application on the first derivative

APPLICATION ON FIRST DERIVATIVEFind the slope of the tangent of the following functions at the indicated points:[1]y = at ( 1 , -1 )[2] y = ( 2x + 3 ) cos 4 x at ( 0 , 3 )[3] y = at ( 1 , [4] y = 3 at ( -1 , 343 )[5] y = cos ( 2 ) + tan at ( )Solution[1] y` = 2slope of the tangent===================================================[2] y` = 2 cos 4 x + ( 2x + 3 ) ( - 4 sin 4 x ) , slope of the tangent = 2 (1) + (3) (0) = 2===================================================[3] y` = , slope of the tangent = ===================================================[4] y` = 3 , slope of the tangent = 3 ===================================================[5] y` = - (-1) sin (2 slope of the tangent= 0 - ===========================================================Find the measure of the positive angles, which the tangent to the curves makes with the positive direction of the x – axis to the following function:[1] y = at ( 1 , 0 )[2] y = at ( 0 , -3)[3] y = at ( 1 , 2 )[4] y = y + sin 2 x + cos 3 x = 0 at Solution[1] y ` = 3 ===================================================[2] y` = y `= tan ===================================================[3] y` = tan ===================================================[4] y` + 2 cos 2 x – 3 sin 3 x = 0 y` + 2 cos 180 – 3 sin 270 = 0 y ` -2 + 3 = 0………………………………………………………………………………………………[1] Find the equation of the tangent to the curve of the function f : f (x) = at the point (2,1)SolutionF (x) = slope of the tangent = -1 = m equation of the tangent y – y1 = m ( x – x 1 ) y - 1 = -1 ( x – 2 ) y – 1 = -x + 2 x + y -3 = 0==================================================[2] Find the equation of the tangent to the function f : f ( x ) = sin x at the point SolutionF`( X ) = cos x = cos 90 = 0 = mThe equation of the tangent , Note : If L // x – axis eq n of L : y = y1 [3] Find the equation of the normal to the curve of the function f : f (x) = at the point ( 3 , 2 ).SolutionF`(x) = 2x – 3 = 2 (3) -3 = 3 = slope of the tangent. slope of the normal = , The equation of the normal y – y1 = m ( x – x 1 \ y – 2 = 3 y – 6 = - x + 3 x + 3 y – 9 = 0===================================================[4] Let y=andSolution 1 = (1) 4 = 2 (2)Subs by (2) in (1)1 = a = -2 a = 2subs in (2) - 4 + b = -1 4 + b = 1 b = 3 b= -3 or a = 2 b =3 b =- 3Note : don’t subs. in (1)[5] IF F (x) = f (0) = 3 and f ` (0) = 2 Find the value of a and b.Solutionf (0) = 3 3 = f (x) = f `(x) = 2 = 2 = - a -3 ===================================================[6] Find the equation of the tangent to the curve of the function f : f (x) = at the point of intersection with the y – axis.SolutionAt y – axis x = 0 ( 0 , 2 )F ` (x) = 3The equation of the tangent : y – y1 y – 2 4 x + y – 2 = 0[7] Find the point (s) on the curve y = such that the slope of the tangent at them is equal to – 3.Solution (1) slope of the tangent = -3 (2) x = 1Subs in equation of the curve : y = The point : ( 1 , -1) .==================================================[8] Find the point (s) on the curve y = such that the tangent at them makes an angle of measure 135with the positive direction of the x – axis.Solution (1) slope of the tangent = tan 135= -1 (2) Subs in of the curve y = The point : ( -2 , 0 ) .[9] Find the points on the curve y = and the tangents to this curve are:Parallel to the x – axisParallel to the line y + 9x = 1Perpendicular to the line 15 y = 3 – x Solution(1) y = 3 (1) // x – axisslope of the tangent = 0 (2) 3 X = -3 , x =1Subs in curve y = 0 , y = -32The points are ( -3 , 0 ) and ( 1 , -32 )===================================================(2) slope of the straight line y + 9x =1 Is -9 // line Slope of the tangent = -9 (2) 3 3 x ( x + 2 ) = 0 X = 0 , x = -2 Y = -27 , y = -5The points are : ( 0 , -27 ) and ( -2 , -5 )(3) (1)slpoe of the line 15 y = 3 – x is (2)=15 X = - 4 x = 2Subs in curve y = -7 y = -25The points( -4 , -7 ) and ( 2 , -25)[10] IF f: R where f (x) = cos 3x , then Find the points which lie on this curve , for which the tangents to this curve are parallel to the x – axis. Solution (1)tangent // x – axis slope of the tangent = 0 (2) -3 sin 3 x = 0 Sin 3 x = 03x = 0 3x = 3x = 2 x = 0 x = x =Subs in curve : y =1 y = -1 y = 1The points : ( 0 ,1 )[11] Let y = a, If = 8 at x = 1 , and the average rate of change of y x changes from -1 to 2 equal 7 Find the values of the constants a and b Solution (1) A(h) = (2)– (2) b = 1 subs in (2) 3a+1=7