Chapter 8.4, Problem 12E

Solution

To find: The equation used to convert an equation in terms of $\left(x,y\right)$ to an equation in terms of $\left(u,v\right)$ when the angle of rotation is $\theta ={\tan }^{-1}\left(\frac{5}{12}\right)$ .

The liner equation is $x=\frac{12u}{13}-\frac{5v}{13}$ and $y=\frac{5u}{13}+\frac{12v}{13}$ .

Given Information:

The axis of rotation is $\theta ={\tan }^{-1}\left(\frac{5}{12}\right)$ .

Formula:

The Pythagorean theorem is ${\left(\text{Opposite}\right)}^2+{\left(\text{Adjacent}\right)}^2={\left(\text{Hypotenuse}\right)}^2$ .

The equations for the axis rotation are,

  $\begin{array}{l}x=u\cos \theta -v\sin \theta \\ y=u\sin \theta +v\cos \theta \end{array}$

Calculation:

Take tangent function in both side of the equation, $\theta ={\cos }^{-1}\left(\frac{3}{5}\right)$ and solve using the property, $\tan \left({\tan }^{-1}\theta \right)=\theta$ .

  $\begin{array}{l}\tan \theta =\tan \left({\tan }^{-1}\left(\frac{5}{12}\right)\right)\\ \tan \theta =\frac{5}{12}\end{array}$

Since the tangent function is $\tan \theta =\frac{\text{Opposite}}{\text{Adjacent}}$ , the opposite is 5 and adjacent is 12. Apply the Pythagorean theorem.

  $\begin{array}{c}{\left(\text{Hypotenuse}\right)}^2=5^2+{12}^2\\ {\left(\text{Hypotenuse}\right)}^2=25+144\\ {\left(\text{Hypotenuse}\right)}^2=169\\ \text{Hypotenuse}=13\end{array}$

Substitute 5 for opposite and 13 for hypotenuse into the definition, $\sin \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}$ .

  $\sin \theta =\frac{\text{5}}{\text{13}}$

Substitute 12 for adjacent and 13 for hypotenuse into the definition, $\cos \theta =\frac{\text{Opposite}}{\text{Hypotenuse}}$ .

  $\cos \theta =\frac{\text{12}}{\text{13}}$

Substitute $\sin \theta =\frac{\text{5}}{\text{13}}$ and $\cos \theta =\frac{\text{12}}{\text{13}}$ in above equation to find the equation in terms of $x$ .

  $\begin{array}{c}x=u\cdot \frac{12}{13}-v\cdot \frac{5}{13}\\ =\frac{12u}{3}-\frac{5v}{13}\end{array}$

Similarly, substitute $\sin \theta =\frac{\text{5}}{\text{13}}$ and $\cos \theta =\frac{\text{12}}{\text{13}}$ in above equation to find the equation in terms of $y$ .

  $\begin{array}{c}y=u\cdot \frac{5}{13}+v\cdot \frac{12}{13}\\ =\frac{5u}{13}+\frac{12v}{13}\end{array}$

Therefore, the required equations are $x=\frac{12u}{13}-\frac{5v}{13}$ and $y=\frac{5u}{13}+\frac{12v}{13}$ .