Chapter 3.2, Problem 4QQ

(a)

To determine

To find : $\frac{dy}{dx}$ .

Answer to Problem 4QQ

  $\frac{dy}{dx}=\frac{3x^{2}y-y^2}{\left(2xy-x^3\right)}$

Explanation of Solution

Given information :

A curve $x{y}^2-x^{3}y=6$ .

Formula needed :

  $\frac{d}{dt}\left(ab\right)=a\cdot \frac{d}{dt}\left(b\right)+b\cdot \frac{d}{dt}\left(a\right)$

Differentiate $x{y}^2-x^{3}y=6$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(x{y}^2-x^{3}y\right)=\frac{d}{dx}\left(6\right)\\ \frac{d}{dx}\left(x{y}^2\right)-\frac{d}{dx}\left(x^{3}y\right)=0\\ y^2\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(y^2\right)-y\frac{d}{dx}\left(x^3\right)-x^3\frac{d}{dx}\left(y\right)=0\\ y^2+x\left(2y\right)\frac{dy}{dx}-y\left(3x^2\right)-x^3\frac{dy}{dx}=0\end{array}$

Further simplifying.

  $\begin{array}{c}2xy\frac{dy}{dx}-x^3\frac{dy}{dx}=y\left(3x^2\right)-y^2\\ \left(2xy-x^3\right)\frac{dy}{dx}=3x^{2}y-y^2\\ \frac{dy}{dx}=\frac{3x^{2}y-y^2}{\left(2xy-x^3\right)}\end{array}$

Hence, $\frac{dy}{dx}=\frac{3x^{2}y-y^2}{\left(2xy-x^3\right)}$

(b)

To determine

To find : the equation of tangent line at $x$ − coordinate $1$ .

Answer to Problem 4QQ

The equation of tangent line at point $\left(1,-2\right)$ is $y=2x-4$ .

The equation of tangent line at point $\left(1,3\right)$ is $y=3$ .

Explanation of Solution

Given information :

A curve $x{y}^2-x^{3}y=6$ .

Formula needed :

Equation of tangent line is $y=mx+c$ where $m$ is slope and $c$ is constant.

Substitute $x=1$ in the curve $x{y}^2-x^{3}y=6$ .

  $\begin{array}{c}1y^2-1^{3}y=6\\ y\left(y-1\right)=6\\ y=3,-2\end{array}$

Thus, the points of contact are $\left(1,-2\right)$ and, $\left(1,3\right)$ .

Substitute $\left(1,-2\right)$ in $\frac{dy}{dx}=\frac{3x^{2}y-y^2}{\left(2xy-x^3\right)}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{\left(1,-2\right)}\\ =\frac{3{\left(1\right)}^2\left(-2\right)-{\left(-2\right)}^2}{\left(2\left(1\right)\left(-2\right)-1^3\right)}\\ =\frac{-6-4}{\left(-4-1\right)}\\ =\frac{-10}{-5}\\ =2\end{array}$

Plug the value of $m=2$ in $y=mx+c$ .

  $y=2x+c$

Plug the value of $\left(1,-2\right)$ in $y=2x+c$ .

  $\begin{array}{c}-2=2+c\\ c=-4\end{array}$

So, the equation of tangent line at point $\left(1,-2\right)$ is $y=2x-4$ .

Substitute $\left(1,3\right)$ in $\frac{dy}{dx}=\frac{3x^{2}y-y^2}{\left(2xy-x^3\right)}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{\left(1,3\right)}\\ =\frac{3{\left(1\right)}^2\left(3\right)-{\left(3\right)}^2}{\left(2\left(1\right)\left(3\right)-1^3\right)}\\ =\frac{9-9}{\left(6-1\right)}\\ =0\end{array}$

Plug the value of $m=0$ in $y=mx+c$ .

  $y=c$

Plug the value of $\left(1,3\right)$ in $y=c$ .

  $\begin{array}{c}3=c\\ c=3\end{array}$

So, the equation of tangent line at point $\left(1,3\right)$ is $y=3$ .

(c)

To determine

To find : the $x$ − coordinate where the tangent line is vertical.

Answer to Problem 4QQ

There are two $x$ - coordinate that is $x=-1.888$ and $x=0$ where the tangent line is vertical.

Explanation of Solution

Given information :

A curve $x{y}^2-x^{3}y=6$ .

Formula needed :

Tangent line is vertical implies $\frac{1}{m}=0$ .

From above $\frac{dy}{dx}=\frac{3x^{2}y-y^2}{\left(2xy-x^3\right)}$ .

  $\begin{array}{c}\frac{1}{m}=0\\ \frac{1}{dy/dx}=0\\ \frac{\left(2xy-x^3\right)}{3x^{2}y-y^2}=0\\ \left(2xy-x^3\right)=0\left[3x^{2}y-y^2\ne 0\right]\end{array}$

Simplifying further,

  $\begin{array}{c}\left(2xy-x^3\right)=0\\ x\left(2y-x^2\right)=0\end{array}$

Either $x=0$ or $2y-x^2=0$ .

So, either $x=0$ or $y=\frac{x^2}{2}$ .

Substitute $y=\frac{x^2}{2}$ in $x{y}^2-x^{3}y=6$ .

  $\begin{array}{c}x{\left(\frac{x^2}{2}\right)}^2-x^3\cdot \frac{x^2}{2}=6\\ \frac{x^5}{4}-\frac{x^5}{2}=6\\ -\frac{x^5}{4}=6\\ -x^5=24\end{array}$

Hence, the value of $x=-1.888$ .

There are two $x$ - coordinate that is $x=-1.888$ and $x=0$ where the tangent line is vertical.