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

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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) About Me  (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:

r%5E2=x%5E2%2By%5E2
Taking the derivative of the equation with respect to time yields:
2r%28dr%2Fdt%29=2x%28dx%2Fdt%29%2B2y%28dy%2Fdt%29
Everything is divisible by 2 so divide the whole thing by 2 to make things easier.
highlight%28r%28dr%2Fdt%29=x%28dx%2Fdt%29%2By%28dy%2Fdt%29%29
Now use a little algebra to find the missing components.
Given:
dx%2Fdt=4the rate of the first ant
dy%2Fdt=5the rate of the second ant
x=12 the distance the first ant traveled.

distance = rate*time

We get t from the first ant.
12=4t
12%2F4=4t%2F4
3=t
Therefore the second ant traveled.
y=5%28t-2%29
y=5%28highlight%283%29-2%29
y=5

We use the Pythagorean Theorem to find r

r%5E2=x%5E2%2By%5E2
r%5E2=12%5E2%2B5%5E2
r%5E2=144%2B25
r%5E2=169
r=sqrt%28169%29
r=13
Take all the parts and put them into the equation you took the derivative of.
highlight%28r%28dr%2Fdt%29=x%28dx%2Fdt%29%2By%28dy%2Fdt%29%29
13%28dr%2Fdt%29=12%284%29%2B5%285%29
13%28dr%2Fdt%29=48%2B25
13%28dr%2Fdt%29=73
dr%2Fdt=73%2F13
dr%2Fdt=5.6ft/min
That should be it barring any careless mistakes. Happy Calculating!