# Partial Derivative: Chain Rule

1. Nov 25, 2009

### iamyes

1. The problem statement, all variables and given/known data

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?

3. 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.

2. Nov 25, 2009

### arildno

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:
$$A(t)=(0.5-60*t,0)$$
(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. Nov 25, 2009

### naspek

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. Nov 25, 2009

### arildno

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. Nov 25, 2009

### naspek

if.. i do it my way, i got the final answer is -100km/h..
is that same as yours?

6. Nov 25, 2009

### arildno

Post your way, please.

7. Nov 25, 2009

### naspek

i got dX/dt = -100km/h

8. Nov 25, 2009

### arildno

That is indeed correct!

9. Nov 25, 2009

### naspek

hurmm.. i'm still wondering why did my lecturer told me that there's something
wrong with my solution..

10. Nov 25, 2009

### arildno

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. Nov 25, 2009

### naspek

did u mean by the above D(t)

12. Nov 25, 2009

### arildno

Quite so.
That function is equal to a constant, its derivative is..0

13. Nov 25, 2009

### naspek

just for final confirmation...
my calculation and my answer for this question is totally correct
right?

14. Nov 25, 2009

### arildno

Not if you specified D(t) in the manner YOU did, no.

The answer won't follow from that at all.

15. Nov 25, 2009

### iamyes

could anyone pls give me the solution?
damn im stuck and it is 3.30am over here
:(

16. Nov 25, 2009

### arildno

Then follow my hints in post 2.

They are sufficient.

17. Nov 25, 2009

### iamyes

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. Nov 25, 2009

### iamyes

i cant do it :(
im stupid
and i dont know anything on how to do it

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