Tangent is also a line which touches the curve. To find the equation of tangent, we have to follow the given below.
i) Find the slope of the tangent drawn at the point (x_{1}, y_{1}) from the given equation of curve.
Use the formula,
y - y_{1} = m(x - x_{1})
ii) Here m is the slope of the tangent line at the point of contact. (x_{1}, y_{1}) is the point of contact.
For each problem, find the equation of the line tangent to the function at the given point. Your answer should be in slope – intercept form.
Problem 1 :
y = x^{3} – 3x^{2} + 2 at (3, 2)
Solution :
y = x^{3} – 3x^{2} + 2
Differentiating with respect to x, we get
dy/dx = 3x^{2} - 6x
Slope at (3, 2)
dy/dx = 3(3)^{2} - 6(3)
dy/dx = 27 - 18
dy/dx = 9
Slope of tangent = 9
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - 2) = 9(x - 3)
y - 2 = 9x - 27
y = 9x - 27 + 2
y = 9x - 25
Problem 2 :
Solution :
y = -5(x^{2} + 1)^{-1}
Differentiating with respect to x, we get
dy/dx = -5(-1) (x^{2} + 1)^{-2} (2x)
dy/dx = 10x (x^{2} + 1)^{-2}
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Problem 3 :
y = x^{3} – 2x^{2} + 2 at (2, 2)
Solution :
y = x^{3} – 2x^{2} + 2
Differentiating with respect to x, we get
dy/dx = 3x^{2} - 4x
Slope at (2, 2)
dy/dx = 3(2)^{2} - 4(2)
dy/dx = 12 - 8
dy/dx = 4
Slope of tangent = 4
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - 2) = 4(x - 2)
y - 2 = 4x - 8
y = 4x - 8 + 2
y = 4x - 6
Problem 4 :
Solution :
y = -3(x^{2} - 25)^{-1}
Differentiating with respect to x, we get
dy/dx = -3(-1) (x^{2} - 25)^{-2} (2x)
dy/dx = 6x (x^{2} - 25)^{-2}
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Problem 5 :
Solution :
y = -3(x^{2} - 4)^{-1}
Differentiating with respect to x, we get
dy/dx = -3(-1) (x^{2} - 4)^{-2} (2x)
dy/dx = 6x (x^{2} - 4)^{-2}
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Problem 6 :
Solution :
Differentiating with respect to x, we get
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
Problem 7 :
y = In (-x) at (-2, In 2)
Solution :
y = In (-x)
Differentiating with respect to x, we get
dy/dx = -1/-x
dy/dx = 1/x
Slope at (-2, In 2)
dy/dx = 1/-2
dy/dx = -1/2
Slope of tangent = -1/2
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
(y - In 2) = -1/2(x + 2)
(y - In 2) = -1/2 x - 2/2
y - In 2 = -1/2 x - 1
y = -1/2 x + In 2 - 1
Problem 8 :
y = -2tan (x) at (-π, 0)
Solution :
y = -2tan (x)
Differentiating with respect to x, we get
dy/dx = -2 ⋅ sec^{2}(x)
Slope at (-π, 0)
dy/dx = -2 ⋅ sec^{2}(-π)
= -2 ⋅ 1
= -2
Slope of tangent = -2
Equation of tangent :
(y - y_{1}) = m(x - x_{1})
y - 0 = -2(x + π )
y - 0 = -2x - 2π
y = -2x - 2π
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM