e^(x) - x³ + y² = 10 at the point (0 ; 3)

e^(x) - x³ + y² = 10 → when x is constant: → [2y].dy = 0

e^(x) - x³ + y² = 10 → when y is constant: → [e^(x) - 3x²].dx = 0

[2y].dy + [e^(x) - 3x²].dx = 0

[2y.dy] = - [e^(x) - 3x²].dx

dy/dx = - [e^(x) - 3x²]/[2y] → at point (0 ; 3), i.e. when: x = 0 and when: y = 3

dy/dx = - [e^(0) - 0]/[6]

dy/dx = - 1/6 ← this is the slope of the tangent line to the curve

To go further, let's calculate the equation of the tangent line to the curve at (0 ; 3)

The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept

As the slope of the tangent line is (- 1/6), the equation of this tangent line becomes: y = - (1/6).x + y₀

The tangent line passes through (0 ; 3), so these coordinates must verify the equation of the tangent line.

y = - (1/6).x + y₀

y₀ = y + (1/6).x → you substitute x and y by the coordinates of the point (0 ; 3)

y₀ = 3

→ The equation of the tangent line to the curve at (0 ; 3) is: y = - (1/6).x + 3

y = - (1/6).x + 3

y = (- x + 18)/3

3y = - x + 18

x + 3y = 18

To go further, let's calculate the equation of the normal line to the curve at (0 ; 3)

The normal line to the curve at (0 ; 3) is perpendicular to the tangent line.

Two lines are perpendicular if the product of their slope is (- 1).

As the slope of the tangent line is (- 1/6), the slope of the normal line is (6).

The equation of the normal line becomes: y = 6x + y₀

The normal line passes through (0 ; 3), so these coordinates must verify the equation of the normal line.

y = 6x + y₀

y₀ = y - 6x → you substitute x and y by the coordinates of the point (0 ; 3)

y₀ = 3

→ The equation of the normal line to the curve at (0 ; 3) is: y = 6x + 3

y = 6x + 3

6x - y = - 3