When the height of the water is y, then the radius of the surface is x=log_2(y+1)
Thus, the volume of the water when the depth is h is
∫[0,h] πx^2 dy
Now,
∫π(log_2(y+1))^2 dy
= π(y+1)(ln^2(y+1)-2ln(y+1)+2)/ln^2(2)
That gives me
∫[0,3]π(log_2(y+1))^2 dy = 30.06
That's somewhat different from your answer.
Anyway, we have v as a function of h.
v = F(h), so dv/dt = f(h)
Then we can say
dv/dt = f(h) dh/dt
-4 = f(1) dh/dt
So, plug in your expression for f(h). In this case, it is (I assert) π(log_2(h+1))^2
A water tank is made by rotating f(x)=2^x-1 between [0,2] about the y-axis. The water tank is initially full, when a hole is opened at the bottom tip so that the water drains at a rate of 4 unit^3/second. How fast is the height of the water decreasing when the height has decreased to 1? I did the integration and I got that the volume was 46.35, but I'm not sure what to do after that.
1 answer
1 answer