Does anyone know how to do this!!!!
1 answer:
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Answer:
1/6
Step-by-step explanation:
Given:
- Length of the trough: 9 ft
=> The volume of the trough: V = * (b * h) (1)
- An isosceles right triangle with hypotenuse 2 feet
=> the other two sides of the triangle is:
= tan(45 degrees) = h/(b/2)
<=> b = 2h substitute in (1), we have:
V = *(2h *h) = 9
Take derivative of volume with respect to time to find equation for rate of filling the trough
dV/dt = 2 * 9 *h dh/dt = 18h dh/dt
<=> dh/dt = dV/dt /(18h)
As we know that, dV/dt = 2
So, dh/dt = 2 / 18h = 1/9h
<=> V = t * rate = 2 * 2 = 4
But V = 9
<=> 9 = 4
<=> h = 2/3
The rate is the height h feet of the water in the trough changing 2 minutes after the water begins to flow:
dh/dt = 1/(9h) = 1/(9 * 2/3) = 1/6