The intensity , in arbitrary units, of light from a point source is inversely proportional to the square of the distance , in meters, from the source, so that for some constant . If the intensity of the light is 36 units at a distance of 2 meters, at what distance, in meters, from the source is the intensity 4 units?

Solution available with step-by-step explanation

SAT Nonlinear Functions Question 89 | Free SAT Prep | Aniko

00:00

Question 89·Medium·Nonlinear Functions

0 / 234

The intensity $I$ , in arbitrary units, of light from a point source is inversely proportional to the square of the distance $r$ , in meters, from the source, so that $I = \dfrac{k}{r^2}$ for some constant $k$ .

If the intensity of the light is 36 units at a distance of 2 meters, at what distance, in meters, from the source is the intensity 4 units?

For inverse-variation questions like this, immediately translate the words into an equation: here, "inversely proportional to the square" means $I = k/r^2$ , or equivalently $Ir^2 = k$ is constant. Use the first set of values to find that constant quickly (either compute $k$ explicitly or set up $I_1 r_1^2 = I_2 r_2^2$ ), then plug in the target intensity and solve for the new distance. Always check that your answer is reasonable: if intensity goes down, distance in an inverse-square law should go up, and by the square root of the intensity ratio, not the full ratio.

Hints

Click me!

Understand the relationship

What does "inversely proportional to the square of the distance" tell you about the formula that connects $I$ and $r$ ?

Use the first situation to find the constant

Plug $I = 36$ and $r = 2$ into the formula from the prompt to find the constant $k$ .

Set up the equation for the target intensity

After you find $k$ , write a new equation using $I = 4$ and solve that equation for $r$ .

Check your result logically

As distance increases, intensity should decrease. Make sure your final distance is greater than 2 meters and that it fits the inverse-square pattern.

Desmos Guide

Find the constant using Desmos

In a Desmos expression line, type k = 36*(2^2) and note the value of $k$ that appears (this is the constant in the formula).

Graph the intensity function

In a new line, type I(r) = k/(r^2) to define the intensity as a function of distance $r$ . Make sure the graphing window shows positive $r$ values (e.g., $r$ from 0 to 20).

Mark the target intensity and find the distance

In another line, type y = 4 to draw a horizontal line for intensity 4. Use the intersection tool (tap/click the point where the curves meet) and read off the $r$ -coordinate of the intersection; that $r$ -value is the distance where the intensity is 4 units.

Step-by-step Explanation

Translate the description into an equation

"Inversely proportional to the square of the distance" means the intensity $I$ satisfies

I = \frac{k}{r^2}

for some constant $k$ . Here, $I$ is the intensity and $r$ is the distance from the source.

Use the first data point to find the constant

You are told that when $r = 2$ meters, the intensity $I = 36$ units.

Substitute $I = 36$ and $r = 2$ into $I = k/r^2$ :

36 = \dfrac{k}{2^2}

36 = \dfrac{k}{4}

Multiply both sides by 4:

k = 36 \times 4 = 144

So the intensity formula is

I = \frac{144}{r^2}.

Set up the equation for the new intensity

Now you want the distance $r$ where the intensity is $4$ units.

Use $I = 144/r^2$ and substitute $I = 4$ :

4 = \dfrac{144}{r^2}

Now solve this equation for $r$ .

Solve for the distance

From

4 = \dfrac{144}{r^2}

multiply both sides by $r^2$ :

4r^2 = 144

Divide both sides by 4:

r^2 = 36

Take the positive square root (distance must be positive):

r = \sqrt{36} = 6

So the intensity is 4 units at a distance of 6 meters, which corresponds to choice B.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation