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SashulF [63]
3 years ago
13

The illumination at a point is inversely proportional to the square of the distance of the point from the light source and direc

tly proportional to the intensity of the light source. Suppose two light sources are ð  feet apart and their intensities are ð¼ and ð½, respectively. Let P be the point between them where the sum of their illuminations is a minimum. The distance of P from light source ð¼ will be
Mathematics
1 answer:
Ksju [112]3 years ago
6 0

Answer:

ð/[2^(1/3)] + 1 or ð/2.26

Step-by-step explanation:

Step 1: Let the illumination be denoted as C

Let the two intensities be denoted as M and N respectively

Let the distance from point P to M be x and distance from P to N be (ð-x

Step 2:

Illumination at point P from M: Cm= kM/x^2

Illumination at point P from N: Cn= kN/(ð-x)^2

The sun of the illumination: Q=Cm+Cn = (kM/x^2) + (kN/(ð-x)^2)

Differentiate Q wrt x and equate to zero, we have

(-2kM/x^3) + [2kN/(ð-x)^3] = 0

Simplifying the above equation, we have

(ð-x)^3/x^3 = -2kN/-2kM

[(ð-x)/x]^3 = N/M

[(ð-x)/x] = (N/M)^(1/3)

ð-x = x[(N/M)^(1/3)]

ð = x[((N/M)^(1/3)) + 1]

Therefore, x = ð/[((N/M)^(1/3)) + 1]

Substitute M= ð/4 and N=ð/2 into equation above, we have

x = ð/[2^(1/3)] +1 or x = ð/2.26

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Determine which values of p the following integrals converge. Give your answer in each case by selecting the appropriate inequal
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Answer:

Step-by-step explanation:

a)

\int\limits^2_1 {\frac{1}{x(lnx)^p} } \, dx

this can be done by substitute lnx = u

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When x =1, u =0 and when x =2, u = ln 2

So integral = \int\limits^{ln2} _0 {du/u^p} \\\=\frac{u^{-p+1} }{-p+1}

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When we substitute limits

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b) \int\limits^1_0 {lnx}/x^p \, dx \\\int \frac{\ln \left(x\right)}{x^p}dx=\frac{1}{-p+1}x^{-p+1}\ln \left(x\right)-\frac{x^{-p+1}}{\left(-p+1\right)^2}+C

So not converging for p =1

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