1. ## chain rule. need an idea to start answering

hi there.. can someone just give me an idea to solve this problem?

Two 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?

2. Think of ol' Pythagoras.

The distance between the cars at the time they are .5 and .7 from the intersection, respectively, can be found from

$\displaystyle D=\sqrt{.5^{2}+.7^{2}}=.86 \;\ km$

Just differentiate implicitly w.r.t time, $\displaystyle D^{2}=x^{2}+y^{2}$

$\displaystyle D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$

$\displaystyle \frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}} {D}$

Plug in your knowns on the right side (they are all given) and solve for dD/dt

3. Originally Posted by galactus
Think of ol' Pythagoras.

The distance between the cars at the time they are .5 and .7 from the intersection, respectively, can be found from

$\displaystyle D=\sqrt{.5^{2}+.7^{2}}=.86 \;\ km$

Just differentiate implicitly w.r.t time, $\displaystyle D^{2}=x^{2}+y^{2}$

$\displaystyle D\frac{dD}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$

$\displaystyle \frac{dD}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}} {D}$

Plug in your knowns on the right side (they are all given) and solve for dD/dt
ok got it.. my answer is 100km..
am i got it right?

4. That's what I got.

5. Originally Posted by galactus
That's what I got.
thanks galactus.. problem solved!

6. regarding the solution above, should i get the answer is -100km/h
should take in account the vector of this question?

7. because the distance between the cars getting closer, i think the minus sign is needed.