Partial Derivative: Chain Rule

[iamyes]

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.

 

[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:
[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?

 

[naspek]

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]

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.

 

[naspek]

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

 

[arildno]

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

Post your way, please.

 

[naspek]

i got dX/dt = -100km/h

 

[arildno]

That is indeed correct!

 

[naspek]

That is indeed correct!

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

 

[arildno]

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.

 

[naspek]

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?

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]

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

 

[naspek]

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

 

[arildno]

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

The answer won't follow from that at all.

 

[iamyes]

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

 

[arildno]

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.

 

[iamyes]

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

 

[iamyes]

i can't do it :(
im stupid
and i don't know anything on how to do it