Question

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)
(iii) y = x³ at (1, 1)
(iv) y = x² at (0, 0)
(v) x = cos t, y = sin t at t=π/4

Answer 1

(i) The equation of the curve is y = x⁴ − 6x³ + 13x² − 10x + 5. On differentiating with respect to x, we get:

Thus, the slope of the tangent at (0, 5) is −10. The equation of the tangent is given as:
y − 5 = − 10(x − 0)
⇒ y − 5 = − 10x
⇒ 10x + y = 5

The slope of the normal at (0, 5) is 

Therefore, the equation of the normal at (0, 5) is given as:

(ii) The equation of the curve is y = x⁴ − 6x³ + 13x² − 10x + 5. On differentiating with respect to x, we get:

Thus, the slope of the tangent at (1, 3) is 2. The equation of the tangent is given as:

The slope of the normal at (1, 3) is 

Therefore, the equation of the normal at (1, 3) is given as:

(iii) The equation of the curve is y = x³. On differentiating with respect to x, we get:

Thus, the slope of the tangent at (1, 1) is 3 and the equation of the tangent is given as:

(iv) The equation of the curve is y = x². On differentiating with respect to x, we get:

Thus, the slope of the tangent at (0, 0) is 0 and the equation of the tangent is given as: y − 0 = 0 (x − 0)
⇒ y = 0

The slope of the normal at (0, 0) is 

which is not defined.
Therefore, the equation of the normal at (x₀, y₀) = (0, 0) is given by  x=x₀=0.

(v) The equation of the curve is x = cos t, y = sin t.