{"id":30567,"date":"2022-03-29T12:30:35","date_gmt":"2022-03-29T07:00:35","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=30567"},"modified":"2022-03-29T12:40:03","modified_gmt":"2022-03-29T07:10:03","slug":"ncert-solutions-for-class-12-maths-chapter-6-ex-6-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1\/","title":{"rendered":"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1"},"content":{"rendered":"

These NCERT Solutions for Class 12 Maths<\/a> Chapter 6 Application of Derivatives Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1\/<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nFind the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm
\nSolution:
\nLet A be the area of the circle with radius r.
\nA = \u03c0r\u00b2
\nDifferentiating w.r.t. r, \\(\\frac { dA }{ dr }\\) = 2\u03c0r
\ni. r = 3cm, \\(\\frac { dA }{ dr }\\) = 2\u03c0(3) = 6\u03c0cm\u00b2\/cm
\nii. r = 4cm, \\(\\frac { dA }{ dr }\\) = 2\u03c0(4) = 8\u03c0cm\u00b2 \/cm<\/p>\n

Question 2.
\nThe volume of a cube is increasing at the rate of 8 cm\u00b3\/s. How fast is the surface area increasing when the length of an edge is 12 cm?
\nSolution:
\nLet x be the length of the cube, V the volume and S the surface area at time l
\n\u2234 V = x\u00b3 and S = 6x\u00b2.
\n\\(\\frac { dV }{ dt }\\) = 8cm\u00b3\/s
\nV = x\u00b3
\nDifferentiating both sides, w.r.t. t
\n\"NCERT
\nWhen the edge of the cube is 12 cm, area increases at the rate of \\(\\frac { 8 }{ 3 }\\) cm\u00b2\/s.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nThe radius of a circle is increasing uniformly at the rate of 3 cm\/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
\nSolution:
\nLet r be the radius of the circle.
\nArea of circle = \u03c0r\u00b2 = A
\nalso \\(\\frac { dr }{ dt }\\) = 3 cm\/ sec.
\nr = 10 m
\nDifferentiating A w.r.t. t, we get
\n\\(\\frac { dA }{ dt }\\) = 2\u03c0r.\\(\\frac { dr }{ dt }\\)
\n= 2\u03c0(10)(3) = 60\u03c0 cm\u00b2\/sec<\/p>\n

Question 4.
\nAn edge of a variable cube is increasing at the rate of 3 cm\/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
\nSolution:
\nLet V be the volume and x be the edge of the cube at time \u2018t’.
\nV = x\u00b3
\nGiven \\(\\frac { dx }{ dt }\\) = 3 cm\/s
\nDifferentiating both sides, w.r.t. t
\n\\(\\frac { dV }{ dt }\\) = 3x\u00b2.\\(\\frac { dx }{ dt }\\) = 3x\u00b2(3) = 9x\u00b2
\nWhen x = 10 cm, \\(\\frac { dV }{ dt }\\) = 9 x (10)\u00b2 = 900 cm\u00b3\/s<\/p>\n

Question 5.
\nA stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm\/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
\nSolution:
\nLet r be the radius of a wave circle:
\n\\(\\frac { dx }{ dt }\\) = 5cm\/sec.
\nA = \u03c0r\u00b2
\nDifferentiating w.r.t. t
\n\\(\\frac { dA }{ dt }\\) = 2\u03c0r\\(\\frac { dr }{ dt }\\) = 2\u03c0r(5) = 10\u03c0r
\nWhen r = 8, \\(\\frac { dA }{ dt }\\) = 10\u03c0(8) = 80\u03c0 cm\u00b2\/s<\/p>\n

Question 6.
\nThe radius of a circle is increasing at the rate of 0.7 cm\/s. What is the rate of increase of its circumference?
\nSolution:
\nThe rate of change of circle w.r.t time t is given to be 0.7 cm\/sec. i.e. \\(\\frac { dr }{ dt }\\) = 0.7 cm\/sec.
\nNow, circumference of the circle is c = 2\u03c0r
\n\\(\\frac{d \\mathrm{C}}{d t}=2 \\pi \\cdot \\frac{d r}{d t}\\) = 2\u03c0(0.7) = 1.4\u03c0cm\/s
\nThe rate of increase of circumference is 1.4 \u03c0 cm\/s.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nThe length x of a rectangle is decreasing at the rate of 5 cm\/minute and the width y is increasing at the rate of 4 cm\/minute. When x = 8 cm and y = 6 cm, find the rates of change of
\n(a) the perimeter, and
\n(b) the area of the rectangle.
\nSolution:
\nThe length x of a rectangle is decreasing at dx the rate of 5cm\/min. \u21d2 \\(\\frac { dx }{ dt }\\) = – 5cm min … (i)
\nThe width y is increasing at the rate of 4cm\/min.
\n\\(\\frac { dy }{ dt }\\) = 4 cm\/minute
\n(a) p = 2x + 2y
\nDifferentiating both sides w.r.t., t,
\n\\(\\frac { dP }{ dt }\\) = 2\\(\\frac { dx }{ dt }\\) + 2\\(\\frac { dy }{ dt }\\)
\n= 2(-5) + 2(4) = – 2cm\/minute
\nPerimeter is decreasing at the rate of 2cm\/ minute<\/p>\n

