The rate of flow of water into a dam is given by R=500+20t L h^-1. If there is 15 000L of water initially in the dam, how much water will there be in the dam after 10 hours?
Answer: 21 000L
If R' = 500 +20 t
then:
Rate of flow of water into the dam:
dR / dt = 500 + 20 t
dR = 500 dt + 20 t ∙ dt
Integrating both sides:
R = 500 t + 20 t² / 2 + C
R = 10 t² + 500 t + C
C = integration constant
Initial condition:
t = 0 , R = 15 000
R = 10 t² + 500 t + C
15 000 = 10 ∙ 0² + 500 ∙ 0 + C
15 000 = C
C = 15 000
R = 10 t² + 500 t + C
R = 10 t² + 500 t + 15 000
After 10 hours:
R = 10 ∙ 10² + 500 ∙ 10 + 15 000
R = 10 ∙ 100 + 5 000 + 15 000
R = 1 000 + 5 000 + 15 000
R = 21 000 L
Well, I must admit, this is a rather straightforward question. But don't worry, I'm here to make it entertaining!
Let's get down to business. The rate of flow into the dam is given by the equation R = 500 + 20t L h^-1, where t represents the time in hours (not to be confused with tea time).
Now, to figure out how much water will be in the dam after 10 hours, we need to integrate the rate of flow function. And for those who are not mathematically inclined, integrating is like mixing all the ingredients in a recipe. Just follow along!
Integrating R with respect to t gives us the expression ∫(500 + 20t) dt = 500t + 10t^2.
Since we want to know the amount of water after 10 hours, we substitute t with 10, which gives us 500(10) + 10(10)^2 = 5000 + 1000 = 6000.
Now, hold your laughter because we are not done yet. The question asks for the amount of water in liters, and we need to convert our result to liters. So, put on your conversion hat.
We know that 1 L = 1 dm^3. Therefore, 6000 dm^3 is equivalent to 6000 liters. Ta-da!
So, after 10 hours, there will be 6000 liters of water in the dam. Now, if you'll excuse me, I'm going to go take a dip in some water to cool off from all this math. Jaunty calculations, my friend!
To find out how much water will be in the dam after 10 hours, we need to integrate the rate of flow function R(t) with respect to time from 0 to 10 and then add the initial amount of water in the dam.
First, let's integrate the rate of flow function:
∫(500 + 20t) dt
Using the power rule of integration, we get:
= 500t + 10t^2/2 + C
Now we can substitute the limits of integration and evaluate the definite integral:
[500(10) + 10(10)^2/2 + C] - [500(0) + 10(0)^2/2 + C]
= 5000 + 500 - 0 - 0 + C - C
= 5500
So, the total amount of water that has flowed into the dam after 10 hours is 5500 L.
To find the total amount of water in the dam, we need to add the initial amount of water:
Total water in the dam after 10 hours = 5500 + 15000
= 21,000 L
Therefore, there will be 21,000 L of water in the dam after 10 hours.
To find out how much water will be in the dam after 10 hours, we need to first determine the rate of flow of water at that time and then calculate the total amount of water that has flown into the dam.
Given that the rate of flow of water into the dam is given by R = 500 + 20t L h^-1, we can substitute t = 10 into the equation to find the rate at 10 hours:
R = 500 + 20(10)
R = 500 + 200
R = 700 L h^-1
So, after 10 hours, the rate of flow of water into the dam is 700 L h^-1.
To calculate the total amount of water that has flown into the dam after 10 hours, we need to multiply the rate of flow by the time:
Total amount of water = Rate of flow * Time
Total amount of water = 700 L h^-1 * 10 h
Total amount of water = 7000 L
Finally, we add the initial amount of water to get the final amount of water in the dam:
Final amount of water = Initial amount of water + Total amount of water
Final amount of water = 15,000 L + 7,000 L
Final amount of water = 21,000 L
Therefore, after 10 hours, there will be 21,000 L of water in the dam.