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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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amorfideBest ResponseYou've already chosen the best response.0
try again @akash_809 wait a few seconds before clicking it
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
@amorfide now its opening
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
watching a 50 minute video brb
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
@nexis u want us to look at the videos
 one year ago

nexisBest ResponseYou've already chosen the best response.0
you can just look at the pdf's http://ocw.mit.edu/courses/mathematics/1801scsinglevariablecalculusfall2010/partboptimizationrelatedratesandnewtonsmethod/session31relatedrates/MIT18_01SCF10_Ses31a.pdf
 one year ago

amorfideBest ResponseYou've already chosen the best response.0
this is about to be the longest time i will ever spend on one question
 one year ago

nexisBest ResponseYou've already chosen the best response.0
No need to view the lecture lol. Just take a look at the transcribed pdf
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
@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....
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
@nexis have i made things a bit clear, or u hv some some doubts
 one year ago

nexisBest ResponseYou've already chosen the best response.0
@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
 one year ago

nexisBest ResponseYou've already chosen the best response.0
OK I got the problem. But isn't the police radar underestimating your speed here?
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
@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
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
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 :)
 one year ago

nexisBest ResponseYou've already chosen the best response.0
yeah, thanks for your time @akash_809, i definitely got it now. :)
 one year ago

akash_809Best ResponseYou've already chosen the best response.2
no probelms got a question worthy of solving after a long time here..though joined only a week ago
 one year ago
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