(b) A = xy
\nDifferentiating both sides w.r.t., t,
\n\\(\\frac { dA }{ dt }\\) = x.\\(\\frac { dy }{ dt }\\) + y.\\(\\frac { dx }{ dt }\\)
\nWhen x = 8 cm and y = 6 cm
\n= 32 – 30 = 2 cm\u00b2\/minute
\nArea is increasing at the rate of 2 cm\u00b2\/minute<\/p>\n

Question 8.
\nA balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
\nSolution:
\nLet V be the volume and r be the radius of the balloon at time t.
\nGiven \\(\\frac { dV }{ dt }\\) = 900
\nV = \\(\\frac { 4 }{ 3 }\\) \u03c0r\u00b3
\nDifferentiating both sides w.r.t. t
\n\\(\\frac { dV }{ dt }\\) = \\(\\frac { d }{ dt }\\)(\\(\\frac { 4 }{ 3 }\\) \u03c0r\u00b3) = \\(\\frac { d }{ dr }\\)(\\(\\frac { 4 }{ 3 }\\) \u03c0r\u00b3) \\(\\frac { d }{ dr }\\)
\n900 = 4\u03c0r\u00b2.\\(\\frac { dr }{ dt }\\)
\n\\(\\frac { dr }{ dt }\\) = \\(\\frac{900}{4 \\pi r^{2}}\\) =\\(\\frac{225}{\\pi r^{2}}\\)
\nwhen r = 15 cm
\n\\(\\frac { dr }{ dt }\\) = \\(\\frac{225}{\\pi(15)^{2}}\\) = \\(\\frac{1}{\\pi}\\) cm\/sec.
\nWhen radius is 15 cm, the rate at which the radius increases is \\(\\frac { 1 }{ \u03c0 }\\) cm\/sec.<\/p>\n

Question 9.
\nA balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
\nSolution:
\nLet r be the variable radius of the balloon which is in the form of sphere Vol. of the sphere
\nDifferentiating w.r.t. r, \\(\\frac { dV }{ dr }\\) = 4\u03c0\u00b2
\nWhen r = 10, \\(\\frac { dV }{ dr }\\) = 4\u03c0(10)\u00b2 = 400\u03c0cm\u00b3\/cm
\ni.e., The rate at which volume of the bal\u00acloon is increasing with respect to the radius is 400\u03c0cm\u00b3\/cm<\/p>\n

\"NCERT<\/p>\n

Question 10.
\nA ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm\/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from die wall ?
\nSolution:
\n\"NCERT
\nLet AB be the ladder and OB be the wall. At an instant,
\nlet OA = x, OB = y,
\nx\u00b2 + y\u00b2 = 25 …(i)
\nDifferentiating both sides w.r.t., t,
\n2x.\\(\\frac { dx }{ dt }\\) + 2y.\\(\\frac { dy }{ dt }\\) = 0 … (1)
\nx\u00b2 + y\u00b2 = 25 \u21d2 y = \\(\\sqrt{25-x^{2}}\\)
\nWhen x = 4m, y = \\(\\sqrt{25-16}\\) = 3m
\nWhen x = 400 cm, y = 300 cm
\n(1) \u21d2 (2 x 400 x 2) + (2 x 300 x \\(\\frac { dy }{ dt }\\)) = 0
\n1600 + 600\\(\\frac { dy }{ dt }\\) = 0
\n\u2234\\(\\frac { dy }{ dt }\\) = \\(\\frac { -16 }{ 6 }\\) = \\(\\frac { -8 }{ 3 }\\) cm\/s
\n\u2234 Height of the ladder on the wall is de-creasing at the rate of \\(\\frac { 8 }{ 3 }\\) cm\/s.<\/p>\n

Question 11.
\nA particle moves along the curve 6y = x\u00b3 + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
\nSolution:
\nWe have
\n6y = x\u00b3 + 2 … (i)
\nDifferentiating both sides w.r.t., t,
\n\"NCERT
\nHence the required points on the curve are (4, 11) and (- 4, \\(\\frac { -31 }{ 3 }\\))<\/p>\n

