![]() ![]() Example :4 Water is poured into a conical paper cup at the rate of 2/3 inches3 /sec. Substitute values of x, y and dx /dt in (3) dy / dt = - 5 / 12 (3) dy / dt = - 5/4 ft /sec Thus when the ladder is 5 ft from the wall the top is sliding down at the rate of 5/4 ft /sec. ![]() If the bottom end is drawn away from the wall at 3 ft/sec how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 5 ft from the wall? Solution : At any given instant, let y be the height of the top of the ladder above the floor and x be the distance b/w the base of the wall and bottom of the ladder By hypothesis dx/dt = 3 We must find the value of dy/dt at t where t is the time for which x = 5 ft By pythgoras theorem 13 feet y x 2 + y 2 = 13 2 x 2 + y2 = 169 Differentiating with respect to t 2x dx / dt + 2y dy/ dt = 0 dy / dt = - x / y dx /dt -(3) x Now at time t for which base of the ladder is 5 ft from wall Substitute x = 5 in (2) 5 2 + y 2 = 169 y 2 = 169 - 25 y = ± 12 ![]() How fast does the end of his shadow move? Solution : By similar triangle s - x s - = - 6 16 16 s – 16 x = 6 s 10 s = 16 x Differentiate with respect to t dsdx 10 - = 16 - dt dt ds 10 - = 16 ( 5 ) dt ds/dt = 8 miles/ hour His shadow move at 8 miles/hour 16ft 6ft x s - x sĮxample :3One end of a 13 foot – Ladder is on the floor, and the other end rests on a vertical wall. - 1 /2 π r2 = dr / dt Substitute the values of dS/dr & dr/dt in (1) dS / dt = 8 π r × -1/ 2 π r 2 dS / dt = - 4 / r when r = 12 ft dS/dt = - 1 / 3 ft2/ min The surface of Spherical balloon shrinking at the rate of - 1/3 square feet per minute.Įxample : 2A man 6 ft tall walks away from a lamp post 16 ft high at the rate of 5 miles per hour.How fast is the surface area shrinking when the radius is 12 ft ? Solution: A sphere of radius r has volume v V = 4/3 π r3 As we know surface area of sphere S S = 4 π r 2 We have to find dS /dt when r = 12 ft here we can see S is independent of time but we can find it in this way dS dS dr - = - × - (1) dt dr dt here S = 4 π r 2 dS/ dr = 8 π r Now for dr /dt : dV / dt = - 2 (given) dVdV dr - = - × - (2) dt dr dt V = 4/3 π r3 dV/ dr = 4 π r2 We have many quantities involve rate of change in time shook intensity of earthquakes, inflation of currency, water height of Sea, Stock rates, rate of increase of volume of balloon etc.Įxample: 1Gas is escaping from a spherical balloon at the rate of 2 ft3 / min. Rate of change of velocity is acceleration. Rate of Change : The derivative of function is a mathematical tool that is used to study rates at which quantity changes. Mean Value Theorem for Integrals 32.Applications OF DIFFERENTIAL CALCULUS BY ERUM RIAZ Department of Mathematics D.A Degree College For Women Ph-VII (Ext)Īnalysis : By the end of the lesson students will be able to explain the applications of derivative.Īpplications of differential calculus There are two major applications Rate of Change Extreme Valuesġ. Indefinite Integrals and the Fundamental Theorem 26. The Mean Value Theorem 17 Derivatives and Graphs 18 Derivatives and Graphs 19/20. ![]() Use Firefox to download the files if you have problems. He has kindly donated them for the use of all students in this course. These powerpoint lectures were created by Professor Mario Borelli in Fall 2011. Professor Borelli's Powerpoint Lectures Home Page ![]()
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