A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

A 25 ft ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec. Determine how fast the top of the ladder is sliding when the bottom of the ladder is 15 ft from the wall <----- PLEASE EXPLAIN STEPS!

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1326763970660:dw| \[y=\sqrt{625-x ^{2}}\] \[dy/dt = (1/2)(-2x)/\sqrt{625-x ^{2}}(dx/dt)\] so at dx/dt =3 and x=15 dy/dt= -3/4 ft/s negative sign shows ladder is sliding downwards

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry multyply 3 also i forget to multiply 3...its -9/4 ft/s

  3. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.