{"id":19665,"date":"2020-01-13T15:36:30","date_gmt":"2020-01-13T10:06:30","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=19665"},"modified":"2021-07-02T15:29:40","modified_gmt":"2021-07-02T09:59:40","slug":"2nd-puc-basic-maths-question-bank-chapter-19-ex-19-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-basic-maths-question-bank-chapter-19-ex-19-1\/","title":{"rendered":"2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1"},"content":{"rendered":"

Students can Download Basic Maths Exercise 19.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1<\/h2>\n

Part-A<\/span><\/p>\n

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Two Marks Questions and Answers<\/strong><\/p>\n

Question 1.
\nThe displacement ‘s\u2019 of a particle at time \u2018t\u2019 is given by S = 4t3<\/sup> – 6t2<\/sup> + t – 7. Find the velocity and acceleration when t = 2 sec.
\nAnswer:
\nGiven S = 4t3<\/sub> – 6t + t – 7
\nVelocity = v = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 12t2<\/sup> -12t + 1
\nAt t = 2secs, v = 12(2)2<\/sup> – 12(2) + 1 = 48 – 24 + 1 = 25 units\/sec.
\nAt t = 2 sec, acceleration = 24.2 – 12 = 48 – 12 = 36 units\/sec2<\/sup>.<\/p>\n

Question 2.
\nIf S = 5t2<\/sup> + 4t – 8. Find the initial velocity and acceleration, (s = displacement, t = time).
\nAnswer:
\nGiven s = 5t2<\/sup> + 4t – 8
\nV = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 10t + 4 dt
\nInitial velocity is velocity when t = 0
\ni.e., = 10.0 + 4 = 4 units\/sec.
\nAcceleration = \\(\\frac{\\mathrm{dv}}{\\mathrm{dt}}\\) = 10 units \/sec2<\/sup>.<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nA stone thrown vertically upward rises \u2018s\u2019 ft. in \u2018t\u2019 sec. where s = 80t – 16t2<\/sup>. What its velocity after 2 sec.? Find the acceleration?
\nAnswer:
\nGiven S = 80t -16t2<\/sup>
\nv = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 80 – 32t
\nAt t=2 sec, v = 80 – 32 (2)
\n= 80 – 64 = 16 ft.\/sec.
\nAcceleration = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = -22 ft\/sec2<\/sup> .<\/p>\n

Question 4.
\nA body is thrown vertically upwards its distance S feet is’t’ sec. is given by S = 5 + 12t – t2<\/sup>. Find the greatest highest by the body.
\nAnswer:
\nGiven s = 5 + 12t – t2<\/sup>
\nv = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 12 – 2t dt
\nMaximum height \u21d2 Kinetic energy = 0 \u21d2 v = 0 \u21d2 12 – 2t = 0 \u21d2 t = 6 sec
\n\u2234 Greatest height = s = 5 + 12.6 – 62<\/sup>
\n= 5 + 72 – 36 = 77 – 36 = 41 feet.<\/p>\n

Question 5.
\nIf v = \\(\\sqrt{s^{2}+1}\\) prove that acceleration is \u2018S’ (V = velocity, S = displacement).
\nAnswer:
\nGiven v = \\(\\sqrt{s^{2}+1}\\)
\nAcceleration = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = \\(\\frac{1}{2 \\sqrt{s^{2}+1}} \\cdot 2 s \\frac{d s}{d t}=\\frac{1}{2 v}\\) . 2 . s . v . s units \/ sec2<\/sup>.<\/p>\n

Question 6.
\nIf S = at3<\/sup> + bt. Find a and b given that when t = 3 velocity is ‘O\u2019 and the acceleration is 14 unit. (S = displacement, t = time).
\nAnswer:
\nGiven s = at3<\/sup>\u00a0+ bt; v = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 3at2<\/sup>+b
\nAcceleration = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 6at
\nWhen v = 0 then 3at2<\/sup> + b = 0 \u21d2 27a + b = 0.
\nt = 3 sec
\nWhen acceleration = 14 then 14 = 6a.3, b = -27a = \\(-\\frac{27.7}{9}\\) = -21
\nF = 3 sec. a = \\(\\frac{14}{18}=\\frac{7}{9}\\)
\n\u2234 a = \\(\\frac { 7 }{ 9 }\\) and b = -21.<\/p>\n

