Chapter 3, Problem 17P

(a)

To determine

To show: If two lines intersect at an angle $\alpha$ then $\tan \alpha =\frac{m_2-m_1}{1+m_1{m}_2}$.

Explanation of Solution

Given:

Two lines are $L_1\text{ and }{L}_2$.

The slope of the lines $L_1\text{ and }{L}_2$ are $m_1\text{ and }{m}_2$ respectively.

Calculation:

Let ${\varphi }_1\text{ and }{\varphi }_2$ be an inclination angle of the line $L_1\text{ and }{L}_2$ respectively.

Here, the slopes of the line $L_1\text{ and }{L}_2$ are $\tan {\varphi }_1\text{ and tan}{\varphi }_2$.

The angle between two lines are $\alpha =\left|{\varphi }_2-{\varphi }_1\right|$.

If the angle ${\varphi }_2>{\varphi }_1$,

$\alpha ={\varphi }_2-{\varphi }_1$

Take tangent on both sides,

$\tan \alpha =\tan \left({\varphi }_2-{\varphi }_1\right)$

Since the identity $\tan \left(x-y\right)=\frac{\tan x-\tan y}{1+\tan x\tan y}$,

$\begin{array}{c}\tan \alpha =\frac{\tan {\varphi }_2-\tan {\varphi }_1}{1+\tan {\varphi }_2\tan {\varphi }_1}\\ =\frac{m_2-m_1}{1+m_1{m}_2}\end{array}$

Hence the required result is proved.

(b)

To determine

To find: The angle between each pair of curves at each points.

Answer to Problem 17P

The angle between two curves $y=x^2\text{ and }y={\left(x-2\right)}^2$ are, $53°\text{ and 127}°$ and

the angle between two curves $x^2-y^2\text{=3 and }{x}^2-4x+y^2+3=0$ are, $\alpha =117°$ and $63°$.

Explanation of Solution

Given:

The pair of equation are

(i) $y=x^2\text{ and }y={\left(x-2\right)}^2$.

(ii) $x^2\text{-}{y}^2\text{=3 and }{x}^2-4x+y^2+3=0$.

Calculation:

Obtain the intersection point of $y=x^2\text{ and }y={\left(x-2\right)}^2$.

$\begin{array}{l}{x}^2={\left(x-2\right)}^2\\ x^2=x^2+4-4x\\ 0=-4\left(x-1\right)\\ x=1\end{array}$

Thus, the intersection is $\left(1,1\right)$.

The slope of tangent line to the curve $y=x^2$ is, $\frac{dy}{dx}=2x$.

Thus, the slope of tangent to the cure $y=x^2$ at $\left(1,1\right)$ is, $m_1=2$.

The slope of the line $y={\left(x-2\right)}^2$ at $\left(1,1\right)$ is,

$\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}{\left(x-2\right)}^2\\ =2\left(x-2\right)\frac{d}{dx}\left(x-2\right)\\ =2\left(x-2\right)\left(1-0\right)\\ =2x-4\end{array}$ .

Substitute $\left(1,1\right)$ in $m_2=\frac{dy}{dx}$,

$\begin{array}{c}{m}_2=2\left(1\right)-4\\ =-2\end{array}$

Thus, the slope of the tangent to the curve at $\left(1,1\right)$ is, $m_2=-2$.

Form part (a), $\tan \alpha =\frac{m_2-m_1}{1+m_1{m}_2}$ where $\alpha$ is angle between two lines,

$\begin{array}{c}\tan \alpha =\frac{-2-2}{1+\left(-2\right)\left(2\right)}\\ =\frac{-4}{-3}\\ =\frac{4}{3}\\ \alpha ={\tan }^{-1}\left(\frac{4}{3}\right)\end{array}$

The value $\alpha =53°$.

The another angle is $180-\alpha$, $127°$.

Therefore, the angle between two curves $y=x^2\text{ and }y={\left(x-2\right)}^2$ are, $53°\text{ and 127}°$.

(ii)

Obtain the intersection point of $x^2-y^2\text{=3 and }{x}^2-4x+y^2+3=0$.

$x^2-4x+y^2+3=0$

Substitute $y^2=x^2-\text{3}$,

$\begin{array}{l}{x}^2-4x+x^2-3+3=0\\ 2x^2-4x=0\\ 2x\left(x-2\right)=0\\ x=0\text{ or }x=2\end{array}$

Here, $x=0$ is extraneous.

Substitute $x=2$ in $y^2=x^2-\text{3}$ and obtain y.

$\begin{array}{l}{y}^2=2^2-\text{3}\\ y^2=1\\ y=\pm 1\end{array}$

Thus, the points of intersection are, $\left(2,1\right)\text{ and }\left(2,-1\right)$.

The slope of tangent line to the curve $x^2-y^2=3$ is computed as follows,

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

Therefore, the slope of the tangent is, $m_1=\frac{x}{y}$

Thus, the slope of tangent to the cure $x^2-y^2\text{=3}$ at the point $\left(2,1\right)$ is, $m_1=2$ and at the point $\left(2,-1\right)$ is $m_2=-2$.

The slope of the line $x^2-4x+y^2+3=0$ is computed as follows,

$\begin{array}{l}\frac{d}{dx}\left(x^2-4x+y^2+3=0\right)=0\\ \frac{d}{dx}\left(x^2\right)-4\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(y^2\right)+\frac{d}{dx}\left(3\right)=0\\ 2x-4\left(1\right)+2y\frac{dy}{dx}+0=0\\ 2y\frac{dy}{dx}=-2x+4\end{array}$

That is, $\frac{dy}{dx}=\frac{-2x+4}{2y}$.

Substitute $\left(2,1\right)$ in $\frac{dy}{dx}=\frac{-2x+4}{2y}$,

$\begin{array}{c}{m}_3=\frac{-2\left(2\right)+4}{2\left(1\right)}\\ =0\end{array}$

Substitute $\left(2,-1\right)$ in $\frac{dy}{dx}=\frac{-2x+4}{2y}$,

$\begin{array}{c}{m}_4=\frac{-2\left(2\right)+4}{2\left(-1\right)}\\ =0\end{array}$

Thus, the slope of the tangent to the curve $x^2-4x+y^2+3=0$ at the point $\left(2,1\right)$ is, $m_3=0$ and at the point $\left(2,-1\right)$ is $m_4=0$.

At the point $\left(2,1\right)$,

The angle between two lines at $\tan \alpha =\frac{m_3-m_1}{1+m_1{m}_3}$ .

Substitute $m_1=2$ and $m_3=0$,

$\begin{array}{c}\tan \alpha =\frac{0-2}{1+\left(2\right)0}\\ =-2\\ \alpha ={\tan }^{-1}\left(-2\right)\\ =117°\end{array}$

The value $\alpha =117°$ and another angle is $180-\alpha =63°$.

At the point $\left(2,-1\right)$,

The angle between two lines is ,$\tan \beta =\frac{m_4-m_2}{1+m_2{m}_4}$.

Substitute $m_2=-2$ and $m_4=0$,

$\begin{array}{c}\tan \beta =\frac{0-\left(-2\right)}{1+\left(-2\right)0}\\ =2\\ \beta ={\tan }^{-1}\left(2\right)\\ =63°\end{array}$

The value $\beta =63°$ and another angle is $180-63°=117°$.

Therefore, the angle between two curves $x^2-y^2=3\text{ and }{x}^2-4x+y^2+3=0$ are, $117°$ and $63°$.