Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

virtus

A 6m high vertical street lamp stands on horizontal ground. A 2 metre tall man runs away from the street lamp at a constant speed of 2.5m/s. When he is y metres from the street lamp his shadow has length x metres. Find the rate at which his shadow lengthens

  • one year ago
  • one year ago

  • This Question is Closed
  1. virtus
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1342236117226:dw|

    • one year ago
  2. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1342236213718:dw|

    • one year ago
  3. virtus
    Best Response
    You've already chosen the best response.
    Medals 0

    i don't understand?

    • one year ago
  4. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1342236435888:dw|

    • one year ago
  5. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1342236535853:dw|

    • one year ago
  6. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    keep the light as the origin is 0

    • one year ago
  7. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    and the ground as zero also

    • one year ago
  8. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    there fore slope for the hypotonuse form is dy/dx (6-2) = dy which is fixed i think so 4 4/(0-2*x)

    • one year ago
  9. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1342237451450:dw|

    • one year ago
  10. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1342237577634:dw|

    • one year ago
  11. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    sfx x = -3.75*x1 by x i mean x+y -3.75*x1=2.5x1+unkn unkn = -6.25*x1

    • one year ago
  12. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    |dw:1342238127714:dw| another way to look at it

    • one year ago
  13. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    where m1= -(6-4)/(2.5x) and m2= - 6/ (2.5x+b) set them equal to eachothter find b

    • one year ago
  14. timo86m
    Best Response
    You've already chosen the best response.
    Medals 3

    b=5x so s a b 1 2.5 5 2 5 10 3 7.5 15 It looks like everysecond the shadow gets 5 meters longer so it is a constant growth of 5 m/s i think

    • one year ago
  15. robtobey
    Best Response
    You've already chosen the best response.
    Medals 0

    From the figure,\[\frac{y+x}{6}=\frac{x}{2} \]The Total Derivative of the above equation is:\[\frac{1}{6} (dx+dy)=\frac{dx}{2} \]Solve for dx:\[dx=\frac{dy}{2} \]Replace dy with 2.5 and simplify.\[dx=1.25 m/\sec \]

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.