Question 7.
\nWhen the brakes are applied to moving car, the car travels a distance ‘s’ ft. in ‘t’ see given by s = 8t – 6t2<\/sup> when does the car stop?
\nAnswer:
\nGiven s = 8t – 6t2<\/sup>
\nv = \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = 8 – 12t car stops when v = 0
\n\u2234 8 – 12t = 0 \u21d2 t = \\(\\frac{8}{12}=\\frac{2}{3}\\) sec.<\/p>\n

\"KSEEB<\/p>\n

Part-B<\/span><\/p>\n

2nd PUC Basic Maths Application of Derivatives Ex 19.1 Three Marks Questions and Answers.<\/strong><\/p>\n

Question 1.
\nThe radius of sphere is increasing at the rate of 0.5 mt\/sec. Find the rate of increase of its surface area and volume after 3 sec.
\nAnswer:
\nGiven \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = 0.5 , t = 3 sec, \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = ? \\(\\frac{\\mathrm{dv}}{\\mathrm{dt}}\\) = ?
\nsurface area = s = 4\u03c0r2
\n\\(\\frac{d s}{d t}\\) = 4\u03c0 . 2r . \\(\\frac{d r}{d t}\\)<\/sup>
\n= 4\u03c0 \u00d7 2 \u00d7 1.5 \u00d7 0.5
\n= 6\u03c0 m2<\/sup>\/sec
\ndr = 0.5 \u00d7 dt
\n\u21d2 r = 0.5 t
\n= 0.5 \u00d7 3
\n= 1.5<\/p>\n

Question 2.
\nThe surface area of a spherical bubble is increasing at the rate of a 0.8cm2<\/sup> \/ sec. Find at what rate is its volume increasing when r = .25cm [r = radius of the sphere].
\nAnswer:
\nGiven \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = 0.8 cm2<\/sup> \/ sec. r = 2.5 cm, \\(\\frac{\\mathrm{dv}}{\\mathrm{dt}}\\) = ?
\ns = 4\u03c0r2<\/sup>
\n\\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = 4\u03c0 . 2r \u00d7 \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\)
\n0.8 = 4\u03c0 \u00d7 2 \u00d7 2.5. \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) \u21d2 \\(\\frac{0.8}{8 \\pi \\times 2.5}=\\frac{0.1}{2.5 \\pi}\\)
\nv = \\(\\frac { 4 }{ 3 }\\)\u03c0r3<\/sup>
\n\\(\\frac{d v}{d t}=\\frac{4}{3} \\pi 3 r^{2} \\cdot \\frac{d r}{d t}=4 \\pi \\times(2.5)^{2} \\times \\frac{0.1}{2.5 \\pi}=1 \\mathrm{cc} \/ \\mathrm{sec}\\)<\/p>\n

Question 3.
\nA spherical balloon is being inflated at the rate 35cc\/sec. Find the rate at which the surface area of the balloon increases when its diameter is 14cm.
\nAnswer:
\nGiven \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = 35cc \/sec 2r = 14 \u21d2 r = 7, \\(\\frac{\\mathrm{ds}}{\\mathrm{dt}}\\) = ?, \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = ?
\nv = \\(\\frac { 4 }{ 3 }\\)\u03c0r3<\/sup> s = 4\u03c0r2<\/sup>
\n\\(\\frac{d V}{d t}=\\frac{4}{3} \\pi 3 r^{2} \\frac{d r}{d t} \\quad \\frac{d S}{d t}=4 \\pi 2 r \\frac{d r}{d t}\\)
\n35 = 4\u03c0 . 72<\/sup> \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = 4\u03c0 \u00d7 2 \u00d7 7 \u00d7 \\(\\frac{5}{28 \\pi}\\)
\n\\(\\frac{d r}{d t}=\\frac{35}{196 \\pi}=\\frac{5}{28 \\pi}=10 \\mathrm{cm}^{2} \/ \\mathrm{sec}\\)<\/p>\n

