Question

light intensity (i) is proportional (∝) to the inverse square of distance (d) of a subject from a light source. the relationship in intensities for subjects at different distances from the same source can be likewise seen as a ratio or proportional relationship. more on these concepts will come later, but for now:

$$i ∝ \frac{1}{d^2}$$

one could, therefore, determine the distance a subject was from a light source based on an intensity presented in the above fashion. moreover, and though beyond the immediate context of this course but useful for understanding linear equations, this relationship can be represented on a graph with each axis of the graph replacing these variables with limits placed on the values to keep them positive. thus,

$$y = i$$

$$x = d$$

$$x ≥ 0$$

$$y = \frac{1}{x^2}$$

using these limitations, solve for x when y = 0.015.

round to the nearest hundredth if necessary. you do not need to use labels for your answer. enter only the number.

Explanation:

Step1: Substitute \( y = 0.015 \) into the equation \( y=\frac{1}{x^{2}} \)

\( 0.015=\frac{1}{x^{2}} \)

Step2: Solve for \( x^{2} \)

Cross - multiply: \( x^{2}=\frac{1}{0.015} \)
Calculate \( \frac{1}{0.015}=\frac{1}{\frac{15}{1000}}=\frac{1000}{15}=\frac{200}{3}\approx66.67 \)

Step3: Solve for \( x \)

Take the square root of both sides. Since \( x\geq0 \), \( x = \sqrt{\frac{200}{3}}=\sqrt{\frac{200\times3}{3\times3}}=\frac{\sqrt{600}}{3}=\frac{10\sqrt{6}}{3}\approx\frac{10\times2.45}{3}\approx8.17 \)

Answer:

\( 8.17 \)