Find the equations of the tangent and normal to the given curves at th

1.	Find the equations of the tangent and normal to the given curves at the indicated points: (i) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)
Answer» (i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) dy \(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx slope at (0,5) = -10 ∴ Equation of tangent at (0, 5) is y – 5 = -10 (x – 0) y – 5 = -10 x 10 x + y – 5 = 0 slope of the normal at (0,5) (0, 5) = \(\frac{-1}{-10} = \frac{1}{10}\) ∴ equation of normal is y – 5 = \(\frac{1}{10}\)(x – 0) = 10y – 50 = x x – 10y + 50 = 0 (ii) \(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx slope of the tangent at x = 1

Find the equations of the tangent and normal to the given curves at the indicated points: (i) y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)(ii) y = x4 – 6x3 + 13x2 – 10x + 5 at (1, 3)

Answer»

(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0, 5) dy

\(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx

slope at (0,5) = -10

∴ Equation of tangent at (0, 5) is

y – 5 = -10 (x – 0)

y – 5 = -10 x 10 x + y – 5 = 0

slope of the normal at (0,5)

(0, 5) = \(\frac{-1}{-10} = \frac{1}{10}\)

∴ equation of normal is

y – 5 = \(\frac{1}{10}\)(x – 0) = 10y – 50 = x

x – 10y + 50 = 0

(ii) \(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10 dx

slope of the tangent at x = 1