Question 4.
\nThe radius of a circular plate is increasing at the rate of \\(\\frac{2}{3 \\pi}\\) cm\/sec. Find the rate of change of its area when the radius is 6cm.
\nAnswer:
\nGiven \\(\\frac{d r}{d t}=\\frac{2}{3 \\pi} r=6 \\mathrm{cm} \\frac{d A}{d t}=?\\)
\nA = \u03c0r2<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nA circular patch of oil spreads on water the area growing at the rate of 16cm2<\/sup>\/min. How fast are radius and the circumference increasing when the diameter is 12cm.
\nAnswer:
\nGiven \\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = 16cm2<\/sup>\/min, d = 2r = 12cm, \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = ? \\(\\frac{\\mathrm{dc}}{\\mathrm{dt}}\\) = ? r = 6cm
\nA = \u03c0r2<\/sup> c = 2\u03c0r
\n\\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = \u03c0 . 2r . \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) \\(\\frac{\\mathrm{dc}}{\\mathrm{dt}}\\) = 2\u03c0 \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\)
\n16 = 2\u03c0 .6 . \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) 2\u03c0 . \\(\\frac{4}{3 \\pi}=\\frac{8}{3}\\)cm \/ min
\n\u21d2 \\(\\frac{d r}{d t}=\\frac{16}{12 \\pi}=\\frac{4}{3 \\pi} \\mathrm{cm} \/ \\mathrm{min}\\)<\/p>\n

Question 6.
\nA stone is dropped into a pond waved in the form of circles are generated and the radius of the outer most ripple increases at the rate 2 inches\/sec. How fast is the area increasing when the (a) radius is 5 inches (b) after 5 sec.?
\nAnswer:
\nGiven \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = 2inch\/sec, r = 5 inch, \\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = ?
\n(a) A = \u03c0r2<\/sup>
\n\\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = \u03c02r. \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = \u03c0 \u00d7 2 \u00d7 5 \u00d7 2 = 20\u03c0 sq. inches \/ sec.<\/p>\n

(b) After 5 sec, dr = 2dt
\nr = 2t
\n\u21d2 when t = 5, r = 10 inches
\n\\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = \u03c0 . 2r \\(\\frac{\\mathrm{dr}}{\\mathrm{dt}}\\) = \u03c0 \u00d7 2 \u00d7 20 \u00d7 2 = 40\u03c0 square inches\/sec.<\/p>\n

Question 7.
\nThe side of an equilateral triangle is increasing at the rate \\(\\sqrt{3}\\) cm.\/sec. Find the rate at which its area is increasing when its side is 2 meters.
\nAnswer:
\nGiven \\(\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\) = \\(\\sqrt{3}\\) cm\/sec., x = 2 meters, \\(\\frac{\\mathrm{dA}}{\\mathrm{dt}}\\) = ?
\nArea of equilateral \u0394le<\/sup> = A = \\(\\frac{\\sqrt{3}}{4}\\) x2<\/sup>
\n\\(\\frac{d A}{d t}=\\frac{\\sqrt{3}}{4} \\cdot 2 x \\frac{d x}{d t}\\)
\n= \\(\\frac{\\sqrt{3}}{4}\\) . 2. 200. \\(\\sqrt{3}\\) = 300cm2<\/sup> \/ sec.<\/p>\n

Question 8.
\nWater is being poured at the rate of 30 mt3<\/sup>\/min. into a cylindrical vessel whose base is a circle of radius 3 mt. Find the rate at which the level of water is rising?
\nAnswer:
\nGiven r = 3mts, \\(\\frac{\\mathrm{dv}}{\\mathrm{dt}}\\) = 30 m3<\/sup>\/min, \\(\\frac{\\mathrm{dh}}{\\mathrm{dt}}\\) = ?
\nV = \u03c0r2<\/sup>h, r = constant
\n\\(\\frac{\\mathrm{dv}}{\\mathrm{dt}}\\) = \u03c0 . (3)2<\/sup> . \\(\\frac{\\mathrm{dh}}{\\mathrm{dt}}\\)
\n30 = 9\u03c0 \\(\\frac{\\mathrm{dh}}{\\mathrm{dt}}\\) \u21d2 \\(\\frac{\\mathrm{dh}}{\\mathrm{dt}}\\) = \\(\\frac{10}{3 \\pi}\\) meter\/min.<\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nSand is being dropped at the rate of 10 mt3<\/sup>\/sec. into a conical pile. If the height of the pile twice the radius of the base, at what rate is the height to the pile is increasing when the sand in the pile is 8mt high.
\nGiven
\n\\(\\frac{d v}{d t}\\) = 10m3<\/sup>\/sec, h = 2r, h = 8, \\(\\frac{d h}{d t}\\) = ?
\nr = \\(\\frac{\\mathrm{h}}{2}\\) \\(\\mathrm{v}=\\frac{1}{3} \\pi \\mathrm{r}^{2} \\mathrm{h}=\\frac{1}{3} \\pi \\cdot\\left(\\frac{\\mathrm{h}}{2}\\right)^{2} \\cdot \\mathrm{h}=\\frac{1}{3} \\pi \\frac{\\mathrm{h}^{3}}{4}\\)
\n\\(\\mathrm{v}=\\frac{1}{12} \\pi \\mathrm{h}^{3} \\Rightarrow \\frac{\\mathrm{dv}}{\\mathrm{dt}}=\\frac{\\pi}{12} \\cdot 3 \\mathrm{h}^{2} \\frac{\\mathrm{dh}}{\\mathrm{dt}} ; \\quad 10=\\frac{\\pi}{4} \\cdot 8^{2} \\cdot \\frac{\\mathrm{dh}}{\\mathrm{dt}}\\)
\n\\(\\frac{d h}{d t}=\\frac{40}{64 \\pi}=\\frac{5}{8 \\pi} \\mathrm{m} \/ \\mathrm{sec}\\)<\/p>\n

