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nexis

  • 3 years ago

In the lecture "Introduction to related rates" ( http://goo.gl/g2iXs), the rate of change of D is defined as -80. Shouldn't x be changing at a rate of -80 as that is the line this cars is progressing along?

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  1. akash_809
    • 3 years ago
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    page not found

  2. nexis
    • 3 years ago
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    This should work

  3. amorfide
    • 3 years ago
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    try again @akash_809 wait a few seconds before clicking it

  4. akash_809
    • 3 years ago
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    @amorfide now its opening

  5. amorfide
    • 3 years ago
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    watching a 50 minute video brb

  6. akash_809
    • 3 years ago
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    @nexis u want us to look at the videos

  7. amorfide
    • 3 years ago
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    this is about to be the longest time i will ever spend on one question

  8. nexis
    • 3 years ago
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    No need to view the lecture lol. Just take a look at the transcribed pdf

  9. akash_809
    • 3 years ago
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    @nexis i think it is right, the radar is measuring the change of hypotenuseD wrt to time and that's what is being mentioned, and this is -80 and this has to be compared with the spped of the car i.e dx/dt which has to be less than 95 , and realtion between D and x is D^2=x^2+900....

  10. akash_809
    • 3 years ago
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    @nexis have i made things a bit clear, or u hv some some doubts

  11. nexis
    • 3 years ago
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    @akash_809 thank you for your help. I'm still failing to completely grasp why the rate of change of x and D would be the same here. I get the rest of the problem

  12. nexis
    • 3 years ago
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    OK I got the problem. But isn't the police radar underestimating your speed here?

  13. akash_809
    • 3 years ago
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    @nexis rate of change of x and rate of chage of d is not the same, but the concept here is while the refrence point from which police is measuring rate/velocity is not the same as refrence point used for measuring speed of driver and since only driver can control only his speed which is dx/dt he has to corellate it with police measurement i.e dD/dt\[D^2 =x^2+900\] differntiating it will give \[2D dD/dt= 2x dx/dt\] = \[D dD/dt=x dx/dt\] , now dD/dt =-80 , put values of x and D to get dx/dt and see if it is less than 95

  14. akash_809
    • 3 years ago
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    but pracitaclly speaking i dont think speed guns use this technique and they should be having the same refrence while measuring rates...... MIT teaching about obselete speed guns :)

  15. nexis
    • 3 years ago
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    yeah, thanks for your time @akash_809, i definitely got it now. :)

  16. akash_809
    • 3 years ago
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    no probelms got a question worthy of solving after a long time here..though joined only a week ago

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