Question

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?

Answer 1

page not found

Answer 2

This should work

Answer 3

try again @akash_809 wait a few seconds before clicking it

Answer 4

@amorfide now its opening

Answer 5

watching a 50 minute video brb

Answer 6

@nexis u want us to look at the videos

Answer 7

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

Answer 8

this is about to be the longest time i will ever spend on one question

Answer 9

No need to view the lecture lol. Just take a look at the transcribed pdf

Answer 10

@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....

Answer 11

@nexis have i made things a bit clear, or u hv some some doubts

Answer 12

@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

Answer 13

OK I got the problem. But isn't the police radar underestimating your speed here?

Answer 14

@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

Answer 15

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 :)

Answer 16

yeah, thanks for your time @akash_809, i definitely got it now. :)

Answer 17

no probelms got a question worthy of solving after a long time here..though joined only a week ago