Question 10.
\nA ladder of 15ft. long leans against a smooth vertical wall. If the top slides downwards at the rate of 2ft sec. Find how fast the lower and is moving when the lower end is 12ft. from the wall.
\nAnswer:
\n\"2nd
\nGiven \\(\\frac{\\mathrm{d} y}{\\mathrm{dt}}\\) = 2ft \/sec, x = 12
\nFrom fig, x2<\/sup> + y2<\/sup> = 152<\/sup>
\n122<\/sup> + y2<\/sup> = 152<\/sup>
\ny2<\/sup> = 152<\/sup> – 122<\/sup>
\ny = \\(\\sqrt{225-144}\\)
\ny = \\(\\sqrt{81}\\) = 9
\nx2<\/sup> + y2<\/sup> = 152<\/sup> \u21d2 2x \\(\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\) + 2y\\(\\frac{\\mathrm{dy}}{\\mathrm{dt}}\\) = 0; 2.12.\\(\\frac{\\mathrm{dx}}{\\mathrm{dt}}\\) + 2 .9 . 2 = 0
\n\u21d2 \\(\\frac{d x}{d t}=\\frac{-36}{24}=\\frac{-3}{2} f t \/ \\sec\\)<\/p>\n

Question 11.
\nAn edge of a variable cube is increasing at the rate of 10cm\/sec. How fast the volume and also its surface area is increasing when the edge is 5cm long.
\nAnswer:
\nGiven \\(\\frac{d x}{d t}\\) = 10 cm\/sec, x = 5 cm, \\(\\frac{d v}{d t}\\) = ? \\(\\frac{d s}{d t}\\) = ?
\n(i) V = x3<\/sup>
\n\\(\\frac{d v}{d t}\\) = 3x2<\/sup> \\(\\frac{d x}{d t}\\) = 3 \u00d7 (52<\/sup>) \u00d7 10 = 750 cm3<\/sup>\/sec.<\/p>\n

(ii) S = 6x2<\/sup> .
\n\\(\\frac{d s}{d t}\\) = 12 \u00d7 .\\(\\frac{d x}{d t}\\) = 12.5 .10 = 600 cm2<\/sup>\/sec.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nA man 6ft. tall is moving directly away from a lamp post of height 10ft. above the ground. If he is moving at the rate 3ft.\/sec. Find the rate at which the length of his shadow is increasing and also the tip of his shadow is moving?
\nAnswer:
\n\"2nd
\nLet the shadow be x & y be the distance the man walks
\nGiven \\(\\frac{d y}{d t}\\) = 3ft\/sec From a similar \u0394les<\/sup> we have
\n\\(\\frac{6}{10}=\\frac{x}{x+y}\\)
\n6x + 6y = 10x
\n6y = 4x \u21d2 3y = 2x
\n\u21d2 \\(3 \\frac{d y}{d t}=2 \\frac{d x}{d t} \\Rightarrow 2 \\frac{d x}{d t}=3.3=9\\)
\n\u2234 the shadow is increasing at the rate.
\n\u2234 \\(\\frac{d x}{d t}=\\frac{9}{2}\\) ft\/sec & the tip of the shadow moves is
\n\\(\\frac{d x}{d t}+\\frac{d y}{d t}=\\frac{9+6}{2}=\\frac{15}{2} \\mathrm{ft} \/ \\mathrm{sec}\\)<\/p>\n

Question 13.
\nThe height of circular cone is 30 cm. and it is constant. The radius of the base is increasing at the rate of 0.25cm\/sec. Find the rate of increase of volume of the cone when the radius of base is 10cm.
\nAnswer:
\ndr
\nGiven h = 30 cm, \\(\\frac{d r}{d t}\\) = 0.25cm\/sec. r = 10cm. dt
\nV = \\(\\frac { 1 }{ 3 }\\) \u03c0r2<\/sup>h
\n\\(\\frac{d v}{d t}\\) = \\(\\frac{\\pi}{3}\\)h.2r. \\(\\frac{d r}{d t}\\) = \u03c0. \\(\\frac { 30 }{ 3 }\\) . 20.(0.25) = 50\u03c0 cm2<\/sup> \/ sec<\/p>\n

Question 14.
\nThe volume of a spherical ball in increasing at the rate 4\u03c0cc\/sec. Find the rate of increase of the radius of the ball when the volume is 288\u03c0CC.
\nAnswer:
\nGiven V = 288\u03c0 C.C., \\(\\frac{d v}{d t}\\) = 4\u03c0cc\/ sec \\(\\frac{d r}{d t}\\) = ?
\nV = \\(\\frac { 4 }{ 3 }\\) \u03c0r3<\/sup>
\n\"2nd
\nr = 6cm
\n\\(\\frac{d v}{d t}=\\frac{4}{3} \\pi \\cdot 3 r^{2} \\frac{d r}{d t} 4 \\pi=\\frac{4}{3} \\pi \\times 3 \\times 36 \\times \\frac{d r}{d t} \\Rightarrow \\frac{d r}{d t}=\\frac{1}{36} \\mathrm{cm} \/ \\mathrm{sec}\\)<\/p>\n

Question 15.
\nA drop of ink spreads over a blotting paper so that the circumferences of the blot is 4\u03c0cm and it changes 3cm\/sec. Find the rate of increase of its radius and also find the rate of increase of its area?
\nAnswer:
\nGiven c = 4\u03c0, \\(\\frac{d C}{d t}\\) = 3cm \/ sec \\(\\frac{d A}{d t}\\) = ? \\(\\frac{d r}{d t}\\) = ?
\nCircumference = c = 2\u03c0r
\n4\u03c0 = 2\u03c0r \u21d2 r = 2
\nAgain
\nC = 2\u03c0r & A = \u03c0r2<\/sup>
\n\\(\\frac{d c}{d t}\\) = 2\u03c0 . \\(\\frac{d r}{d t}\\) \\(\\frac{d A}{d t}\\) = \u03c0 . 2r. \\(\\frac{d r}{d t}\\)
\n3 = 2\u03c0 . \\(\\frac{d r}{d t}\\) = \u03c0 . 2. 2. \\(\\frac{3}{2 \\pi}\\)
\n\u21d2 \\(\\frac{d r}{d t}\\) = \\(\\frac{3}{2 \\pi}\\) cm\/ sec \\(\\frac{d A}{d t}\\) = 6cm2<\/sup> \/ sec<\/p>\n

Question 16.
\nA circular plate of metal is heated so that its radius increase at the rate of O.lmm\/min. At what rate is the [plate\u2019s area increasing when the radius is 25cm [1cm = 10mm].
\nAnswer:
\nGiven \\(\\frac{d r}{d t}\\) = 0.1 mm\/min, r = 25 cm, \\(\\frac{d A}{d t}\\) A = \u03c0r2<\/sup>
\n\\(\\frac{d A}{d t}\\) = \u03c0. 2r . \\(\\frac{d r}{d t}\\) = \u03c0 . 2. \/250 (0.1) = 50\u03c0mm2<\/sup> \/min.<\/p>\n

\"KSEEB<\/p>\n

Question 17.
\nThe surface area of a spherical soap bubble increasing at the rate of 0.6cm2<\/sup>\/sec. Find the rate at which its volume is increasing when its radius is 3cm.
\nAnswer:
\nGiven \\(\\frac{ds}{d t}\\) = 0.6 cm2<\/sup> \/ sec, r = 3cm, \\(\\frac{d v}{d t}\\) = ?
\ns = 4 \u03c0r2<\/sup> &
\n\\(\\frac{d s}{d t}\\) = 4\u03c0. 2r. \\(\\frac{d r}{d t}\\)
\n0.6 = 4\u03c0 \u00d7 3 \u00d7 3 \u00d7 \\(\\frac{d r}{d t}\\)
\n\u2234 \\(\\frac{d r}{d t}=\\frac{0.6}{6 \\times 4 \\pi}=\\frac{0.1}{4 \\pi} \\mathrm{cm} \/ \\mathrm{sec}\\)
\n\\(\\frac { 1 }{ 40 }\\) \u03c0cm\/sec.
\nv = \\(\\frac { 4 }{ 3 }\\) \u03c0r3<\/sup>
\n\\(\\frac{d v}{d t}=\\frac{4}{3} \\pi \\cdot 3 r^{2} \\frac{d r}{d t}\\)
\n= 4\u03c0r2<\/sup> \\(\\frac{d r}{d t}\\)
\n= 4\u03c0(3)2<\/sup> . \\(\\frac{0.1}{4 \\pi}\\)
\n= 0.9 cm3<\/sup> \/ sec.<\/p>\n

Question 18.
\nA rod 13 feet long slides with it end A and B as two straight lines at right angles which meet at \u2018O\u2019. If A is moved away from O with a uniform speed at 4ft.\/sec., find the speed of the end B move when A is 5 feet from O.
\nAnswer:
\n\"2nd
\nFrom fig we have
\nx2<\/sup> + y2<\/sup> = 132<\/sup>
\ny2<\/sup> = 132<\/sup> – 152<\/sup> = 144
\ny = \\(\\sqrt{144}\\) = 12
\nAls0
\nx2<\/sup> + y2<\/sup> = 132<\/sup> \u21d2 2x \\(\\frac{d x}{d t}\\) + 2y\\(\\frac{d y}{d t}\\) = 0
\n\\(5 \\times 4=-12 \\frac{\\mathrm{dy}}{\\mathrm{dt}} \\Rightarrow \\frac{\\mathrm{dy}}{\\mathrm{dt}}=-\\frac{20}{12}=\\frac{-5}{3}=\\frac{\\mathrm{dy}}{\\mathrm{dt}}=\\frac{-5}{3} \\mathrm{ft.} \/ \\mathrm{sec}\\)<\/p>\n

Question 19.
\nA street lamp is hung 12 feet above a straight horizontal floor on which a man of 5 feet is walking how fast his shadow lengthening when he is walking away from the lamp post at the rate of 175ft.\/min.
\nAnswer:
\n\"2nd
\nLet the shadow be y & the distance from the man walks is x.
\nGiven \\(\\frac{d x}{d t}\\) = 175 ft\/min.
\nFrom figure we have
\n\\(\\frac{12}{5}=\\frac{x+y}{y}\\) \u21d2 12y = 5x + 5y \u21d2 7y = 5x
\n\u21d2 \\(\\frac{7 \\mathrm{dy}}{\\mathrm{dt}}=5 \\frac{\\mathrm{dx}}{\\mathrm{dt}}\\)
\n\u2234 the shadow is lengthening \u21d2 \\(\\frac{d y}{d t}=\\frac{5}{7} \\times 175\\) = 125 ft\/ min.<\/p>\n

Question 20.
\nFind a point on the parabola y2<\/sup> = 4x at which the ordinate increases at twice the rate of the abscissa [Ordinate = y, abscissa = x].
\nAnswer:
\nGiven y2<\/sup> = 4x diff. w.r.t. x
\n2y \\(\\frac{d y}{d x}\\) = 4 \\(\\frac{d x}{d t}\\)
\nAlso given \\(\\frac{d y}{d x}\\) = 2. \\(\\frac{d x}{d t}\\) \u21d2 2y .2 \\(\\frac{d x}{d t}\\) = 4. \\(\\frac{d x}{d t}\\)
\n\u21d2 y = 1 \u21d2 12<\/sup> = 4x
\n\u21d2 x = \\(\\frac { 1 }{ 4 }\\)
\n\u2234 the point on the parabola is (\\(\\frac { 1 }{ 4 }\\), 1 ).<\/p>\n

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

Students can Download Basic Maths Exercise 19.1 Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Basic Maths Question Bank Chapter 19 Application of Derivatives Ex 19.1 Part-A 2nd PUC Basic …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19665"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=19665"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/19665\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=19665"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=19665"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=19665"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}