Samantha's question at Yahoo Answers involving related rates

In summary, the problem involves designing a security camera with a variable rate of rotation and a constant rate of movement along a hallway. To accomplish this, we need to find the necessary rate of rotation, which can be solved by using the relationship between the angle of rotation and the position of the camera's beam on the hallway. The angle of rotation is confined to a specific interval, and by differentiating, we can find the necessary rate of rotation. This can be verified by using specific constant rates of movement. Using the given data, we can determine that the angle of rotation will vary between -π/4 and π/4, and the rate of rotation will be a function of the camera's distance from the center of the hallway.
  • #1
MarkFL
Gold Member
MHB
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Here is the question:

A security camera is centered 50 ft above a 100 foot hallway... Related Rates Question! HELP?

A security camera is centered 50 ft above a 100 foot hallway. It rotates back and forth to scan from one end of the hallway to the other. It is easiest to design the camera with a constant angular rate of rotation, but this results in a variable rate at which the images of the surveillance area are recorded. Therefore, it is desirable to design a system with a variable rate of rotation and a constant rate of movement of the scanning beam along the hallway. What rate of rotation is necessary to accomplish this? Verify you solution with some specific constant rates of movement.

Here is a link to the question:

A security camera is centered 50 ft above a 100 foot hallway... Related Rates Question! HELP? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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  • #2
Hello Samantha,

I would solve this problem in more general terms, derive a formula, then plug the given data in.

Suppose we let the camera be $h$ units above the hallway, and the hallway extends $L_1$ units to the left of the camera and $L_2$ units to the right. Let the angle $\theta$ of the camera's angle of rotation be zero radians directly below the camera.

Please refer to the following diagram:

View attachment 4087

Hence the angle of rotation will be confined to the interval:

$\displaystyle -\tan^{-1}\left(\frac{L_1}{h} \right)\le\theta\le\tan^{-1}\left(\frac{L_2}{h} \right)$

We can see that for any positive value of $h$ and any finite values of $L_1$ and $L_2$, we must have:

$\displaystyle -\frac{\pi}{2}<\theta<\frac{\pi}{2}$

Now, letting the position of the camera's beam on the floor of the hallway be $(x,0)$, we have the following relationship between this position and the angle of rotation:

$\displaystyle \tan(\theta)=\frac{x}{h}$

Differentiating with respect to time $t$, we find:

$\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{h}\frac{dx}{dt}$

We require $\displaystyle \left|\frac{dx}{dt} \right|=k$ where $0<k\in\mathbb{R}$. Hence:

$\displaystyle \frac{d\theta}{dt}=\frac{k}{h}\cos^2(\theta)=\frac{k}{h}\frac{h^2}{x^2+h^2}=\frac{kh}{x^2+h^2}$

Using the given data, we find the angle of rotation will vary from:

$\displaystyle -\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$

where:

$\displaystyle \left|\frac{d\theta}{dt} \right|=\frac{50k}{x^2+50^2}$

and:

$\displaystyle -50\le x\le50$
 

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  • #3
MarkFL said:
Hello Samantha,

I would solve this problem in more general terms, derive a formula, then plug the given data in.

Suppose we let the camera be $h$ units above the hallway, and the hallway extends $L_1$ units to the left of the camera and $L_2$ units to the right. Let the angle $\theta$ of the camera's angle of rotation be zero radians directly below the camera.

