Partial Derivative: Chain Rule

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Homework Help Overview

The problem involves two cars approaching an intersection at right angles, with given speeds and distances from the intersection. The objective is to determine the rate at which the distance between the two cars is changing at a specific moment.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss setting up position functions for both cars based on their distances and speeds. There is exploration of how to derive a distance function between the two cars using the Pythagorean theorem. Questions arise regarding the correctness of the derived functions and the method to find the rate of change of distance.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify their approaches and questioning the validity of their calculations. Some have expressed confusion and seek further guidance, while others have provided hints and corrections to the setup. There is no explicit consensus on the correct method yet.

Contextual Notes

Some participants express frustration with their understanding and the time constraints they are under. There are indications of differing interpretations of the distance function and its derivative, leading to confusion about the correct procedure.

iamyes
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Homework Statement



2 straight roads intersect at right angles. Car A, moving on one of the roads, approaches the intersection at 60km/h and car B moving on the other road, approaches the intersection at 80km/h. At what rate is the distance between the cars changing when A is 0.5km from the intersection and B is 0.7km from the intersection?


The Attempt at a Solution



None. Dont know how. Dont give me the solution, give me a clue i want to try to solve it.
 
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1. Let the intersection be the origin in your coordinate system.

2. Now, let A's position at t (hours)=0 be 0.5 (km) on the right-hand side of the origin.
Then, A's position vector as a FUNCTION of time, A(t) will be:
[tex]A(t)=(0.5-60*t,0)[/tex]
(Agreed?)
Can you set up a similar vectorial position function for B?


3. Now that you have both position functions, how can you find the DISTANCE function, D(t) between them? (Hint: Pythagoras..)

4. You are asked for the rate of change for D(t).
How can you find that?
 
so..
2) B(t) = (.7 - 80*t , 0)

3) D(t) = (.5^2 + .7^2)^1/2

i can't figure out this one...--->> 4)You are asked for the rate of change for D(t).
How can you find that?
 
2. is wrong, it should be: B(t)=(0, .7-80t)
3. is doubly wrong, since it
a) only gives D(0), rather than D(t)
and
b) calculates it incorrectly from the A(t) and (incorrect) B(t) you set up.
 
if.. i do it my way, i got the final answer is -100km/h..
is that same as yours?
 
naspek said:
if.. i do it my way, i got the final answer is -100km/h..
is that same as yours?

Post your way, please.
 
i got dX/dt = -100km/h
 
That is indeed correct! :smile:
 
arildno said:
That is indeed correct! :smile:

hurmm.. I'm still wondering why did my lecturer told me that there's something
wrong with my solution..:frown:
 
  • #10
naspek said:
hurmm.. I'm still wondering why did my lecturer told me that there's something
wrong with my solution..:frown:

Well, if you used YOUR version of D(t), then it is a totally incorrect procedure.

The rate of change of THAT expression is just 0.
 
  • #11
naspek said:
so..
2) B(t) = (.7 - 80*t , 0)

3) D(t) = (.5^2 + .7^2)^1/2

i can't figure out this one...--->> 4)You are asked for the rate of change for D(t).
How can you find that?

arildno said:
Well, if you used YOUR version of D(t), then it is a totally incorrect procedure.

The rate of change of THAT expression is just 0.
did u mean by the above D(t)
 
  • #12
Quite so.
That function is equal to a constant, its derivative is..0
 
  • #13
just for final confirmation...
my calculation and my answer for this question is totally correct
right?
 
  • #14
Not if you specified D(t) in the manner YOU did, no.

The answer won't follow from that at all.
 
  • #15
could anyone pls give me the solution?
damn I am stuck and it is 3.30am over here
:(
 
  • #16
iamyes said:
could anyone pls give me the solution?
damn I am stuck and it is 3.30am over here
:(

Then follow my hints in post 2.

They are sufficient.
 
  • #17
yeah i read all this thread..but still can't do it (im stupid :( )
if anyone could go through everything once again i would be thankful
 
  • #18
i can't do it :(
im stupid
and i don't know anything on how to do it
 

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