Calculus: Tangent Line

Question

The function is 3y² + y³ = x³ + x² .

Find equation of tangent line (1, -1).

Answer 1

Greetings! Lets solve this shall we ?

We must apply Implicit Differentiation to solve this problem respectively.

So,

3y²+y³ = x³+x²

6y(dy/dx) + 3y²(dy/dx) = 3x² + 2x

(dy/dx) (6y + 3y²) = 3x² + 2x ......................Then divide the (6y+3y²) to get

dy/dx = (3x²+2x) / (6y+3y²) .................Then we plug in the point (1,-1) to get

dy/dx = (-5/3)..........................Now we use the Point-Slope Formula to obtain the equation of a tangent line such that

y - y₁ = (dy/dx)(x - x₁)

y - (-1) = (-5/3)(x - 1)

y + 1 = (-5/3)(x - 1)

y + 1 = (-5/3)x + (5/3)

-1 = -1

y = (-5/3)x + (2/3)

I hope this helped!