Question

Use implicit differentiation to find the negative slope of a tangent to the circle x^2+y^2=16 at x=-2

Answer 1

Answer:

slope = - $\frac{\sqrt{3} }{3}$

Step-by-step explanation:

Differentiating implicitly with respect to x

2x + 2y $\frac{dy}{dx}$ = 0

2y $\frac{dy}{dx}$ = - 2x

$\frac{dy}{dx}$ = - $\frac{2x}{2y}$ = - $\frac{x}{y}$

$\frac{dy}{dx}$ is the measure of the slope of the tangent

rearrange equation to find corresponding y-coordinate of x = - 2

y² = 16 - 4 = 12 = 2$\sqrt{3}$ ⇒ y = ± 2$\sqrt{3}$

using x = - 2, y = - 2$\sqrt{3}$, then

$\frac{dy}{dx}$ = - $\frac{1}{\sqrt{3} }$ = - $\frac{\sqrt{3} }{3}$