Question

Find an equation for the line tangent to the curve at the point defined by the given value of t. x = sin t, y = 2 sin t, t = wa y = 2x - 213 y = 2x y = 2x + 13 Oy=-2x+ 2/3

Answer 1

dt dx dy Q1. Here given x= sin(t), y = 2 sin(t) then {sin(t)}= cos(t) and dt dt dt dy dy dt 13 = 2. Now at t = we get, x= sin and dx dt dx cos(t) 2 :{2 sin(t)}= 2 cos(t) then 13 = 13. 2 cos(t) TT X y = 2 sin = 2x 3 2 TT TT Now at t= 1 dy dt = 2. Then the equation of tangent line to x= sin(t), y = 2 sin(t) at t= is 3 3 y-y dy dx 1(x-4)3-15=2 = 2 x - 13 2 BY- V3 = 2x – V3 = y= 2x... This is the tangent line. dx d Q2. Here given x=t, y = 7 2t. Then = d -{t} = 1 and dx dy dt = v 2V1 dt dx Then dydydt dx dt dx х v 2 -X1= 217 2V7 Now at t = 18 we get, x = 18 , y=v2x18 = 136 = 6 and dy dx 2 VI 212x3x32x377 2V18 Jt=18 Then the equation of tangent line to x=t, y= V2t at t=18 is y-Y:18 = dy dx Jt=18 (x – X7-18) = y-6= --18) 1 1 18 >>y-6=-x- 6 6 > y= Ex-3+6 > y=-x+3.... 6 This is the required tangent line. TT Q3. In polar coordinate a coordinate (r,0) = 4, is given then in Cartesian coordinate we get, 4 7 1 71 1 x = r cos(O)= = 4 cos = 4 = 4x VE 2 then the coordinate is 4 4 = 4x and y = r sin (0) = 4 sin 2 12 4. 4 in polar > in Cartesian. VIT 4. 4 TT Then 4, VIT 4 TT 4 But if the polar coordinate be (r, 0) -4, then x=r cos(0)=-4cos 6 =-4x and 4 2 7 y = r sin (O)=-4 sin =-4x 4 V then the coordinate is 4 va 4 4 JE 7 4 4 2 Then -4, -4 in polar > ☺ 0 in Cartesian. 4 VZ 2 So 4. π 4. л and -4, 4 are not same in polar coordinate.

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