SOLUTION: I need help on this calculus word problem: Two ants are at a common point at time t=0, the first ant starts crawling along a straight line at the rate of 4 ft/min. Two minutes l

Question 682777: I need help on this calculus word problem:
Two ants are at a common point at time t=0, the first ant starts crawling along a straight line at the rate of 4 ft/min. Two minutes later, the second ant starts crawling in a direction perpendicular to that of the first, at a rate of 5 ft/min. How fast is the distance between them changing when the first insect has traveled 12 feet?
Thank you!
Answer by funmath(2933) (Show Source): You can put this solution on YOUR website!
Two ants are at a common point at time t=0, the first ant starts crawling along a straight line at the rate of 4 ft/min. Two minutes later, the second ant starts crawling in a direction perpendicular to that of the first, at a rate of 5 ft/min. How fast is the distance between them changing when the first insect has traveled 12 feet?

Let r be the distance between them, x be the distance of the first and y be the distance of the second. Then according to the Pythagorean theorem:

Taking the derivative of the equation with respect to time yields:

Everything is divisible by 2 so divide the whole thing by 2 to make things easier.

Now use a little algebra to find the missing components.
Given:
the rate of the first ant
the rate of the second ant
the distance the first ant traveled.

distance = rate*time

We get t from the first ant.

Therefore the second ant traveled.

We use the Pythagorean Theorem to find r

Take all the parts and put them into the equation you took the derivative of.

ft/min
That should be it barring any careless mistakes. Happy Calculating!
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