Find the time at which the depth of the water in the tank is eual to half its initial depth. D(t)=sqrt(100-2t)
the tank is empty when t=50 min
at t=0, D is sqrt100=10
so what time is D=5
5=sqrt(100-2t)
solve that for t.
So what is the depth when full, when t=0?
We want to make sqrt(100-2t) = half of that, which is 5.
sqrt(100-2t) = 5
Square both sides.
Subtract 25 from both sides.
Now t is easy.
To find the time at which the depth of the water in the tank is equal to half its initial depth, we can set up an equation using the given function D(t) = sqrt(100-2t) and solve for t.
Since the tank is empty when t = 50 minutes, we can substitute that value into the equation:
D(50) = sqrt(100 - 2(50))
D(50) = sqrt(100 - 100)
D(50) = sqrt(0)
D(50) = 0
This tells us that when the tank is empty at t = 50 minutes, the depth of the water is 0.
Now, we can set up the equation to find the time at which the depth of the water is equal to half its initial depth:
D(t) = sqrt(100-2t)
We know that the initial depth is at t = 0, so we substitute that into the equation:
D(0) = sqrt(100-2(0))
D(0) = sqrt(100-0)
D(0) = sqrt(100)
D(0) = 10
This tells us that the initial depth of the water in the tank is 10.
Now, we set up the equation for when the depth is half the initial depth:
D(t) = 10/2
sqrt(100-2t) = 5
Squaring both sides of the equation:
100-2t = 25
Simplifying the equation:
2t = 100-25
2t = 75
t = 75/2
t = 37.5
Therefore, the time at which the depth of the water in the tank is equal to half its initial depth is at t = 37.5 minutes.