3 Analysis :By the end of the lesson students will be able to explain the applications of derivative.
4 Applications of differential calculus There are two major applicationsRate of ChangeExtreme Values
5 1. Rate of Change :The derivative of function is a mathematical tool that is used to study rates at which quantity changes.Rate of change of position is velocity.Rate of change of velocity is acceleration.Rate of change of momentum is force .We have many quantities involve rate of change in timeshook intensity of earthquakes , inflation of currency , water height of Sea , Stock rates , rate of increase of volume of balloon etc.
6 Example: 1 Gas is escaping from a spherical balloon at the rate of 2 ft3 / min .How fast is the surface area shrinking when the radius is 12 ft ?Solution:A sphere of radius r has volume vV = 4/3 π r3As we know surface area of sphere SS = 4 π r 2We 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 waydS dS dr= × (1)dt dr dthere S = 4 π r 2dS/ dr = 8 π rNow for dr /dt :dV / dt = (given)dV dV dr= × (2)dt dr dtV = 4/3 π r3dV/ dr = π r2
7 Substitute the values in (2) = π r 2 × dr /dt- 1 /2 π r2 = dr / dtSubstitute the values of dS/dr & dr/dt in (1)dS / dt = π r × -1/ 2 π r 2dS / dt = / rwhen r = 12 ftdS/dt = / 3 ft2/ minThe surface of Spherical balloon shrinking at the rate of /3 square feet per minute.
8 Example : 2 A man 6 ft tall walks away from a lamp post 16 ft high at the rate of 5 miles per hour. How fast does the end of his shadow move?Solution :By similar triangles - x s=16 s – 16 x = 6 s10 s = xDifferentiate with respect to tds dx=dt dtds= ( 5 )dtds/dt = miles/ hourHis shadow move at 8 miles/hour16ft6ftxs - xs
9 Example :3 One end of a 13 foot – Ladder is on the floor , and the other end rests on a vertical wall. 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 ladderBy hypothesisdx/dt = 3We must find the value of dy/dt at t where t is the time for which x = 5 ftBy pythgoras theorem13 feetyx y 2 =x y = 169Differentiating with respect to t2x dx / dt y dy/ dt = 0dy / dt = - x / y dx /dt (3)xNow at time t for which base of the ladder is 5 ft from wallSubstitute x = 5 in (2)y 2 = 169y 2 =y = ± 12
10 substitute values of x , y and dx /dt in (3) dy / dt = / 12 (3)dy / dt = /4 ft /secThus when the ladder is 5 ft from the wall the top is sliding down at the rate of 5/4 ft /sec.Example :4Water is poured into a conical paper cup at the rate of 2/3 inches3 /sec . if the cup is 6 inches tall and top of the cup has radius of 2 inches , how fast does the water level rise when the water is 4 inches deep ?Solution :At any time t let h be the height of water , V be volume of water and r is the radius of top surface of waterAs we know volume of cone is V = 1/3 π r2 hSince water enters the cup at the rate of 2/3 inches3/ secdV/dt = 2/ (1)2 inchesWe have to find dh/dt at time t for which h = 4Using similar triangle in figurer / h = 2 /6r / h = 1 / 3r = h / 3r2r6 inchesh64 inchesh
11 V = 1 / 3 π r2 hV = 1 / 3 π ( h/3)2 hV = 1/27 π h 3dv/dh = 3 π h2 /27dv/dh = π h2 / (2)Now dv /dt = dv /dh × dh /dtSubstitute the values from ( 1 ) and (2)2/3 = πh2 / 9 × dh/dtdh /dt = / 3 π h2dh /dt = 6 / π h 2when the water is 4 inches deepdh / dt = 6 / π (4)2dh / dt = 3/8π inches / secThus the water level is rising at the rate of 3/8π inches/sec when the water is 4 inches deep
12 2 .Extreme Values Relative Maximum : Relative Minimum : A function f is said to have an extreme value at a if it has either a relative maximum or a relative minimum at a in such a situation a is called an extreme point of f .Relative Maximum :Let f be a function , defined on some neighbourhood of a point a ,then f is said to haverelative maximum value at a if there exist δ – neighbourhood A of a such thatf ( a ) > f ( x ) for all a € Aand as such f ( a ) is called a relative maximum of f .Relative Minimum :f is said to have a relative minimum value at a if there exist δ – neighbourhood A of a such thatf ( a ) < f ( x ) for all a € Aand such f ( a ) is called relative minimum of f.
