Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1) (astroid)

Answer 1

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula :$\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x=$-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m=$\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

Answer 2

Applying implicit differentiation, the equation of the tangent line is given by:

$y - \sqrt{3} = \sqrt{3}(x + 3)$

The equation of a line tangent to a function f(x) at a point $(x_0, y_0)$ is given by:

$y - y_0 = f^{\prime}(x_0,y_0)(x - x_0)$

In this problem, point $(-3, \sqrt{3})$, thus $x_0 = -3, y_0 = \sqrt{3}$

The function is:

$\frac{x^2}{3} + \frac{y^2}{3} = 4$

To find the derivative, we have to apply implicit differentiation, thus:

$\frac{2x}{3}\frac{dx}{dx} + \frac{2y}{3}\frac{dy}{dx} = 0$

$\frac{2y}{3}\frac{dy}{dx} = -\frac{2x}{3}$

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

Thus:

$f^{\prime}(-3,\sqrt{3}) = -\frac{-3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \sqrt{3}$

Then, the equation to the line tangent is:

$y - y_0 = f^{\prime}(x_0,y_0)(x - x_0)$

$y - \sqrt{3} = \sqrt{3}(x + 3)$

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