# Partial Derivative: Chain Rule

## 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.

Related Calculus and Beyond Homework Help News on Phys.org
arildno
Homework Helper
Gold Member
Dearly Missed
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?

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?

arildno
Homework Helper
Gold Member
Dearly Missed
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?

arildno
Homework Helper
Gold Member
Dearly Missed
if.. i do it my way, i got the final answer is -100km/h..
is that same as yours?

i got dX/dt = -100km/h

arildno
Homework Helper
Gold Member
Dearly Missed
That is indeed correct!

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

arildno
Homework Helper
Gold Member
Dearly Missed
hurmm.. i'm still wondering why did my lecturer told me that there's something
wrong with my solution..
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.

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)

arildno
Homework Helper
Gold Member
Dearly Missed
Quite so.
That function is equal to a constant, its derivative is..0

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

arildno
Homework Helper
Gold Member
Dearly Missed
Not if you specified D(t) in the manner YOU did, no.

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

arildno
Homework Helper
Gold Member
Dearly Missed
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.

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

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