An ant on the corner of the desk heads west at 6steps per sec. An ant 60 ant steps below on the floor heads south at 9 ant per sec. After 20 sec, what is the rate of change of the distance between the two ants?
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Can you use calculus?
distance = sqrt(w^2 + s^2 + z^2)
z is up/down distance. w is distance west, s is distance south. By distance, I mean the distance between the two ants.
(d/dt)distance = (2w)(dw/dt)(2s)(ds/dt)(1/2)(w^2 + s^2 + z^2)^(-1/2)
We find this derivative using the chain rule. Remember that w and s are functions of t, while z is constant. If this is confusing to you, imagine that w = 1 + t. and just try to find (d/dt)(w^2 + c)^(1/2) for some constant c. You'll get to a similar answer as the one shown above for the time derivative of distance.
Now, we just need to find functions for w and s. (z=60) They're just rate*time, so
w = 6t
s = 9t
If we plug that into the above equation, you get
(d/dt)distance = (2(6t))*(6)*(2(9t))*(9)*(1/2)( (6t)^2 + (9t)^2 + 60^2)^(-1/2)
(d/dt)distance = 5832(t^2)*( 36t^2 + 81t^2 + 3600)^(-1/2)
I don't have a calculator on me, but it's just a matter of plugging in t at that point.
(thanks for the interesting problem!)
distance = sqrt(w^2 + s^2 + z^2)
z is up/down distance. w is distance west, s is distance south. By distance, I mean the distance between the two ants.
(d/dt)distance = (2w)(dw/dt)(2s)(ds/dt)(1/2)(w^2 + s^2 + z^2)^(-1/2)
We find this derivative using the chain rule. Remember that w and s are functions of t, while z is constant. If this is confusing to you, imagine that w = 1 + t. and just try to find (d/dt)(w^2 + c)^(1/2) for some constant c. You'll get to a similar answer as the one shown above for the time derivative of distance.
Now, we just need to find functions for w and s. (z=60) They're just rate*time, so
w = 6t
s = 9t
If we plug that into the above equation, you get
(d/dt)distance = (2(6t))*(6)*(2(9t))*(9)*(1/2)( (6t)^2 + (9t)^2 + 60^2)^(-1/2)
(d/dt)distance = 5832(t^2)*( 36t^2 + 81t^2 + 3600)^(-1/2)
I don't have a calculator on me, but it's just a matter of plugging in t at that point.
(thanks for the interesting problem!)