Question

1. Find the equation of the tangent line to the parabola y = x^3 at the point P(2,8).

Use x = 2.1, 2.01, 2.001, 2.0001, 2.00001; 1.9,1.99, 1.999,1.9999, 1.99999 to find slope.

Answer 1

a) Equation of the tangent of the parabola

12 x - y -16=0

b)

slope of the tangent at x=2.1

`((dy)/(dx)) = 3(2.1)^(2) = 13.23`

slope of the tangent at x=2.01

`((dy)/(dx)) = 3(2.01)^(2) = 12.1203`

Explanation:

Step(i):-

Given parabola y = x³ ....(i)

Differentiating equation (i) with respective to 'x'

`(dy)/(dx) = 3x^(2)`

slope of the tangent

`((dy)/(dx))_((2,8)) = 3x^(2) = 3(2)^(2) =12`

Step(ii):-

Equation of the tangent of the parabola

`y-y_(1) = m (x-x_(1) )`

y-8 = 12 (x -2)

12 x - 24 -y +8 =0

12 x - y -16=0

b)

slope of the tangent at x=2.1

`((dy)/(dx)) = 3(2.1)^(2) = 13.23`

slope of the tangent at x=2.01

`((dy)/(dx)) = 3(2.01)^(2) = 12.1203`