Question

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 1

WELCOME TO OS!!!!

Answer 2

That is not a vaild answer

Answer 3

lol

Answer 4

its a regulation most OS'ers say this when someone new comes on xD

Answer 5

hmm..

Answer 6

@sammixboo

Answer 7

@sleepyjess

Answer 8

@DanJS

Answer 9

Who are they?

Answer 10

?

Answer 11

Let's draw a diagaram

Answer 12

We have ant 1 moving at a rate of 4 feet of every minute of time, hence the multiplication. The t+2 comes from the 2 minute headstart that ant 1 had.

Using pythagoras, we can get an expression for the distance between the two ants with respect to time.
\[distance = \sqrt{(4\times(t+2))^2+(5\times t)^2}\]
\[D (t) = \sqrt{(4t+8)^2+(5t)^2}\]
\[D(t) = \sqrt{14t^2+64t+64+25t^2}\]
\[D(t) = \sqrt{39t^2+64t+64}\]

Answer 13

The question asks for D'(t) after ant 1 (represented by y) has travelled 12 feet. So let's let y = 12

y = 4(t+2)
12 = 4(t+2)
(12/4)-2 = t
t = 1

So the question is asking for D'(1). I'm assuming you know how to derive this so:
\[D'(x) = \frac{ 78t+64 }{ \sqrt{39t^2+64t+64} }\]

Now we just have to substitute it in. Yay!
\[D'(1) = \frac{ 78 \times 1+64 }{ \sqrt{39\times1^2+64\times1+64} }\]
\[D'(1) = 10.99feet/s\]

Hope this helps