"A tank holds 1000L of water, which takes an hour to drain from the bottom of the tank, then the volume V of water remaining in the tank after t minutes is: V = 1000(1-(t/60))^2 where t is between 0 and 60 inclusive.
Find the rate at which the water is flowing out of the tank (the instantaneous rate of change of V with respect to t) after 10 min."
I found the volume at 10 min and at 0 min, then did the slope formula, but I'm not getting answer I'm supposed to. Am I reading the question incorrectly?
This is clearly a Calculus question.
You have taken the average rate of change from t=0 to t=10
It asked for the "instantaneous rate of change", which is the derivative dV/dt
So find dV/dt, then sub in t=10
Oh! I haven't been taught derivatives yet. Thank you for pointing that out. I'm getting ahead of myself here. :)
Your approach seems correct, but let's go through the problem step by step to ensure we reach the correct answer.
The volume of water remaining in the tank after t minutes is given by the equation:
V = 1000(1 - (t/60))^2
To find the rate at which the water is flowing out of the tank after 10 minutes (the instantaneous rate of change of V with respect to t), we need to calculate the derivative of V with respect to t at t = 10.
Let's differentiate the equation V = 1000(1 - (t/60))^2 with respect to t:
dV/dt = d/dt[1000(1 - (t/60))^2]
= 1000 * 2 * (1 - (t/60)) * (-1/60)
= -2000 * (1 - (t/60)) / 60
Now, we can substitute t = 10 into this derivative expression:
dV/dt at t = 10 = (-2000 * (1 - (10/60)) / 60
= (-2000 * (1 - 1/6)) / 60
= (-2000 * (5/6)) / 60
= (-5/6) * (2000/60)
= (-5/6) * (100/3)
= -500/3
Therefore, the rate at which the water is flowing out of the tank after 10 minutes is approximately -166.67 L/min (since we have negative flow rate, we consider it as flowing out of the tank).
I hope this clarifies your confusion. Let me know if you need any further assistance!
To find the rate at which the water is flowing out of the tank after 10 minutes, we need to calculate the derivative of the volume function V = 1000(1- (t/60))^2 with respect to time.
The first step is to differentiate the function with respect to t. In this case, we can use the chain rule to differentiate the function.
Applying the chain rule, we differentiate the outer function (1000(1- (t/60))^2) with respect to the inner function (1- (t/60)). The derivative of (1- (t/60))^2 with respect to (1- (t/60)) is 2(1- (t/60))^(2-1) multiplied by the derivative of (1- (t/60)).
Differentiating (1- (t/60)), we get -1/60.
So, the derivative of V with respect to t is:
dV/dt = 2(1- (t/60))^2 * (-1/60)
Next, we substitute t = 10 into the derivative function to find the instantaneous rate of change at that specific time, after 10 minutes.
dV/dt = 2(1- (10/60))^2 * (-1/60)
= 2(1- 1/6)^2 * (-1/60)
= 2(5/6)^2 * (-1/60)
= 2(25/36) * (-1/60)
= -50/1800
Simplifying the expression, we have:
dV/dt = -1/36
So, the rate at which the water is flowing out of the tank after 10 minutes is -1/36 L per minute.
It's possible that your calculation or interpretation of the formula might have led to a different result. Double-check your work to ensure you've correctly applied the chain rule and substituted the value of t into the derivative function.