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Gradients and Tangents e.g. Find the gradient of the line joining the points with coordinates ( 1, 1 ) and ( 3, 7 ) Solution: m y 2 y1 x 2 x1 m 71 31 6 3 2 difference in the y-values ( 3, 7 )x 7-1=6 ( 1, 1 )x 3-1=2 difference in the x-values Gradients and Tangents The gradient of a straight line is given by the difference in the y - values m the difference in the x values We use this idea to get the gradient at a point on a curve Gradients are important as they measure the rate of change of one variable with another. For the graphs in this section, the gradient measures how y changes with x This branch of Mathematics is called Calculus Gradients and Tangents The Gradient at a point on a Curve Definition: The gradient of a point on a curve equals the gradient of the tangent at that point. e.g. y x 2 12 (2, 4)x 3 Tangent at ( 2, 4) The gradient of the tangent at (2, 4) is m So, the gradient of the curve at (2, 4) is 4 12 3 4 Gradients and Tangents The gradient changes as we move along a curve e.g. y x m 6 2 Gradients and Tangents y x 2 m 4 Gradients and Tangents y x 2 m 2 Gradients and Tangents The Rule for Differentiation y x 2 m0 Gradients and Tangents The Rule for Differentiation y x 2 m2 Gradients and Tangents y x 2 m4 Gradients and Tangents y x 2 m6 Differentiation from first principles 400 360 320 P 280 f(x+h) 240 200 160 120 80 A f(x)40 0 0 0.5 x1 1.5 2 2.5 f( x h ) f( x ) Gradient of AP = h 3 x3.5 +h 4 Differentiation from first principles 400 360 320 280 240 200 P f(x+h) 160 120 80 A f(x)40 0 0 0.5 x1 1.5 2 2.5 f( x h ) f( x ) Gradient of AP = h x +3 h 3.5 4 Differentiation from first principles 400 360 320 280 240 200 160 P 120 f(x+h) 80 A f(x)40 0 0 0.5 x1 1.5 2 2.5 x+ h f( x h ) f( x ) Gradient of AP = h 3 3.5 4 Differentiation from first principles 400 360 320 280 240 200 160 120 P 80 f(x+h) A f(x)40 0 0 0.5 x1 1.5 2 x+ h 2.5 f( x h ) f( x ) Gradient of AP = h 3 3.5 4 Differentiation from first principles 400 360 320 280 240 200 160 120 80 f(x+h) f(x)40 A P 0 0 0.5 x1 x 1.5 +h 2 2.5 f( x h ) f( x ) Gradient of AP = h 3 3.5 4 Differentiation from first principles 400 360 320 280 240 200 160 120 80 A 40 f(x) 0 0 0.5 x1 1.5 2 2.5 3 L im it f( x h ) f( x ) Gradient of tangent at A = h 0 h 3.5 4 Differentiation from first principles f(x) = x2 f( x h ) f( x ) ( x h) x 2 2 x 2 xh h x 2 h h (2 x h ) h h 2x h h f '( x ) L im it f( x h ) f( x ) h 0 h 2 2x 2 2 xh h h 2 Differentiation from first principles f(x) = x3 f( x h ) f( x ) ( x h) x 3 3 x 3 x h 3 xh h x 3 2 h h h (3 x 3 xh h ) h 2 2 3 x 3 xh h 2 h f '( x ) L im it f( x h ) f( x ) h 0 h 2 3x 2 3 3 3 x h 3 xh h 2 2 h 2 Generally with L im it f( x h ) f( x ) h 0 h h is written as δx And f(x+ δx)-f(x) is written as δy Limit x 0 x dy y dx Generally f ( x ) kx y kx n f ' ( x ) nkx n 1 dy n nkx n 1 dx f ( x ) kx y kx n f ' ( x ) nkx n 1 dy dx n nkx n 1 Gradients and Tangents Points with a Given Gradient e.g. Find the coordinates of the points on the curve 3 y x 8 x 7 where the gradient equals 4 Gradient of curve = gradient of tangent = 4 We need to be able to find these points using algebra Gradients and Tangents Gradients and Tangents Gradients and Tangents Exercises Find the coordinates of the points on the curves with the gradients given 1. y x 4x 3 2 where the gradient is -2 Ans: (-3, -6) 3 2 2. y x 3 x 21 x 20 where the gradient is 3 ( Watch out for the common factor in the quadratic equation ) Ans: (-2, 2) and (4, -88) Gradients and Tangents Increasing and Decreasing Functions • An increasing function is one whose gradient is always greater than or equal to zero. dy 0 dx • for all values of x A decreasing function has a gradient that is always negative or zero. dy dx 0 for all values of x Gradients and Tangents e.g.1 Show that y x 3 4 x is an increasing function dy 2 3 3x 4 Solution: y x 4 x dx dy is the sum of dx • a positive number ( 3 ) a perfect square ( which is positive or zero for all values of x, and • a positive number ( 4 ) dy dx 0 for all values of x so, y x 3 4 x is an increasing function Gradients and Tangents e.g.2 Show that y 1 x 3 3 x 2 9 x is an 3 increasing function. Solution: y 1 3 dy dx x 3x 9x 3 2 dy x 6x 9 dx 0 for all values of x 2 Gradients and Tangents The graphs of the increasing functions 3 2 3 y x 4 x and y 13 x 3 x 9 x y x 4x 3 and y 1 3 x 3x 9x 3 2 are Gradients and Tangents Exercises 1. Show that y x 3 is a decreasing function and sketch its graph. 2. Show that y 1 x 3 2 x 2 5 x is an increasing 3 function and sketch its graph. Solutions are on the next 2 slides. Gradients and Tangents Solutions 1. Show that y x 3 is a decreasing function and sketch its graph. Solution: dy dx 3 x . This is the product of a 2 square which is always 0 and a negative number, dy so 0 for all x. Hence y x 3 is a dx decreasing function. y x 3 Gradients and Tangents Solutions 2. Show that function Solution: y dy 1 3 3 2 x 2 x 5 x is an increasing x 4x 5 . dx dy 2 Completing the square: ( x 2) 1 dx which is the sum of a square which is 0 2 and a positive number. Hence y is an increasing function. y 1 3 x 2x 5x 3 2 Gradients and Tangents The equation of a tangent e.g. 1 Find the equation of the tangent at the point (-1, 3) on the curve with equation 3 2 y x 3x 2x 1 3 2 Solution: y x 3 x 2 at x a1 point and the gradient of The gradient of a curve dy the tangent Gradient =equal -5 2 at that point are 3x 6x 2 dx At x = 1 (-1, 3) x m 3( 1) 6( 1) 2 2 5 y mx c y 5 x c (-1, 3) on line: 3 5 ( 1 ) c 2c So, the equation of the tangent is y 5 x 2