@katiepemberton1,
Related rates were the most difficult part of Calc A for me...
On the first problem you are right about the answer being too big.
Good intuition!
katiepemberton1 wrote:
1.) The volume of a cube decreases at the rate of .5ft^3/min. What is the rate of change of the side length when the lengths are 12 ft.?
How I did this was:
I know: ds/dt=.5ft^3/1 min <-- should be: dV/dt = .5
V=s^3 dV/dt=3s^2 <-- the complete formula is dV/dt = 3s^2 * ds/dt
(Remember to use the chain rule, since you're differentiating s^3 with respect to t, not with respect to s.
dv/dt x .5ft^3/1min giving me 216ft^3/min <-- see below for correction
That seems too high...did I do it right? <-- it is too high
So the first thing to do is to derive the related rates formula, which is (
dV/dt) = 3
s^2 * (
ds/dt).
You were more or less correct about that formula, but not quite.
Next, what information are you looking for? You're looking for the rate at which side length changes, so solve for ds/dt:
(
ds/dt) = (
dV/dt) / 3
s^2
You know the values of
dV/dt (which is .5) and
s (which is 12), so
(
ds/dt) = .5 / 3(12)^2 = .00115740740...
(That may seem too small, but it's correct.)
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