April 8th, 2018, 10:54 PM  #1 
Newbie Joined: Feb 2018 From: Burnaby Posts: 5 Thanks: 0  Chain rule
Car A is traveling north on Highway 152 and car B is traveling west on Highway 251. Each car is approaching the intersection of these two highways. At a certain moment, car A is 0.7 km from the intersection and traveling 95 km/h while car B is 0.7 km from the intersection and traveling at 80 km/h. How fast is the distance between the cars changing at that moment? Please could i get help on this, I don't know what I'm doing wrong i got an answer of 102.9kmph apparently somehow i must use partial derivatives Last edited by moody; April 8th, 2018 at 11:31 PM. 
April 9th, 2018, 03:31 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
102.9 is the correct answer! Why would you think you are doing this wrong?

April 12th, 2018, 03:03 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Since moody has not responded You don't use "partial derivatives" since there is only one independent variable, the time, t. But you do use the "chain rule". Let x be the distance from the intersection to car A and let y be the distance from the intersection to car B. Then the straight line distance between cars, z, satisfies $\displaystyle z^2= x^2+ y^2$. Differentiating but sides by t (using the chain rule), $\displaystyle 2d\frac{dz}{dt}= 2x\frac{dx}{dt}+ 2y\frac{dy}{dt}$. The "2"s cancel to give $\displaystyle d\frac{dz}{dt}= x\frac{dx}{dt}+ y\frac{dy}{dt}$. You are told the values for x, y, [/math]\frac{dx}{dt}[/math], and $\displaystyle \frac{dy}{dt}$.


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