Question

please solve Q1 and Q2

3 ths Homework - Rate My Maths Homework - X Differentiating parametrics k/differentiating parametric functions 10 Q2 - Tangents and normals A curve is defined parametrically as 12 Q2 + x= 1, y = 13 Find dy dx at the point 1 = 3. dy dx 10 Computer and Mathematical Sciences Find the gradient of the tangent to the curve 1 Find the equation of the tangent to the graph at t = 3. at the point t = 2. .X= [2] Total , y = 3-1 10 gradient = (3) Find the equation of the normal to the graph at / = 2. Write your answer in the form + by = 0 [2] X + [3] Mark

Answer 1

1 x=t, y=t dy d (+3) - 342 dt dt 342 1 dos d+ 27 dy dx 2 slope of tangent 27 t=3 slop of Normal 27 E at t=3 x = 3, y = 33 y = 27 (L, y) = (3, 27) equation Line of tangent yu ), = m (X - x) 6 at=3 slope y-27 = 27(x-3) Y = 278 27x3 + 27 y = 27x- 54 Ars

(2) 2 X = 1 y = 3-t t then t= 1 thin y = 30 1 X X Y = = 32-1 f(x,y) 3% - xy-120 of= gradiant df of dy dx (3x - xy-1) (3x-xy-1) j dy (3-X)i + (-) j (3-7) 1 - xj ut Now at t=2 X = 1 You 1 Y = 3-2 x = 11 y=1 then uf Vei (3-1)- 2 i Ij 2 Ans

با ولا differentiate Wir. ticho d ( 3x-xy-1) = 0 3-dXY -0 =0 > 3- [x du [ x dy + y com] CO ſ 3- x dy – Y = 0 x dy 3-Y dx dy da 3-1 at t=2 (+= 1/2, 4 = 1) then da ) ( . ) 3-1 2 2. 2. slope of Normal = { 1 h then equation of Normal at t= 2 ( + 1) yol = 4Y - 4 = -x + 1 by - 8 = - 2x + 1 2x+8y Ans

00 let bed limit of u changed 3x-I U differentiale it whin xro us -1 6x dx whin x 12U 2 du lxdx xdx = du 6 Now integral become :- 12 (1) du 2 fu du 2 21 15 (16-1] Ane 2 (2) X (X-114 dx let X-1 = u changed limit otte dx = du When x1 = uzo Wheny= 2 =) U = 1 thin :- Ju+1) (91+ du s cur+ut) du 0 0 И 5+ 6 1 + 1 6 IL 30 30 Ans

3 xx-2 dx 2 limit of u let (x-2)= u when x= 2 = 40 dx = du when x = 3 = U=1 ((1+2) Su cu O [uru su +254) de (u's +262 Jau 0 3,41 We t 2 3+1 17. 27. + 6+20 26 15 15 Am