Please refer to the following diagram:
Hence the angle of rotation will be confined to the interval:

$\displaystyle -\tan^{-1}\left(\frac{L_1}{h} \right)\le\theta\le\tan^{-1}\left(\frac{L_2}{h} \right)$

We can see that for any positive value of $h$ and any finite values of $L_1$ and $L_2$, we must have:

$\displaystyle -\frac{\pi}{2}<\theta<\frac{\pi}{2}$

Now, letting the position of the camera's beam on the floor of the hallway be $(x,0)$, we have the following relationship between this position and the angle of rotation:

$\displaystyle \tan(\theta)=\frac{x}{h}$

Differentiating with respect to time $t$, we find:

$\displaystyle \sec^2(\theta)\frac{d\theta}{dt}=\frac{1}{h}\frac{dx}{dt}$

We require $\displaystyle \left|\frac{dx}{dt} \right|=k$ where $0<k\in\mathbb{R}$. Hence:

$\displaystyle \frac{d\theta}{dt}=\frac{k}{h}\cos^2(\theta)=\frac{k}{h}\frac{h^2}{x^2+h^2}=\frac{kh}{x^2+h^2}$

Using the given data, we find the angle of rotation will vary from:

$\displaystyle -\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$

where:

$\displaystyle \left|\frac{d\theta}{dt} \right|=\frac{50k}{x^2+50^2}$

and:

$\displaystyle -50\le x\le50$

Mark,How did you arrive at
$\displaystyle \frac{k}{h}\frac{h^2}{x^2+h^2}$
From $\displaystyle \frac{k}{h}\cos^2(\theta)$
 
  • #4
MSW said:
Mark,

How did you arrive at
$\displaystyle \frac{k}{h}\frac{h^2}{x^2+h^2}$
From $\displaystyle \frac{k}{h}\cos^2(\theta)$

If we have defined:

\(\displaystyle \tan(\theta)=\frac{x}{h}\)

This means in the right triangle the side opposite from $\theta$ is x and the side adjacent is $x$. By Pythagoras, the hypotenuse must then be \(\displaystyle \sqrt{x^2+h^2}.\)

As cosine is defined to be the ratio of adjacent to hypotenuse, we then find:

\(\displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{h}{\sqrt{x^2+h^2}}\)

Hence:

\(\displaystyle \cos^2(\theta)=\frac{h^2}{x^2+h^2}\)

Does this make sense?
 
  • #5
MarkFL said:
If we have defined:

\(\displaystyle \tan(\theta)=\frac{x}{h}\)

This means in the right triangle the side opposite from $\theta$ is x and the side adjacent is $x$. By Pythagoras, the hypotenuse must then be \(\displaystyle \sqrt{x^2+h^2}.\)

As cosine is defined to be the ratio of adjacent to hypotenuse, we then find:

\(\displaystyle \cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{h}{\sqrt{x^2+h^2}}\)

Hence:

\(\displaystyle \cos^2(\theta)=\frac{h^2}{x^2+h^2}\)

Does this make sense?

Perfect sense! Thank you!
 

FAQ: Samantha's question at Yahoo Answers involving related rates

What are related rates in mathematics?

Related rates is a concept in mathematics that involves finding the rate of change of one quantity with respect to another related quantity. It is often used in calculus to solve problems involving changing variables.

How do you approach a related rates problem?

To solve a related rates problem, you need to identify the variables involved and their rates of change. Then, you can use the chain rule and implicit differentiation to find the relationship between the variables and their rates of change. Finally, you can plug in the given values to solve for the unknown rate.

What are some real-life examples of related rates?

Real-life examples of related rates include the changing volume of a balloon as it is inflated, the rate at which the distance between two moving objects is changing, and the rate at which the water level in a tank is changing as it is being filled or drained.

Can you give a step-by-step example of solving a related rates problem?

Sure, let's say we have a triangle with a fixed base of 10 cm and an increasing height of 5 cm/s. To find the rate at which the area is changing, we can use the formula A = 1/2 * base * height. Taking the derivative with respect to time, we get dA/dt = 1/2 * 10 * dh/dt = 5 cm^2/s.

Are there any tips for solving related rates problems?

One tip for solving related rates problems is to carefully read the problem and draw a diagram to help visualize the situation. It is also important to clearly define the variables and their rates of change. Additionally, practicing with different types of related rates problems can help improve problem-solving skills.

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