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Calculus Related Rates Problems

  1. Oct 31, 2011 #1
    A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.1 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall?
     
  2. jcsd
  3. Oct 31, 2011 #2

    Dick

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    You can't just ask a question here. You need to show an attempt. Label some variables and try it.
     
  4. Oct 31, 2011 #3
    dx/dt=1.1 ft/sec
    x=8ft
    cos(theta)=x/10
    -sin(theta)*dtheta/dt=1/10 * dx/dt
    -sin(.644) * dtheta/dt=1/10 * 1.1
    dtheta/dt=-.183

    I'm not getting the correct answer. I don't know what I'm doing wrong. Help!
     
  5. Nov 1, 2011 #4

    Mark44

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    Did you draw a picture of the situation?
    What does x represent? You have that x = 8ft. Does this mean that x is always 8 ft or just at a particular moment?
     
    Last edited: Nov 1, 2011
  6. Nov 1, 2011 #5

    Dick

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    That actually looks pretty good unless I am missing something too. There is an easier and more accurate way to find sin(theta) than the way you did it, though. Do you know what the answer you are supposed to get is?
     
    Last edited: Nov 1, 2011
  7. Nov 1, 2011 #6
    Mark44- x represents the length of the ground. I think it means that x=8 at that particular moment

    Dick- What is the more accurate way to find sin(theta)? I don't know what the correct answer is supposed to be. It's an online problem so I would assume it needs to be as specific as possible. My previous answers include: -.183 , -.2 , and -.1375 all in rad/s. Would it be possible for the answer to be positive?
     
  8. Nov 1, 2011 #7

    Dick

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    You found theta by using cos(theta)=8/10 since the adjacent side is 8 and the hypotenuse is 10. What's the opposite side? Use it to find sin(theta) directly.
     
  9. Nov 1, 2011 #8
    I did what you said and got the answer of -.183, which I already tried and it was wrong. I tried .183 and it is still incorrect. I don't know what I'm doing wrong. Do you have any suggestions?
     
  10. Nov 1, 2011 #9

    Dick

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    I would say your answer SHOULD be correct. My only other suggestion is that you can write sin(theta)=6/10 exactly. Do you see why?
     
  11. Nov 1, 2011 #10
    I tried that and it made no difference. I am going to approach my Professor about it and hopefully she will help me.
     
  12. Nov 1, 2011 #11

    Mark44

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    Maybe they're looking for an answer in degrees/sec.
     
  13. Nov 1, 2011 #12

    blB

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    I think you should try leaving the base of the triangle as 1.1t and not replace it with 8 until after you have derived.

    So you should first get a function of θ(t), then derive and then replace time t with the time that the ladder would be 8 feet from the ground.
     
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