Chapter 3.2, Problem 50E

(a)

To determine

To find : the points where the tangents is parallel to x − axis.

Answer to Problem 50E

The points where tangents are parallel to x − axis are $\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)$ and $\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)$ .

Explanation of Solution

Given information :

The curve is $x^2+xy+y^2=7$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

The tangent is parallel to x − axis means slope is zero.

For slope, differentiate $x^2+xy+y^2=7$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(x^2+xy+y^2\right)=\frac{d}{dx}\left(7\right)\\ \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(7\right)\\ 2x+y\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(y\right)+2y\frac{dy}{dx}=0\\ 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\end{array}$

Solve further,

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

Now slope is zero.

  $\begin{array}{c}-\frac{\left(2x+y\right)}{\left(x+2y\right)}=0\\ \left(2x+y\right)=0\left[\left(x+2y\right)\ne 0\right]\\ y=-2x\end{array}$

Plug $y=-2x$ in the curve $x^2+xy+y^2=7$ .

  $\begin{array}{c}{x}^2+x\left(-2x\right)+{\left(-2x\right)}^2=7\\ x^2-2x^2+4x^2=7\\ 3x^2=7\\ x^2=\frac{7}{3}\end{array}$

Take square root on both sides,

  $\begin{array}{c}\sqrt{x^2}=\pm \sqrt{\frac{7}{3}}\\ x=-\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\end{array}$

The points where tangents are parallel to x − axis are $\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)$ and $\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)$ .

(b)

To determine

To find : the points where the tangents is parallel to y − axis.

Answer to Problem 50E

The points where tangents are parallel to y − axis are $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right)$ and $\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$ .

Explanation of Solution

Given information :

The curve is $x^2+xy+y^2=7$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

Product rule of derivative: $\frac{d}{dx}\left(u\cdot v\right)=u^{\prime }\cdot v+u\cdot v^{\prime }$

The tangent is parallel to y − axis means reciprocal of slope is zero.

For slope, differentiate $x^2+xy+y^2=7$ with respect to $x$ .

  $\begin{array}{c}\frac{d}{dx}\left(x^2+xy+y^2\right)=\frac{d}{dx}\left(7\right)\\ \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(xy\right)+\frac{d}{dx}\left(y^2\right)=\frac{d}{dx}\left(7\right)\\ 2x+y\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(y\right)+2y\frac{dy}{dx}=0\\ 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\end{array}$

Solve further,

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

Now reciprocal of slope is zero.

  $\begin{array}{c}-\frac{\left(x+2y\right)}{\left(2x+y\right)}=0\\ \left(x+2y\right)=0\left[\left(2x+y\right)\ne 0\right]\\ x=-2y\end{array}$

Plug $x=-2y$ in the curve $x^2+xy+y^2=7$ .

  $\begin{array}{c}{y}^2+\left(-2y\right)y+{\left(-2y\right)}^2=7\\ y^2-2y^2+4y^2=7\\ 3y^2=7\\ y^2=\frac{7}{3}\end{array}$

Take square root on both sides,

  $\begin{array}{c}\sqrt{y^2}=\pm \sqrt{\frac{7}{3}}\\ y=-\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\end{array}$

The points where tangents are parallel to y − axis are $\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right)$ and $\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)$ .