Partial Derivative: Chain Rule

  • Thread starter iamyes
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  • #1
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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.
 

Answers and Replies

  • #2
arildno
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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?
 
  • #3
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so..
2) B(t) = (.7 - 80*t , 0)

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

i cant figure out this one...--->> 4)You are asked for the rate of change for D(t).
How can you find that?
 
  • #4
arildno
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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.
 
  • #5
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if.. i do it my way, i got the final answer is -100km/h..
is that same as yours?
 
  • #6
arildno
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if.. i do it my way, i got the final answer is -100km/h..
is that same as yours?
Post your way, please.
 
  • #7
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i got dX/dt = -100km/h
 
  • #8
arildno
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That is indeed correct! :smile:
 
  • #9
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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
arildno
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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
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so..
2) B(t) = (.7 - 80*t , 0)

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

i cant figure out this one...--->> 4)You are asked for the rate of change for D(t).
How can you find that?
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
arildno
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Quite so.
That function is equal to a constant, its derivative is..0
 
  • #13
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just for final confirmation...
my calculation and my answer for this question is totally correct
right?
 
  • #14
arildno
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Not if you specified D(t) in the manner YOU did, no.

The answer won't follow from that at all.
 
  • #15
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could anyone pls give me the solution?
damn im stuck and it is 3.30am over here
:(
 
  • #16
arildno
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could anyone pls give me the solution?
damn im stuck and it is 3.30am over here
:(
Then follow my hints in post 2.

They are sufficient.
 
  • #17
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yeah i read all this thread..but still cant do it (im stupid :( )
if anyone could go through everything once again i would be thankful
 
  • #18
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i cant do it :(
im stupid
and i dont know anything on how to do it
 

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