# Math Help - Road and Railroad Related Rates Problem

1. ## Road and Railroad Related Rates Problem

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2. ## Re: Road and Railroad Related Rates Problem

Originally Posted by Chaim

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You have the right idea, but when $t=0$ you train is negative 15 units from the crossing still. The way you set up the equations, the trains crosses when $t=10$.

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

P.S. I didn't check your algebra or calculus.

3. ## Re: Road and Railroad Related Rates Problem

Originally Posted by TheEmptySet
You have the right idea, but when $t=0$ you train is negative 15 units from the crossing still. The way you set up the equations, the trains crosses when $t=10$.

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

P.S. I didn't check your algebra or calculus.
So when t = 0, it is negative 15. Was that right when I did 90(t - 10/60).
Cause I'm kind of still confused how my answer is still pretty far off from the actual answer :O

4. ## Re: Road and Railroad Related Rates Problem

Originally Posted by Chaim
So when t = 0, it is negative 15. Was that right when I did 90(t - 10/60).
Cause I'm kind of still confused how my answer is still pretty far off from the actual answer :O
What you have is a right triangle

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

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

Then taking the derivative with respect to time gives

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

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

$2(15)\frac{dc}{dt}=2(12)(60)+2(9)(90) \iff 30\frac{dc}{dt}=3060 \iff \frac{dc}{dt}=102$

5. ## Re: Road and Railroad Related Rates Problem

Originally Posted by TheEmptySet
What you have is a right triangle

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

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

Then taking the derivative with respect to time gives

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

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

$2(15)\frac{dc}{dt}=2(12)(60)+2(9)(90) \iff 30\frac{dc}{dt}=3060 \iff \frac{dc}{dt}=102$
Thanks for replying and showing how you got the answer
But just wondering, how did you get 12 and 9 :O