72 - 74 Light intensity of illumination and theory of attraction | Differential Calculus Review at MATHalino

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

$I = \dfrac{k \sin \theta}{d^2}$

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

Solution:

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$I = \dfrac{k \sin \theta}{d^2}$

From the figure:
$\sin \theta = h/d$

$I = \dfrac{kh}{d^3}$

$\dfrac{dI}{dd} = \dfrac{kd^3 \dfrac{dh}{dd} - 3kd^2h}{d^6} = 0$

$\dfrac{dh}{dd} = \dfrac{3h}{d}$

$h^2 + a^2 = d^2$

$2h\dfrac{dh}{dd} = 2d$

$2h \left( \dfrac{3h}{d} \right) = 2d$

$d^2 = 3h^2$

Thus,
$h^2 + a^2 = 3h^2$

$h^2 = \frac{1}{2}a^2$

$h = \frac{1}{\sqrt{2}}a = \frac{1}{\sqrt{2}}a \times \frac{\sqrt{2}}{\sqrt{2}}$

$h = \frac{1}{2}\sqrt{2}a$ 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

$\dfrac{h}{(a^2 + h^2)^{3/2}}$

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:

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Attraction:
$A = \dfrac{h}{(a^2 + h^2)^{3/2}}$

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

$(a^2 + h^2)^{3/2} - 3h^2(a^2 + h^2)^{1/2} = 0$

$a^2 + h^2 = 3h^2$

$2h^2 = a^2$

$h^2 = \frac{1}{2}a^2$

$h = \frac{1}{\sqrt{2}}a = \frac{1}{\sqrt{2}}a \times \frac{\sqrt{2}}{\sqrt{2}}$

$h = \frac{1}{2}\sqrt{2}a$ answer

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

$\dfrac{h}{(h^2 + l^2)\sqrt{h^2 + 2l^2}}$

Find h for maximum attraction.

Solution:

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$A = \dfrac{h}{(h^2 + l^2)\sqrt{h^2 + 2l^2}}$

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

$(h^2 + l^2)\sqrt{h^2 + 2l^2} - \dfrac{h^2(h^2 + l^2)}{\sqrt{h^2 + 2l^2}} - 2h \sqrt{h^2 + 2l^2} = 0$

$(h^2 + l^2)(h^2 + 2l^2) - h^2(h^2 + l^2) - 2h^2(h^2 + 2l^2) = 0$

$(h^4 + 3l^2h^2 + 2l^4) - (h^4 + l^2h^2) - (2h^4 + 4l^2h^2) = 0$

$-2h^4 - 2l^2h^2 + 2l^4 = 0$

$h^4 + l^2h^2 - l^4 = 0$

$h^2 = \dfrac{-l^2 \pm \sqrt{l^4 + 4l^4}}{2}$

$h^2 = \dfrac{-l^2 \pm \sqrt{5}\,l^2}{2}$

$h^2 = \dfrac{-1 \pm \sqrt{5}}{2}l^2$

$h = \sqrt{\dfrac{-1 - \sqrt{5}}{2}} \, l \,\, \text{ (imaginary)}$

$h = \sqrt{\dfrac{-1 + \sqrt{5}}{2}} \, l \,\, \text{ (ok!)}$

Use
$h = \sqrt{\dfrac{-1 + \sqrt{5}}{2}} \, l$

$h = 0.7862l$ answer

72 - 74 Light intensity of illumination and theory of attraction

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