# chain rule. need an idea to start answering

• Nov 23rd 2009, 02:10 AM
nameck
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?
• Nov 23rd 2009, 02:45 AM
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
• Nov 23rd 2009, 03:04 AM
nameck
Quote:

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?
• Nov 23rd 2009, 03:05 AM
galactus
That's what I got. (Hi)
• Nov 23rd 2009, 03:10 AM
nameck
Quote:

Originally Posted by galactus
That's what I got. (Hi)

thanks galactus.. problem solved! (Rofl)
• Nov 25th 2009, 05:05 AM
nameck
regarding the solution above, should i get the answer is -100km/h
should take in account the vector of this question?
• Nov 25th 2009, 05:46 AM
dedust
because the distance between the cars getting closer, i think the minus sign is needed.