RATE OF CHANGE

SLOPE OF A CURVEDEFINITION OF TERMSTANGENT LINESECANT LINE

SLOPE (m)-a line that intersects the curve at only one point.-a line that intersects the curve in two or more distinct points.-the ratio of the change in vertical distance(rise) to the change in horizontal distance (run)-tangent of its angle of inclination

rise run m = tan =SLOPE (m)-the ratio of the change in vertical distance(rise) to the change in horizontal distance (run)-tangent of its angle of inclination xy y1 y2x1x2riserunP1P2 y2 y1 x2 x1 y x==

-the slope of a curve is not constant and must be determined for each particular point of interest.xy y1 y2x1x2Tangent LineSecant LineP1P2SLOPE OF A CURVElet the points P1 (x1,y1) and P2 (x 2,y2) be any two points on the curve.=yy2 y1 x2 x1 xmsec =y =f(x)=yy2 y1 x2 x1 xmsec =lim x0lim x0mtan =lim x0First derivative is the slope of the curve at that point

SLOPE OF A CURVE=yy2 y1 x2 x1 xmsec =lim x0lim x0mtan =lim x0also y1 = f(x1) ; y2 = f(x2)Let mtan= f(x)= , x = x2 x1 ; x2 = x1 + x dydxmtan= f(x)= =dydxf(x2) f(x1) limx 0xm= f(x)= =dydxlimx 0f(x +x) f(x) xThe slope of the curve y = f(x) at (x, f(x)) is equal to the slope of its tangent line at (x,f(x)), and defined by

RATE OF CHANGEDEFINITIONThe average rate of change of y with respect to x, as x changes from x1 to x2, is the ratio of the change in output to the change in input.y2 y1 x2 x1 average rate of change =

RATE OF CHANGEDEFINITIONConsider a function y = f(x) and two inputs x1 & x2 . The change in input or the change in x is x2 x1The change in output or the change in y is y2 y1 y2 y1 x2 x1 xy= f(x2) f(x1)x2 x1 x1x2 y2 = f(x2) y1 = f(x1)Q P f(x2) f(x1)Slope of the line from P(x1 ,y1) to Q(x2 ,y2) PQ is a secant line

RATE OF CHANGEFind the average rate of change as x changes from 1 to 5 in the function f(x) = 2x3.when x = 1f(1) = 2when x = 5f(5) = 250y2 y1 x2 x1 = f(x2) f(x1)x2 x1 250 2 5 = 62 1

RATE OF CHANGEFind the average rate of change as x changes from 2 to 3 in the function f(x) = 2x + 5.when x = 2f(2) = 9when x = 3f(3) = 11y2 y1 x2 x1 = f(x2) f(x1)x2 x1 11 9 3 2 = 2

Remarks:If the functional relationship between x and y is given by y = f(x) and if x changes from the value x1 to x2 , where x = x2 x1 and x2 = x1 + x, then y changes from f(x1) to f(x1 + x). Change in y is denoted by y (when the change in x is x) and defined by f(x1+ x) f(x1)f(x1+ x) f(x1) xyx=let x1= x and h = x

f(x1+ x) f(x1) xyx=let x1= x and h = x fromf(x + h) f(x) h, where h 0Difference QuotientDefinition: The average of change f with respect to h is also called difference quotient. It is given by the formula

Find the difference quotient for f(x) = x2 + 2 when x = 4 and h = 0.001.=f(x + h)=(x + h)2f(x) = x2 + 2+ 2f(x + h)= x2 + 2xh + h2 + 2f(x) = x2 + 2f(x + h)= x2 + 2xh + h2 + 2f(x + h) f(x) hx2 + 2xh + h2 + 2 (x2 + 2)h2xh + h2=h= 8.001= 2x + h

The instantaneous rate of change for a function f at x = x0 isVelocitySuppose a car leaves the town at time t = 0 and travels due north. Let s(t) represent the position of the car(its distance from the town in km) at time (t). provided this limit exists.INSTANTANEOUS RATE OF CHANGEf(x0 + h) f(x0) hlimh0 Assume that s(t) = t2 5t + 6Ave.Velocity =change in distancechange in timebet. t1= 4 and t2 = 5

INSTANTANEOUS RATE OF CHANGEs(t) = t2 5t + 6Ave.Velocity =change in distancechange in timet1= 4 ; t2 = 5Ave.Velocity =[(5)2 5(5) + 6]s(5) s(4) t2 t1 Ave.Velocity = [(4)2 5(4) + 6]5 4 Ave.Velocity = 4

The instantaneous velocity at t = 3s(3 + h) s(3) hlimh0s(t) = t2 5t + 6hlimh0[(3 + h)2 5(3+h) + 6] [(3)2 5(3) + 6] limh0[9 + 6h +h2 15 5h + 6] [9 15 + 6] hlimh0h2 + hhs(3 + h) s(3) hlimh0(h + 1)limh0= 1

VelocityIf v(t) represents the velocity at time t of an object moving in a straight line with position s(t), thens(t + h) s(t) h, provided this limits exists.limh0v(t) =

Example:A toy rocket fired a straight up into the air reaches a height of s(t) = 48t 5t2 meters. What is the rockets initial velocity (when t = 0)?s(t + h) s(t) hlimh0v(t) =v(t) =[48(t + h) 5(t + h)2] [ 48t 5t2]limh0hlimh0[48t + 48h 5t2 10ht 5h2] [ 48t 5t2]hlimh048h 10ht + 5h2hlimh0(48 10t + 5h)v(t) =v(0) =limh0(48 10t)= 48 m/s