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- Nov 1st 2012, 03:46 PMChaimRoad and Railroad Related Rates Problem
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Click on the picture to enlarge it. - Nov 1st 2012, 04:42 PMTheEmptySetRe: Road and Railroad Related Rates Problem
You have the right idea, but when $\displaystyle t=0$ you train is negative 15 units from the crossing still. The way you set up the equations, the trains crosses when $\displaystyle t=10$.

So ten minutes later from then is when $\displaystyle t=20$ in your model

P.S. I didn't check your algebra or calculus. - Nov 1st 2012, 05:31 PMChaimRe: Road and Railroad Related Rates Problem
- Nov 1st 2012, 06:39 PMTheEmptySetRe: Road and Railroad Related Rates Problem
What you have is a right triangle

$\displaystyle c^2=a^2+b^2$

What you want to find is the rate of change of the hypotenuse.

Then taking the derivative with respect to time gives

$\displaystyle 2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$

So we need to know a,b and c.

The car has traveled 12 miles (a=12) and the train has traveled 9 miles (b=9) and by the Pythagorean theorem c=15.

Plugging all this in gives

$\displaystyle 2(15)\frac{dc}{dt}=2(12)(60)+2(9)(90) \iff 30\frac{dc}{dt}=3060 \iff \frac{dc}{dt}=102 $ - Nov 1st 2012, 06:48 PMChaimRe: Road and Railroad Related Rates Problem