Consider the curve: xy^2 - x^3y = 6, dy/dx = (3x^2y - y^2)/(2xy - x^3) a) Find all points on the...

Question:

Consider the curve: {eq}xy^2 - x^3y = 6, \frac{\mathrm{d} x}{\mathrm{d} y} = \frac{3x^2y - y^2}{2xy - x^3} {/eq}

a) Find all points on the curve whose x-coordinate is 1 and write an equation for the tangent line of each of these points.

b) Find the x-coordinate of each point on the curve where the tangent line is vertical.

Tangent Line:

In the form, {eq}y-y_1 = m(x-x_1) {/eq}, the tangent line shows where the line passes through a point on the function with a defined slope. In the tangent line, {eq}m {/eq} symbolizes the slope, while {eq}y_1 {/eq} and {eq}x_1 {/eq} are components that represent the initial condition.

Answer and Explanation:

Given: {eq}xy^2-x^3y = 6 \\ \frac{dy}{dx} = \frac{3x^2y-y^2}{2xy-x^3} \text{ [Interpretation of misaligned code and directions]} {/eq}


a. To find the tangent lines at {eq}x = 1 {/eq}, the initial {eq}y {/eq} values and the derivative at each condition are needed to determine the corresponding tangent lines.


Finding the {eq}y {/eq} values:

{eq}\begin{align*} (1)y^2-(1)^3y = 6 &\Rightarrow y^2-y = 6 \\ &\Rightarrow y^2-y-6 = 6-6 \\ &\Rightarrow y^2-y-6 = 0 \\ &\Rightarrow (y-3)(y+2) = 0 \\ &\Rightarrow y-3 = 0, y+2 = 0 \\ &\Rightarrow y-3+3 = 0+3, y+2-2 = 0-2 \\ &\Rightarrow y = 3, y = -2 \\ \end{align*} {/eq}


Now the tangent lines can be found for the initial conditions {eq}(1,3) {/eq} and {eq}(1,-2) {/eq}.


For the initial condition {eq}(1,3) {/eq}:

{eq}\begin{align*} m &= \frac{dy}{dx} \\ &= \frac{3(1)^2(3)-(3)^2}{2(1)(3)-(1)^3} \\ &= \frac{9(1)-9}{6-1} \\ &= \frac{9-9}{5} \\ &= \frac{0}{5} \\ &= 0 \\ \end{align*} {/eq}


Now the tangent line can be determined for the initial condition {eq}(1,3) {/eq} using the form: {eq}y-y_1 = m(x-x_1) {/eq}.


{eq}\begin{align*} y-y_1 = m(x-x_1) &\Rightarrow y-3 = 0(x-1) \\ &\Rightarrow y-3 = 0-0 \\ &\Rightarrow y-3 = 0 \\ &\Rightarrow y-3+3 = 0+3 \\ &\Rightarrow y = 3 \\ \end{align*} {/eq}


For the initial condition {eq}(1,-2) {/eq}:

{eq}\begin{align*} m &= \frac{dy}{dx} \\ &= \frac{3(1)^2(-2)-(-2)^2}{2(1)(-2)-(1)^3} \\ &= \frac{-6(1)-4}{-4-1} \\ &= \frac{-6-4}{-5} \\ &= \frac{-10}{-5} \\ &= 2 \\ \end{align*} {/eq}


Now the tangent line can be determined for the initial condition {eq}(1,-2) {/eq} using the form: {eq}y-y_1 = m(x-x_1) {/eq}.


{eq}\begin{align*} y-y_1 = m(x-x_1) &\Rightarrow y-(-2) = 2(x-1) \\ &\Rightarrow y+2 = 2x-2 \\ &\Rightarrow y+2-2 = 2x-2-2 \\ &\Rightarrow y = 2x-4 \\ \end{align*} {/eq}


b. For a tangent line to be vertical, the slope or derivative of the function will have to undefined, or equal to form {eq}\frac{\text{number}}{0} {/eq}. Therefore, the strategy is to set the derivative {eq}\frac{3x^2y-y^2}{2xy-x^3} {/eq} equal to this value and solve for the {eq}x {/eq}-coordinates.


{eq}\begin{align*} \frac{3x^2y-y^2}{2xy-x^3} = \frac{\text{number}}{0} &\Rightarrow 2xy-x^3 = 0 \\ &\Rightarrow x(2y-x^2) = 0 \\ &\Rightarrow x = 0, 2y-x^2 = 0 \\ &\Rightarrow x = 0, 2y-x^2+x^2 = 0+x^2 \\ &\Rightarrow x = 0, 2y = x^2 \\ &\Rightarrow x = 0, \frac{1}{2}\cdot 2y = x^2\cdot \frac{1}{2} \\ &\Rightarrow x = 0, y = \frac{x^2}{2} \\ \end{align*} {/eq}


Although we know one of the {eq}x {/eq}-coordinates that show the tangent line is vertical, the other condition requires substitution into the equation {eq}xy^2-x^3y = 6 {/eq}.


{eq}\begin{align*} x(\frac{x^2}{2})^2-x^3(\frac{x^2}{2}) = 6 &\Rightarrow x(\frac{x^4}{4})-x^3(\frac{x^2}{2}) = 6 \\ &\Rightarrow \frac{x^5}{4}-\frac{x^5}{2} = 6 \\ &\Rightarrow \frac{x^5}{4}-(\frac{x^5}{2})\cdot \frac{2}{2} = 6 \\ &\Rightarrow \frac{x^5}{4}-\frac{2x^5}{4} = 6 \\ &\Rightarrow \frac{-x^5}{4} = 6 \\ &\Rightarrow \frac{-x^5}{4}\cdot -4 = 6\cdot -4 \\ &\Rightarrow x^5 = -24 \\ &\Rightarrow \sqrt[5]{x^5} = \sqrt[5]{-24} \\ &\Rightarrow x = -1.88815023 \end{align*} {/eq}


Therefore, the {eq}x {/eq}-coordinate of each point on the curve where the tangent line is vertical is {eq}x = 0 {/eq} and {eq}x = -1.88815023 {/eq}.


Learn more about this topic:

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Tangent Line: Definition & Equation

from NY Regents Exam - Geometry: Tutoring Solution

Chapter 1 / Lesson 11
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