The angle is increased in approximately 0.095 radians (5.443°) for [tex]x = 0.45\,m[/tex].
Based on the statement we construct the geometric diagram, by definition of tangent we have expressions for the initial and final angles ([tex]\theta_{1}[/tex], [tex]\theta_{2}[/tex]), in radians, of the figure:
Initial triangle
[tex]\tan \theta_{1} = \frac{y_{1}}{x_{1}}[/tex] (1)
Final triangle
[tex]\tan \theta_{2} = \frac{y_{2}}{x_{2}}[/tex] (2)
By using (2), the equivalence [tex]\theta_{2} = \theta_{1}+\Delta \theta[/tex] and trigonometric identities we have the following expression:
[tex]\frac{y_{2}}{x_{2}} = \frac{\tan \theta_{1}+\tan \theta_{2}}{1-\tan \theta_{1}\cdot \tan \theta_{2}}[/tex] (3)
By (1), we simplify the expression:
[tex]\frac{y_{2}}{x_{2}} = \frac{\frac{y_{1}}{x_{1}} + \tan \Delta \theta}{1-\frac{y_{1}}{x_{1}}\cdot \tan \Delta \theta}[/tex] (3b)
If [tex]0 \le \Delta \theta \le \frac{\pi}{6}[/tex], then we can use the following approximation:
[tex]\tan \Delta\theta \approx \Delta \theta[/tex] (4)
Then, we reduce (3b) into an entirely algebraic expression:
[tex]\frac{y_{2}}{x_{2}} = \frac{\frac{y_{1}}{x_{1}}+\Delta \theta }{1-\frac{y_{1}}{x_{1}}\cdot \Delta \theta }[/tex] (3c)
Where [tex]y_{2} = \sqrt{r^{2}-x_{2}^{2}}[/tex].
Now we clear [tex]\Delta \theta[/tex] within the formula:
[tex]\Delta \theta = \frac{\frac{y_{2}}{x_{2}}-\frac{y_{1}}{x_{1}}}{\frac{y_{1}}{x_{1}}\cdot \left(1+\frac{y_{2}}{x_{2}} \right) }[/tex] (5)
If we know that [tex]x_{1} = y_{1} = 0.5[/tex], [tex]r = 0.707[/tex] and [tex]x_{2} = 0.45[/tex], then we estimate the angle change:
[tex]y_{2} = \sqrt{0.707^{2}-0.45^{2}}[/tex]
[tex]y_{2} \approx 0.545[/tex]
[tex]\Delta \theta = \frac{\frac{0.545}{0.45}-1 }{1\cdot \left(1+\frac{0.545}{0.45} \right)}[/tex]
[tex]\Delta \theta = 0.095[/tex]
As [tex]\Delta \theta < \frac{\pi}{6}[/tex], then the result seems to be reasonable. The angle is increased in approximately 0.095 radians (5.443°) for [tex]x = 0.45\,m[/tex].
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