Find the tangent and normal of the following curves at the given points on the curve :
Problem 1 :
y = x^{2} - x^{4 }at (1, 0)
Solution :
Given that, y = x^{2} - x^{4 }
dy/dx = 2x - 4x^{3}
Slope at (1, 0)
dy/dx = 2(1) - 4(1)^{3}
= 2 - 4
Slope = -2
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Point is at (1, 0)
y - 0 = -2(x - 1)
y = -2x + 2
2x + y = 2
Equation of normal :
(y - y_{1}) = -1/m(x - x_{1})
Point is at (1, 0)
y - 0 = (1/2)(x - 1)
2y = x - 1
x - 2y = 1
Problem 2 :
y = x^{4} + 2e^{x} at (0, 2)
Solution :
y = x^{4} + 2e^{x}
dy/dx = 4x^{3} + 2e^{x}
Slope at (0, 2)
dy/dx = 4(0)^{3} + 2e^{0}
= 0 + 2(1)
Slope of tangent = 2
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Point is at (0, 2)
y - 2 = 2(x - 0)
y - 2 = 2x
2x - y = - 2
Equation of normal :
(y - y_{1}) = -1/m(x - x_{1})
(y - 2) = (-1/2)(x - 0)
2(y - 2) = -x
2y - 4 = -x
x + 2y = 4
Problem 3 :
y = x sin x at (π/2, π/2)
Solution :
Using product rule, finding the derivative
dy/dx = x(cos x) + sin x (1)
= x cos x + sin x
Slope at (π/2, π/2)
= π/2 cos (π/2) + sin (π/2)
= π/2 (0) + 1
Slope of tangent = 1
Slope of normal = -1
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - π/2) = 1(x - π/2)
y = x - π/2 + π/2
y = x
Equation of normal:
(y - y_{1}) = -1/m(x - x_{1})
(y - π/2) = -1(x - π/2)
y = -x + π/2 + π/2
y = -x + π
Problem 4 :
x = cos t, y = 2 sin^{2} 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 sin^{2} t at t = π/3 y = 2 sin^{2}π/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 - y_{1}) = m(x - x_{1})
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 - y_{1}) = (-1/m)(x - x_{1})
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
Problem 5 :
Find the equations of the tangents to the curve
y = (x + 1) / (x - 1)
which are parallel to the line x + 2y = 6.
Solution :
Find the derivative of the given function to figure out slope.
x + 2y = 6
2y = -x + 6
y = (-1/2)x + 6/2
y = (-1/2)x + 3
Slope = -1/2 -----(2)
(1) = (2)
x - 1 = 2 x = 2 + 1 x = 3 |
x - 1= -2 x = -2 + 1 x = -1 |
Applying x = 3, we get y = (3+1)/(3-1)
y = 4/2
y = 2
Applying x = -1, we get y = (-1+1)/(-1-1)
y = 0
So, the required points are (3, 2) and (-1, 0)
Equation of tangents :
(y - y_{1}) = m(x - x_{1})
(3, 2) and slope = -1/2
(y - 2) = -1/2 (x - 3)
2(y - 2) = -1(x - 3)
2y - 4 = -x + 3
x + 2y - 4 - 3 = 0
x + 2y - 7 = 0
(-1, 0) and slope = -1/2
(y - 0) = -1/2 (x + 1)
2y = -x - 1
x + 2y = -1
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM