Question

Reflecting Surface Assume that when the plane curve C shown in Figure 1.3.23 is revolved about the x-axis, it generates a surface of revolution with the property that all light rays L parallel to the x-axis striking the surface are reflected to a single point O (the origin). Use the fact that the angle of incidenceis equal to the angle of reflection to determine a differential equation that describes the shape of the curve C. Such a curve C is important in applications ranging from construction of telescopes to satellite antennas, automobile headlights, and solar collectors. [Hint: Inspection of the figure shows that wecan write l. Why? Now use an appropriate trigonometric identity.

Short answer

Differential Equation that determines the shape of the curve C is $y^{\text{'}}=\frac{-x+\sqrt{x^2+y^2}}{y}$.

Step-by-step solution

Step 1:Definition of Newton’s law of cooling with mathematical statement.

Newton’s law of cooling/warming translates into the mathematical statement

$\frac{dT}{dt}\propto T-T_m\text{or}\frac{dT}{dt}=k\left(T-T_m\right)$

Where k is a constant of proportionality. In either case, cooling or warming, if $T_m$is a constant, it stands to reason that $k<0$.

Compute Differentiation.

Given the ray is parallel to the x axis the angle between L and PO equals the complement of $\varphi$.

Then we have $2\theta +\pi -\varphi =\pi$which implies$\varphi =2\theta$

We then have by definition that $\tan \varphi =\frac{y}{-x}$.

So,$-\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{y}{-x}$

We also have that the slope of the tangent of $C$at $P(x,y)$equals $\tan \theta ,$

Substitution.

Substituting $y^{\text{'}}=\tan \theta$, which we can reorganize into the quadratic equation

$\frac{2y^{\text{'}}}{1-{\left(y^{\text{'}}\right)}^2}=\frac{y}{x}$.

$y{\left(y^{\text{'}}\right)}^2+2x{y}^{\text{'}}-y=0.$

Solving for $y^{\text{'}}$via the quadratic formula gives us

$$\begin{gathered} y^{\text{'}}=\frac{-2x\pm \sqrt{4x^2+4y^2}}{2y} \\ =\frac{-x\pm \sqrt{x^2+y^2}}{y} \end{gathered}$$

Since we are looking for solutions with $y^{\text{'}}>0$.

It can be concluded that the Differential Equation that determines the shape of the curve C is $y^{\text{'}}=\frac{-x+\sqrt{x^2+y^2}}{y}$.