Problem 72

A light is to be placed above the center of a circular area of radius a. What height gives the best illumination on a circular walk surrounding the area? (When light from a point source strikes a surface obliquely, the intensity of illumination is

where θ is the angle of incidence and d the distance from the source.)

Solution:
[math]I = \dfrac{k \sin \theta}{d^2}[/math]

From the figure:
[math]\sin \theta = h/d[/math]

[math]I = \dfrac{kh}{d^3}[/math]

[math]\dfrac{dI}{dd} = \dfrac{kd^3 \dfrac{dh}{dd} - 3kd^2h}{d^6} = 0[/math]

[math]\dfrac{dh}{dd} = \dfrac{3h}{d}[/math]

[math]h^2 + a^2 = d^2[/math]

[math]2h\dfrac{dh}{dd} = 2d[/math]

[math]2h \left( \dfrac{3h}{d} \right) = 2d[/math]

[math]d^2 = 3h^2[/math]

Thus,
[math]h^2 + a^2 = 3h^2[/math]
[math]h^2 = \frac{1}{2}a^2[/math]
[math]h = \frac{1}{\sqrt{2}}a = \frac{1}{\sqrt{2}}a \times \frac{\sqrt{2}}{\sqrt{2}}[/math]
[math]h = \frac{1}{2}\sqrt{2}a \,\,[/math]            answer

Problem 73

It is shown in the theory of attraction that a wire bent in the form of a circle of radius a exerts upon a particle in the axis of the circle (i.e., in the line through the center of the circle perpendicular to the plane) an attraction proportional to

where h is the height of the particle above the plane of the circle. Find h, for maximum attraction. (Compare with Problem 72 above)

Solution:
Attraction:
[math]A = \dfrac{h}{(a^2 + h^2)^{3/2}}[/math]

[math]\dfrac{dA}{dh} = \dfrac{(a^2 + h^2)^{3/2} - \frac{3}{2}(a^2 + h^2)^{1/2}(2h)h}{(a^2 + h^2)^3} = 0[/math]

[math](a^2 + h^2)^{3/2} – 3h^2(a^2 + h^2)^{1/2} = 0[/math]
[math]a^2 + h^2 = 3h^2[/math]
[math]2h^2 = a^2[/math]
[math]h^2 = \frac{1}{2}a^2[/math]
[math]h = \frac{1}{\sqrt{2}}a = \frac{1}{\sqrt{2}}a \times \frac{\sqrt{2}}{\sqrt{2}}[/math]
[math]h = \frac{1}{2}\sqrt{2}a \,\,[/math]            answer

Problem 74
In Problem 73 above, if the wire has instead the form of a square of side [math]2l[/math], the attraction is proportional to

Find h for maximum attraction.

Solution:
[math]A = \dfrac{h}{(h^2 + l^2)\sqrt{h^2 + 2l^2}}[/math]

[math]\dfrac{dA}{dh} = \dfrac{(h^2 + l^2)\sqrt{h^2 + 2l^2}(1) - h\,\left[ (h^2 + l^2)\dfrac{2h}{2\sqrt{h^2 + 2l^2}} + 2h \sqrt{h^2 + 2l^2} \right]}{(h^2 + l^2)^2(h^2 + 2l^2)} = 0[/math]

[math](h^2 + l^2)\sqrt{h^2 + 2l^2} - \dfrac{h^2(h^2 + l^2)}{\sqrt{h^2 + 2l^2}} - 2h \sqrt{h^2 + 2l^2} = 0[/math]

[math](h^2 + l^2)(h^2 + 2l^2) – h^2(h^2 + l^2) – 2h^2(h^2 + 2l^2) = 0[/math]
[math](h^4 + 3l^2h^2 + 2l^4) – (h^4 + l^2h^2) – (2h^4 + 4l^2h^2) = 0[/math]
[math]–2h^4 – 2l^2h^2 + 2l^4 = 0[/math]
[math]h^4 + l^2h^2 – l^4 = 0[/math]

[math]h^2 = \dfrac{-l^2 \pm \sqrt{l^4 + 4l^4}}{2}[/math]

[math]h^2 = \dfrac{-l^2 \pm \sqrt{5}\,l^2}{2}[/math]

[math]h^2 = \dfrac{-1 \pm \sqrt{5}}{2}l^2[/math]

[math]h = \sqrt{\dfrac{-1 - \sqrt{5}}{2}} \, l \,\, \text{ (imaginary)}[/math]

[math]h = \sqrt{\dfrac{-1 + \sqrt{5}}{2}} \, l \,\, \text{ (ok!)}[/math]

Use
[math]h = \sqrt{\dfrac{-1 + \sqrt{5}}{2}} \, l[/math]

[math]h = 0.7862l \,\, [/math]            answer