Chapter 4.2, Problem 4QQ

(a)

To determine

To Find: Differentiation of the given function.

Answer to Problem 4QQ

The differentiation is $\frac{3x^{2}y-y^2}{2xy-x^3}$ .

Explanation of Solution

Given information: Given function is $x{y}^2-x^{3}y=6x$ .

Formula used: $\frac{d(x^n)}{dx}=n{x}^{n-1}\text{and}\frac{d(\text{constant})}{dx}=0$ with product, chain and subtraction rule.

Calculation:

  $x{y}^2-x^{3}y=6$

Differentiate with respect to x ,

  $\begin{array}{l}\left(x\times \frac{d(y^2)}{dy}\times \frac{dy}{dx}+\frac{d(x)}{dx}\times y^2\right)-\left(x^3\times \frac{dy}{dx}+\frac{d(x^3)}{dx}\times y\right)=0\\ \left(x\times 2y\times \frac{dy}{dx}+1\times y^2\right)-\left(x^3\frac{dy}{dx}+3x^2\times y\right)=0\\ 2xy\frac{dy}{dx}+y^2-x^3\frac{dy}{dx}-3x^{2}y=0\\ 2xy\frac{dy}{dx}-x^3\frac{dy}{dx}=3x^{2}y-y^2\\ \frac{dy}{dx}\times (2xy-x^3)=3x^{2}y-y^2\\ \frac{dy}{dx}=\frac{3x^{2}y-y^2}{2xy-x^3}\end{array}$

(b)

To determine

To Find: Equation of tangent on the given curve.

Answer to Problem 4QQ

The required equations are $y=3\text{and}2x-y-4=0$

Explanation of Solution

Given information: Given function is $x{y}^2-x^{3}y=6$ with x-coordinate is 1.

Formula used: For equation $y-y_1=m(x-x_1)\text{where}m\text{is}\text{the}\text{slope}$

For derivatives $\frac{d(x^n)}{dx}=n{x}^{n-1}\text{and}\frac{d(\text{constant})}{dx}=0$ with product, chain and subtraction rule.

Calculation:

Let the intersection points of the curve and tangent is $P(x_1,y_1)$

Since x-coordinate is 1 and points lie on the curve, we get

  $\begin{array}{l}{y}_1^2-y_1=6\\ y_1^2-y_1-6=0\\ y_1^2-3y_1+2y_1-6=0\\ y_1(y_1-3)+2(y_1-3)=0\\ (y_1-3)(y_1+2)=0\\ y_1=3,y_1=-2\\ \text{Thus}\text{required}\text{points}\text{are}(1,3)\text{and}(1,-2)\end{array}$

also, the slope of tangent at given points

  $x{y}^2-x^{3}y=6$

Differentiate with respect to x

  $\begin{array}{l}\left(x\times \frac{d(y^2)}{dy}\times \frac{dy}{dx}+\frac{d(x)}{dx}\times y^2\right)-\left(x^3\times \frac{dy}{dx}+\frac{d(x^3)}{dx}\times y\right)=0\\ \left(x\times 2y\times \frac{dy}{dx}+1\times y^2\right)-\left(x^3\frac{dy}{dx}+3x^2\times y\right)=0\\ 2xy\frac{dy}{dx}+y^2-x^3\frac{dy}{dx}-3x^{2}y=0\\ 2xy\frac{dy}{dx}-x^3\frac{dy}{dx}=3x^{2}y-y^2\\ \frac{dy}{dx}\times (2xy-x^3)=3x^{2}y-y^2\\ \frac{dy}{dx}=\frac{3x^{2}y-y^2}{2xy-x^3}\\ {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{3x_1^2{y}_1-y_1^2}{2x_1{y}_1-x_1^3}\end{array}$

and slope of tangent at these points

  $\begin{array}{l}\text{Slope}\text{of}\text{tangent}\text{at}(1,3)\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=\frac{3{(1)}^2(3)-{(3)}^2}{2(1)(3)-{(1)}^3}\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=\frac{9-9}{6-1}\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=0\\ \text{Slope}\text{of}\text{tangent}\text{at}(1,-2)\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=\frac{3{(1)}^2(-2)-{(-2)}^2}{2(1)(-2)-{(1)}^3}\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=\frac{-6-4}{-4-1}\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=\frac{-10}{-5}\\ {\left(\frac{dy}{dx}\right)}_{(1,-2)}=2\end{array}$

Now,

  $\begin{array}{l}\text{Equation}\text{of}\text{tangent}\text{at}(1,3)\\ y-3=0(x-1)\\ y-3=0\\ y=3\\ \text{Equation}\text{of}\text{tangent}\text{at}(1,-2)\\ y+2=2(x-1)\\ y+2=2x-2\\ 2x-y-4=0\end{array}$

(c)

To determine

To Find: Finding the x-coordinate of the given curve whose tangent is parallel to y-axis.

Answer to Problem 4QQ

The required x -coordinate is

  ${(-24)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

Explanation of Solution

Given information: Slope of tangent parallel to y-axis.

Formula used: $\frac{d(x^n)}{dx}=n{x}^{n-1}\text{and}\frac{d(\text{constant})}{dx}=0$ with product, chain and subtraction rule.

Calculation:

Let the intersection points of the curve and tangent is $P(x_1,y_1)$

also, the slope of tangent at given points

  $x{y}^2-x^{3}y=6$

Differentiate with respect to x

  $\begin{array}{l}\left(x\times \frac{d(y^2)}{dy}\times \frac{dy}{dx}+\frac{d(x)}{dx}\times y^2\right)-\left(x^3\times \frac{dy}{dx}+\frac{d(x^3)}{dx}\times y\right)=0\\ \left(x\times 2y\times \frac{dy}{dx}+1\times y^2\right)-\left(x^3\frac{dy}{dx}+3x^2\times y\right)=0\\ 2xy\frac{dy}{dx}+y^2-x^3\frac{dy}{dx}-3x^{2}y=0\\ 2xy\frac{dy}{dx}-x^3\frac{dy}{dx}=3x^{2}y-y^2\\ \frac{dy}{dx}\times (2xy-x^3)=3x^{2}y-y^2\\ \frac{dy}{dx}=\frac{3x^{2}y-y^2}{2xy-x^3}\\ {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{3x_1^2{y}_1-y_1^2}{2x_1{y}_1-x_1^3}\end{array}$

Since tangent is parallel to y-axis

  $\begin{array}{l}\therefore \text{Slope}\text{of}\text{tangent}=\text{Slope}\text{of}y\text{-axis}\\ {\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\text{not}\text{defined}\\ \text{the}\text{given}\text{slope}\text{is}\text{not}\text{defined}\text{when}\\ 2x_1{y}_1-x_1^3=0\\ x_1(2y_1-x_1^2)=0\\ x_1=0,y_1=\frac{x_1^2}{2}\\ \text{Since}\text{point}\text{lie}\text{also}\text{on}\text{the}\text{curve,}\text{we}\text{have}\\ x_1{y}_1^2-x_1^3{y}_1=6\\ at{x}_1=0,y_1\text{is}\text{not}\text{defined}\\ at{y}_1=\frac{x_1^2}{2}\\ x_1{\left(\frac{x_1^2}{2}\right)}^2-x_1^3\left(\frac{x_1^2}{2}\right)=6\\ \frac{x_1^5}{4}-\frac{x_1^5}{2}=6\\ \frac{-x_1^5}{4}=6\\ x_1^5=-24\\ $ x_1={(-24)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$\text{Thus}\text{the}\text{required}x\text{-coordinate}\text{is}{(}^{-}\end{array}$