Problem 1 :
Find the equation of tangent and normal to the curve given by x = 7 cos t and y = 2 sin t at any point on the curve.
Solution :
x = 7 cos t and y = 2 sin t
To find equation of tangent, we have to find the slope of the tangent line by finding the derivative of the given curves.
dx/dt = -7 sin t
dy/dt = 2 cos t
dy/dx = 2 cos t / (-7 sin t)
Slope = (-2/7) (cos t / sin t)
Equation of the tangent line :
(y - y1) = m(x - x1)
(y - 2 sin t) = (-2/7) (cos t / sin t)(x - 7 cos t)
sin t(y - 2 sin t) = -2 cos t(x - 7 cos t)
7y sin t - 14 sin2 t = -2 x cos t + 14 cos2 t
2x cos t + 7y sin t = 14 cos2 t + 14 sin2 t
2x cos t + 7y sin t = 14 (cos2 t + sin2 t)
2x cos t + 7y sin t = 14 (1)
2x cos t + 7y sin t = 14
Equation of the tangent line :
(y - y1) = -1/m(x - x1)
(y - 2 sin t) = (7/2) (sin t / cost t)(x - 7 cos t)
2 cos t (y - 2 sin t) = 7 sin t (x - 7 cost)
2y cos t - 4 sin t cos t = 7x sin t - 49 sin t cos t
7x sin t - 2y cos t - 49 sin t cos t + 4 sin t cos t = 0
7x sin t - 2y cos t - 45 sin t cos t = 0
Problem 2 :
x = cos t, y = 2 sin2 t at t = π/3
Solution :
Applying the value of t in the given function, we get the value of x and y.
x = cos t at t = π/3 x = cos π/3 x = 1/2 |
y = 2 sin2 t at t = π/3 y = 2 sin2π/3 y = 2 (√3/2)2 y = 3/2 |
dx/dt = -sin t, dy/dt = 4 sin t cost t
dy/dx = (dy/dt) / (dx/dt)
= (4 sin t cost t) / (-sin t)
dy/dx = -4 cos t
Slope at t = π/3
dy/dx = -4 cos (π/3)
= - 4(1/2)
Slope of tangent = -2
Slope of normal = 2
Equation of tangent :
(y - y1) = m(x - x1)
x = 1/2 and y = 3/2
Slope = -2
(y - (3/2)) = -2 (x - (1/2))
2y - 3 = -2(2x - 1)
2y - 3 = -4x + 2
4x + 2y -3 - 2 = 0
4x + 2y -5 = 0
Equation of normal :
(y - y1) = (-1/m)(x - x1)
x = 1/2 and y = 3/2
Slope = 1/2
(y - (3/2)) = 1/2(x - (1/2))
2y - 3 = 1/2(x - 1)
4y - 6 = x - 1
x - 4y - 1 + 6 = 0
x - 4y + 5 = 0
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM