Question

Find the Tangent and Normal line to the given curve. 9x^3 - y^3 = 1 at (1,2)

Answer 1

Solution:

Tangent will have a slope of $\tfrac{dy}{dx}$

Normal will have a slope of $- \tfrac{dx}{dy}$

$9x^3-y^3=1$

Let's confirm that it does pass through (1 , 2)

$9*1^3-2^3=1$

9 - 8 = 1

1 = 1

Confirmed

$9x^3-y^3=1$

Derive implicitly

$27x^2dx-3y^2dy=0$

$9x^2dx-y^2dy=0$

$9x^2dx=y^2dy$

$x=1,y=2$

$9*1^2dx=2^2dy$

$9dx=4dy$

$\tfrac{9}{4}=\tfrac{dy}{dx}$

$-\tfrac{4}{9}=-\tfrac{dx}{dy}$

Point-slope form:  y - k = m * (x - h), where m = slope and (h , k) is a point on the line.

Tangent: m = $\tfrac{9}{4}$ , (1 , 2)

y - 2 = ($\tfrac{9}{4}$ ) * (x - 1)

Normal:  m = -$\tfrac{4}{9}$ , (1 , 2)

y - 2 = (-$\tfrac{4}{9}$ ) * (x - 1)