Consider the equation x2 – xy? + y³ = 13, where y is a differentiable function of r.First, show that the point (-1, 2) is on the graph of the equation, Then find an equation forthe line L tangent to the graph of the equation at (-1,2).

Answered: Consider the equation x2 – xy? + y³ =…

Trigonometry (MindTap Course List)

10th Edition

ISBN:9781337278461

Author:Ron Larson

Publisher:Ron Larson

Chapter6: Topics In Analytic Geometry

Section6.2: Introduction To Conics: parabolas

Problem 4ECP: Find an equation of the tangent line to the parabola y=3x2 at the point 1,3.

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Question

Consider the equation x2 – xy? + y³ = 13, where y is a differentiable function of r.
First, show that the point (-1, 2) is on the graph of the equation, Then find an equation for
the line L tangent to the graph of the equation at (-1,2).

Expert Solution

Step 1

The given equation is:

$x^{2} - x y^{2} + y^{3} = 13$

where $y$ is a differentiable function of $x$ .

Check whether $(- 1, 2)$ satisfies the equation. Substitute $(x, y) = (- 1, 2)$ in the equation:

${(- 1)}^{2} - (- 1) {(2)}^{2} + {(2)}^{3} = 13 1 + 4 + 8 = 13 13 = 13$

The point $(- 1, 2)$ satisfies the equation. Hence the point $(- 1, 2)$ is on the graph of the equation.

Step 2

Result: The slope of a tangent line of a function at a point is the value of the derivative of the function at the same point.

Result: The derivative of a constant is zero.

Find the derivative $(y')$ of the function $y$ using implicit differentiation:

$x^{2} - x y^{2} + y^{3} = 13 x^{2} - x y^{2} (x) + y^{3} (x) = 13$

Differentiate the equation both sides with respect to $x$ :

$\begin{array}{rcl} (x^{2} - x y^{2} (x) + y^{3} (x))' & = & (13)' \\ (x^{2})' - (x y^{2} (x))' + (y^{3} (x))' & = & 0 \end{array}$

Step 3

Formula (Product of derivatives): Let $u and v$ be differentiable function. Then, $(u v)' = u v' + v u'$

Formula (Chain rule): Let $f (x)$ and $g (x)$ be differentiable function. Then, $(f (g (x)))' = f' (g (x)) g' (x)$

Formula: $(x^{n})' = n x^{n - 1}$

$\begin{array}{rcl} (x^{2}) \end{array} \begin{array}{rcl} ' \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} (x y^{2} (x)) \end{array} \begin{array}{rcl} ' \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} (y^{3} (x)) \end{array} \begin{array}{rcl} ' \end{array} \begin{array}{rcl} = \end{array} \begin{array}{rcl} 0 \end{array}$

Use Product rule to differentiate the middle term:

$(x^{2})' - [x {(y^{2} (x))}^{'} + (x') y^{2} (x)] + (y^{3} (x))' = 0 2 x - [x {(y^{2} (x))}^{'} + (1) y^{2} (x)] + (y^{3} (x))' = 0$

Use Chain rule to differentiate the $y$ terms:

$2 x - [x {(y^{2} (x))}^{'} + y^{2} (x)] + (y^{3} (x))' = 0 2 x - [x (2 y (x)) y' (x) + y^{2} (x)] + 3 y^{2} (x) y' (x) = 0 2 x - x (2 y (x)) y' (x) - y^{2} (x) + 3 y^{2} (x) y' (x) = 0$

Step 4

$2 x - x (2 y (x)) y' (x) - y^{2} (x) + 3 y^{2} (x) y' (x) = 0$

Isolate $y' (x)$ :

$2 x - x (2 y (x)) y' (x) - y^{2} (x) + 3 y^{2} (x) y' (x) = 0 - x (2 y (x)) y' (x) - y^{2} (x) + 3 y^{2} (x) y' (x) = - 2 x - x (2 y (x)) y' (x) + 3 y^{2} (x) y' (x) = - 2 x + y^{2} (x) y' (x) [- x (2 y (x)) + 3 y^{2} (x)] = - 2 x + y^{2} (x) y' (x) = \frac{- 2 x + y^{2} (x)}{- 2 x y (x)) + 3 y^{2} (x)}$

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Consider the equation x2 – xy? + y³ = 13, where y is a differentiable function of r. First, show that the point (-1,2) is on the graph of the equation, Then find an equation for the line L tangent to the graph of the equation at (-1,2).