Related rates and a lamp

In summary, a man 1.8 m tall is walking towards a building at a rate of 1.5 m/s from a lamp located 10 m away. The rate at which his shadow on the wall is shortening when he is 3.2 m from the building is 0.58 m/s. This is found by using the formula \frac{1}{10}\frac{dh}{dt}=\frac{-1.8}{x^2}\frac{dx}{dt} and setting x= 6.8 and \frac{dx}{dt}= 1.5 m/s. The man's shadow will become taller as he approaches the building, with his shadow being his height when he is right
  • #1
whkoh
29
0
A lamp is located on the ground 10 m from a building. A man 1.8 m tall walks from the light toward the building at a rate of 1.5 m s⁻¹. What is the rate at which the man's shadow on the wall is shortening when he is 3.2 m from the building ? Give your answer correct to two decimal places. [0.58 m/s]
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I tried using similar triangles first to find the height of the lamp but nothing came out of that.
http://img364.imageshack.us/img364/5294/manbuilding9vw.png [Broken]
let h be the height of the lamp
[tex]
{h \over{x+10}} = {1.8 \over {x+3.2}}[/tex]

[tex]
h = { {1.8x+18} \over {x + 3.2}}
[/tex]

How is the velocity of the man related to the height of the shadow? Or, what else should I find?

Also, shouldn't the shadow of the man on the wall become taller as he approaches the building?
 
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  • #2
What is x? The distance the man is from the wall or the distance the man is from the lamp?


First rule for a problem like this is to draw a picture and label it carefully. I took x as the distance the man is from the lamp, because that gave simpler distances. Drawing a line from the lamp to the top of the man's head to the wall (i.e. the light ray that gives the top of the shadow) I see two similar right triangles. the smaller has the man as opposite side, so height 1.7m and base x. The larger has the wall as opposite side. Taking h as the height of the shadow, the height of this right triangle is h and the base is 10m. Since they are similar triangles, [tex]\frac{h}{10}= \frac{1.8}{x}[/tex].

You appear to have tried the same thing but have "10+ x" and "3.2+ x". I see no reason to add the distance from the lamp to the wall, especially since the man is standing between the lamp and the wall. If your x is the distance from the man to the wall (rather than the lamp) then the formula would be [itex]\frac{h}{10}= \frac{1.8}{10-x}[/itex].

I finally got around to looking at your picture. You have the lamp on the opposite side of the wall from the man!? How is the light getting through the wall? Where is it casting the shadow?

Also, the "3.2 m" is (related to) the specific value of x at which you want to find the rate of change. It has nothing to do with the general formula connecting h and x.

Once you have a general "static" formula for two quantities, you can find a formula relating their "rates of change" by differentiating with respect to t (using the chain rule since there is no t in the formula itself).
With my formula [itex]\frac{h}{10}= \frac{1.8}{x}[/itex], the derivative of [itex]\frac{h}{10}[/itex] with respect to x is [itex]\frac{1}{10}\frac{dh}{dt}[/itex] and the derivative of [itex]\frac{1.8}{x}= 1.8x^{-1}[/itex] is [itex]-1.8x^{-2}\frac{dx}{dt}= \frac{-1.8}{x^2}\frac{dx}{dt}[/itex].
That is, from [itex]\frac{h}{10}= \frac{1.8}{x}[/itex] we get [itex]\frac{1}{10}\frac{dh}{dt}=\frac{-1.8}{x^2}\frac{dx}{dt}[/itex]. Now, set x= 10- 3.2= 6.8 (since my x is the distance from the lamp to the man, not from the wall to the man), set [itex]\frac{dx}{dt}= 1.5 m/s[/itex] and solve for [itex]\frac{dh}{dt}[\itex].

And, yes, the man's shadow gets smaller as he walks toward the wall. When he is right at the wall, his shadow is exactly his height. When he is very close to the lamp, he is blocking a lot of the light and his shadow is huge.
 
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  • #3
Many thanks for your help.

I assumed that the lamp had a height and that's why I got everything wrong. Now I understand...

Thanks!
 

What is a related rate and how does it relate to a lamp?

A related rate is a mathematical concept that involves finding the rate of change of one quantity with respect to another. In the context of a lamp, this could involve finding how fast the lamp's height is changing as the lamp's light intensity changes.

What are some real-life examples of related rates involving lamps?

One example is finding how fast the shadow of a lamp post is moving as the sun sets, or determining how quickly the light bulb in a lamp is heating up as it is turned on.

What are the steps for solving a related rate problem involving a lamp?

The first step is to identify the known and unknown quantities and their rates of change. Then, use the appropriate formula or equation to relate the two quantities. Next, take the derivative of the equation with respect to time and substitute in the given rates of change. Finally, solve for the unknown rate of change.

What are some common mistakes made when solving related rate problems involving lamps?

One mistake is not properly relating the known and unknown quantities in the problem. Another is not correctly taking the derivative with respect to time. It is also important to pay attention to units and ensure they are consistent throughout the problem.

How can understanding related rates and lamps be useful in scientific research?

Related rates can help scientists understand the relationships between different variables and how they change over time. In the case of lamps, it can aid in designing more efficient lighting systems or predicting the behavior of light in different environments.

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