Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

hamidic

  • 3 years ago

a right angled triangle with length of the edges of the right angle "20" cm and"15"cm ;the first decrease by rate 2cm/sec . the second increase by 3cm/sec. find at any moment the rate of the change of the area and when the change stop

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

    \[\frac{dy}{dt} = 2 cm/\sec\] \[\frac{dx}{dt} = 3 cm/\sec\] \[determine \frac{dA}{dt}\]

  2. completeidiot
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sorry wrote that wrong, should be -2cm/sec

  3. hamidic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i need the area value at any instant in term of "t"

  4. completeidiot
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    A(t) = (1/2)(20-2t)(15+3t)

  5. completeidiot
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{d}{dt}A(t)= \frac{d}{dt}\frac{1}{2}*(20-2t)(15+3t)\]

  6. hamidic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    that right ; this problem was at my test and i want to check the second part "wiil the variation stop or not?'

  7. completeidiot
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it has to at t=10, because after this point, the area would be negative and negative areas dont exist (sort of)

  8. completeidiot
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i mean you might see negative areas when doing derivatives, but for this problem, i dont think negative areas count

  9. hamidic
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but area at any instant =0.5(20-2t)(15+3t) ;then area=\[-3t ^{2}+15t+150\] then ;the rate of change of area at any instant=-6t+15; at rate stops -6t+15=0;then the variation stops at (15/6) not at 10 ;if i have any mistake tell me please

  10. y2o2
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @completeidiot He asked when will the variation stop, so the idea is about letting the derivative(dA/dt) = 0 , not the the area [A(t)]= 0

  11. 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.