos t)|
\nLength of the sub-normal = |y1<\/sub>.f'(x1<\/sub>)|
\n= |a (sin t – t cos t) tan t|
\n= |a tan t (sin t – t cos t)|<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(c) I. Question 1. Find the lengths of subtangent and sub-normal at a point of the curve … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22154"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22154"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22154\/revisions"}],"predecessor-version":[{"id":22155,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22154\/revisions\/22155"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22154"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22154"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22154"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}