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This should work
try again @akash_809 wait a few seconds before clicking it
@amorfide now its opening
watching a 50 minute video brb
@nexis u want us to look at the videos
you can just look at the pdf's http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-optimization-related-rates-and-newtons-method/session-31-related-rates/MIT18_01SCF10_Ses31a.pdf
this is about to be the longest time i will ever spend on one question
No need to view the lecture lol. Just take a look at the transcribed pdf
@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....
@nexis have i made things a bit clear, or u hv some some doubts
@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
OK I got the problem. But isn't the police radar underestimating your speed here?
@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
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 :)
yeah, thanks for your time @akash_809, i definitely got it now. :)
no probelms got a question worthy of solving after a long time here..though joined only a week ago