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Confusing Related Rates Problem

  1. May 6, 2004 #1
    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.
  2. jcsd
  3. May 6, 2004 #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).
  4. May 6, 2004 #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


    -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*
    Last edited: May 6, 2004
  5. May 6, 2004 #4


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    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.
    Last edited: May 6, 2004
  6. May 6, 2004 #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?
    Last edited: May 6, 2004
  7. May 6, 2004 #6
    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?
  8. May 6, 2004 #7
    I got the same a muzza, and I used x as the distance from the wall.
  9. May 7, 2004 #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?
  10. May 7, 2004 #9
    ratio, this is a classic problem.
  11. May 8, 2004 #10


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    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
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