Question 12.
\nThe radius of an air bubble is increasing at the rate of \\(\\frac { 1 }{ 2 }\\) cm\/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
\nSolution:
\nLet r be the radius then V = \\(\\frac { 4 }{ 3 } \\pi { r }^{ 3 }\\)
\n\\(\\frac { dr }{ dt } =\\frac { 1 }{ 2 } \\)cm\/sec
\n\\(\\frac { dv }{ dt } =\\frac { d }{ dt } \\left( \\frac { 4 }{ 3 } \\pi { r }^{ 3 } \\right) =\\frac { 4 }{ 3 } { \\pi .3r }^{ 2 }.\\frac { dr }{ dt } \\) = 2\u03c0r\u00b2
\nHence, the rate of increase of volume when radius is 1 cm = 2\u03c0 x 1\u00b2 = 2\u03c0 cm3<\/sup>\/sec.<\/p>\n

\"NCERT<\/p>\n

Question 13.
\nA balloon, which always remains spherical, has a variable diameter \\(\\frac { 3 }{ 2 } \\)(2x+1). Find the rate of change of its volume with respect to x.
\nSolution:
\nLet r be the radius and V be the volume at time t.
\n\"NCERT<\/p>\n

Question 14.
\nSand is pouring from a pipe at the rate of 12 cm3<\/sup>\/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
\nSolution:
\nLet V be the volume, r be the base radius and h be the height of the sand cone at time t.
\n\"NCERT
\nWhen h = 4 cm, \\(\\frac { dh }{ dt }\\) = \\(\\frac { 1 }{ 48\u03c0 }\\) cm\/sec
\nWhen the height is 4 cm, the rate at which the height of the sand cone increases is \\(\\frac { 1 }{ 48\u03c0 }\\) cm\/sec.<\/p>\n

Question 15.
\nThe total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007 x3<\/sup> – 0.003 x2<\/sup> + 15x + 4000. Find the marginal cost when 17 units are produced.
\nSolution:
\nC(x) = 0.007x\u00b3 – 0.003x\u00b2 + 15x + 4000
\nMarginal cost, MC = \\(\\frac { d }{ dx }\\) C(x)
\n= 3(0.007)x\u00b2 – 2(0.003)x + 15
\n= 0.021 x\u00b2 – 0.006x + 15
\nWhen x = 17
\nMC = 0.021(17\u00b2) – (0.006)17 + 15
\n= 6.069 – 0.102 + 15 = 20.967
\nWhen 17 units are produced,
\nMarginal cost = \u20b9 20.967
\n\u2234 The required marginal cost \u20b9 30.02 (approx)<\/p>\n

\"NCERT<\/p>\n

Question 16.
\nThe total revenue in rupees received from the sale of JC units of a product is given by
\nR (x) = 13x\u00b2 + 26x + 15.
\nFind the marginal revenue when x = 7.
\nSolution:
\nMarginal Revenue (MR)
\n= Rate of change of total revenue w.r.t. the
\nnumber of items sold at an instant = \\(\\frac { dR }{ dx } \\)
\nWe know R(x) = 13x\u00b2 + 26x + 15,
\nMR = \\(\\frac { dR }{ dx } \\) = 26x + 26 = 26(x+1)
\nNow x = 7, MR = 26 (x + 1) = 26 (7 + 1) = 208
\nHence, Marginal Revenue = Rs 208.<\/p>\n

Question 17.
\nThe rate of change of the area of a circle with respect to its radius r at r = 6 cm is
\n(a) 10\u03c0
\n(b) 12\u03c0
\n(c) 8\u03c0
\n(d) 11\u03c0
\nSolution:
\n(b) \u2235 A = \u03c0r\u00b2 \u21d2 \\(\\frac { dA }{ dr } \\) = 2\u03c0 x 6 = 12\u03c0 cm\u00b2\/radius<\/p>\n

Question 18.
\nThe total revenue in Rupees received from the sale of x units of a product is given by R (x) = 3x\u00b2 + 36x + 5. The marginal revenue, when x = 15 is
\n(a) 116
\n(b) 96
\n(c) 90
\n(d) 126
\nSolution:
\n(d) R(x) = 3x\u00b2 + 36x + 5 ,
\nMR = \\(\\frac { dR }{ dx } \\) = 6x + 36 ,
\nAt x = 15; \\(\\frac { dR }{ dx } \\) = 6 x 15 + 36 = 90 + 36 = 126<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1 Question 1. Find the rate of change of the area of a circle with respect to …<\/p>\n

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","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":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-6-ex-6-1\/ NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.1 Question 1. 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