Asked by Henry
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.
Answers
Answered by
Steve
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
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
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