Confusing Related Rates Problem

In summary, the conversation is about solving a problem involving a spotlight shining on a building and a man walking towards it. Through the use of similar triangles and implicit differentiation, the rate of change of the length of the man's shadow is calculated to be -15/8 ft/sec when he is 12 feet away from the building. This is the correct answer.
  • #1
Caldus
106
0
On the final exam that I just took, the last problem went like this:

A spotlight is shining on a building. It is 36 feet away from the building. A man that is 6 feet tall is walking toward the building. Calculate the rate of change in the length of his shadow if he is walking toward the building at a rate of -5 and he is currently 12 feet away from the building.

I ended up with 100/3. Anywhere close? Thanks.
 
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  • #2
If he's walking towards the building, wouldn't the length of his shadow decrease? I'm getting -15/8 ft/s anyway (I'm assuming the spotlight is placed on the ground).
 
  • #3
This is how I did it:

I used this formula to relate x and y (if y is the length of the shadow):

(dx/dt = -5)

(36-x)^2 + y^2 = c^2
1296 - 72x + x^2 + y^2 = c^2

Derivative:

-72 + 2x dx/dt + 2y dy/dt = 0
-72 + 2(24) (-5) + 2(9) dy/dt = 0
-312 + 12dy/dt = 0
+312 +312
12dy/dt = 312
dy/dt = 26

LOL, now I am getting a different answer. I am a total failure at these kind of problems...*sigh*
 
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  • #4
Why in the world would you be using "(36-x)^2 + y^2 = c^2"
(Were you thinking the "shadow's length" is the straight line distance from the spotlight to the top point on the wall?)

this is simple "similar triangles". There is one triangle with base the line from the spotlight to the building (length 36 feet) , height, the shadow on the building (length y). The other triangle has base the line from the man's feet to the light (length x) and height the man himself (6 feet). The hypotenuse of both right triangles is the ray of light from the spotlight, past the tip of the man's head to the tip of the shadow. Because those are similar triangles, y/36= 6/x ("height over base of large triangle"= "height over base of small triangle"). You can differentiate that directly:
y'/36= -6/x<sup>2</sup> x' or by multiplying and using "implicit differentiation":
xy= 6*36 so x'y+ xy'= 0.

Using the first, y'/36= (-6/x<sup>2</sup>) x' with x'= -5 and x= 12,
y'= (36)(-6/144)(-5)= 7.5
Using the second, x'y+ xy'= 0 with x'= -5 and x= 12 so that y= 216/12= 18,
(-5)(18)+(12)y'= 0, y'= (5)(18)/12= 7.5.

The shadow's length is increasing by 7.5 ft/sec when the man is 12 feet from the spotlight.

Added in "Edit":
It occurs to me, after I wrote this, that the man's shadow extends from his feet to the wall and THEN up the wall so it's length, properly is
36- x (distance from the man to the wall) plus the y I had before.

That is: shadow's length L= 36-x+ 216/x
Then L'= -x'- (216/x2)x' and when x= 12 and x'= -5,
L'= 5- (216/144)(-5)= 5+ 7.5= 11.25 ft/sec.

That is, it the 7.5 ft/sec that his shadow is get higher on the wall plus the 5 ft/sec that he is moving away from the wall.

How many times do I have to edit this?

He is 12 feet away from the wall?? Okay then, my x is 36-12= 24 and y is
9. L'= (216/576)(5)= 1.875 ft/sec counting only height up the wall and
1.875+ 5= 6.875 counting changing length along the ground also.
 
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  • #5
The first thing that I notice is that you think the derivative of -72x with respect to t is -72. It's not, it's -72 * dx/dt.

HallsofIvy, you've taken x to mean "number of ft away from the spotlight", and you let it equal 12, i.e the man is 12 ft away from the spotlight. But the problem was to figure out the rate of change when the man was 12 ft away from the building, so you'd need to take x = 36 - 12 = 24 instead, or am I misunderstanding something?
 
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  • #6
The other triangle has base the line from the man's feet to the light (length x)...

Wouldn't the length be 36 - x since x describes the distance between the man and the building and not the distance betwee the man and the light?
 
  • #7
I got the same a muzza, and I used x as the distance from the wall.
 
  • #8
What is the right answer? I see a lot of different answers. I am still not sure about this.

I just realized something. If a light is being project on someone, then their shadow is going to be big if they are farther from that wall, right? So the rate of change of y will be a negative number then, right?
 
  • #9
ratio, this is a classic problem.
 
  • #10
CORRECT ANSWER : Shadow's length changes at rate -15/8 ft/sec.

Call x the dist from the spotlight.
Similar triangles gives y/36 = 6/x, where y is the shadow's height.
So, y=216/x .
Hence, dy/dt = -(216/x^2) dx/dt
Plug in x=24, and dx/dt = 5

ANSWER : dy/dt = -15/8 ft/sec
 

1. What are related rates problems?

Related rates problems are a type of calculus problem that involves finding the rate of change of one quantity in relation to another. These problems often involve multiple variables and require the use of derivatives to solve.

2. How do I approach solving a related rates problem?

When solving a related rates problem, start by identifying the variables involved and determining which quantities are changing with respect to time. Then, use the given information and the chain rule to set up an equation relating the rates of change of the different variables. Finally, solve for the desired rate of change using algebra and calculus.

3. Why are related rates problems considered confusing?

Related rates problems can be confusing because they require a strong understanding of calculus concepts and the ability to set up and manipulate equations. They also often involve real-world scenarios, which may add an extra layer of complexity.

4. What are some common mistakes to avoid when solving related rates problems?

Some common mistakes to avoid when solving related rates problems include misinterpreting the given information, not properly setting up the equation, and not taking into account the chain rule. It is important to carefully read and understand the problem and to double-check all steps in the solution.

5. Are there any tips or tricks for solving related rates problems?

One helpful tip for solving related rates problems is to draw a diagram or graph to visualize the problem and the relationships between the variables. It can also be useful to label the variables and their rates of change to keep track of them. Additionally, practice and familiarity with calculus concepts can help make solving these problems easier over time.

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