13 Example :The domain of f(x) = x2 is all real numbers and the range is all nonnegative real numbers. The graph in the figure below suggests that the function has no absolute maximum value and has an absolute minimum of 0, which occurs at x = 0.The Absolute Extreme Values on a Restricted DomainIf the domain of f(x) = x2 is restricted to [-2, 3], the corresponding range is [0, 9]. As shown below, the graph on the interval [-2, 3] suggests that f has an absolute maximum of 9 at x = 3 and an absolute minimum of 0 at x = 0.The two examples above show that the existence of absolute maxima and minima depends on the domain of the function.[-5, 5, 1] x [-2, 10, 1]
14 Example : 6Find absolute maximum or minimum of the function f ( x ) = (2 x ) / x in [ 1 , 5 ]Solution :To find absolute maximum and minimum , First we must find critical numbers by taking derivative of the functionx d/ dx (2x2 + 4 ) (2x2 +4 ) d/dx (x)f ‘ (x) =x2x (4x) (2x2 + 4) (1)f ‘ (x) =2xf ‘ (x) =Now for critical numbers f ‘ ( x ) = 02x= 0x22x = 02x2 = 4x = ± √ 2Since - √ 2 does not lie in [1 , 5] , we can not include as a critical number , thus the only critical number is √2.
15 For determining function has maximum or minimum value at √ 2 , we have to find the 2nd derivative f ‘ (x) =x2x2 d/dx (2x2 -4 ) (2x ) d/dx ( x2 )f ‘’ (x) =x4x2 (4x) (2x ) (2x)=- 8f ‘’ (x) = < 0x3Since f ‘’ (x) < 0 ,function has maximum value at √ 22 ( √ 2 )f ( √ 2 ) =( √ 2 )= √ 24 √2 is the absolute maximum of the given function in [1 , 5 ].
16 Example : 7What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fencing?Solution :Let x be the length , y be width and A be area of rectangleA = x . yDifferentiate with respect to x0 = x . dy/dx + ydy / dx = - y / xNow let P be the perimeter of rectangleP = 2x y0 = dy/dx- 2 = 2 dy/dx- 2 = 2 ( - x / y )1 = x / yx = yRectangular field should be square for least amount of fencing.yx
17 Example : 8 Solution : x 9 9 -2x x x x 9 – 2x x 9 -2x 9 -2x 9 Example : 8A box is to be made of a piece of cardboard 9 inches square by cutting equal squares out of the corners and turning up the sides. Find the volume of the largest box that can be made in thisx99 -2xxxx9 – 2xx9 -2x9 -2xSolution :9Let V be the volume of the boxV = (9 -2x )2 .xV = 81x – 36x2 + 4x3Differentiate with respect to xdV / dx = x x 2For finding maximum volume dV/dx = – 72x + 12 x 2 = 04x 2 – 24 x = 0
18 By solving above equation x = or x = 1.5use x = 1.5 ( for maximum volume )V= ( 9 – 2 ( 1.5 ) )2 (1.5)V = 54 cubic inches
19 Example : 9A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing?Solution :Let x be the length and y be the width of the fieldA be the areaA = x.yDifferentiate with respect to xdA/dx = x. dy/dx + y0 = x . dy/dx + ydy / dx = - y / x (1)RiverFieldLet P be the perimeter of the field excluding theRiver sideP = x yDifferentiate with respect to xdP/dx = dy/dxFor least amount of fencing dP/dx =00 = dy/dx0 = ( - y / x ) from (1)y = 1 / 2 xWidth = ½ length for least amount of fencing
20 Example : 10 Solution : c y x If the hypotenuse of the right triangle is given, show that the area is maximum when the triangle is isosceles.Solution :Let x , y be the base and perpendicular and c be the hypotenuseof the triangle , by pythagoras theoremx y 2 = c 2Differentiate with respect to x2x y .dy /dx = 0dy/dx = - x /y (1)cyxlet A be the area of triangle A = ½ x . yDifferentiate with respect to xdA / dx = ½ [ x . dy/dx + y ]= ½ [ x . dy/dx + y ]x . dy/dx + y = 0x . ( - x / y ) + y = 0y = x 2 / yx 2 = x 2y = xThis shows that area is maximum when the triangle is an isosceles.
21 Conclusion :Above Examples shows that differential calculus is a mathematical tool that solve a lot of problems of daily life , Engineering and other sciences.
22 References : (1 ) Calculus by Robert Ellis & Denny Gulick Applied Calculusby TanCalculusby Frank Ayres , JR. Elliott MendelsonMathematics XIISindh text book board Jamshoro(5) www. Mathalino.com