Question

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?

Answer 1

what formula can we use to determine the distance between them?

Answer 2

Pythagorean theorem

Answer 3

good, implicit that, what do we get?

Answer 4

d^2 = x^2 + y^2

2d d' = 2x x' + 2y y'

d d' = x x' + y y'

d' = x x' + y y'
--------
d

we are told x' and y', and the information is all given to determine x,y,d

Answer 5

dx/dt=4the rate of the first ant
dy/dt=5the rate of the second ant
x=12 the distance the first ant traveled.

Answer 6

distance = rate*time

Answer 7

good, and 3 minutes has passed to get to 12 feet

Answer 8

3 min - 2 min = 1 min @ 5ft/min, y = 5

Answer 9

For 1st Ant t=3

Answer 10

for 2nd t=5

Answer 11

t=3 regardless :)

Answer 12

if we assume the 2nd ant started at the same time, then it was simply at -10 and didnt get to the crossing yet

Answer 13

We use the Pythagorean Theorem to find r

Answer 14

yep

Answer 15

r=13

Answer 16

thanks for help

Answer 17

youre welcome