Question

A trough is 9 feet long, and its cross section is in the shape of an isosceles right triangle with hypotenuse 2 feet, as shown above. Water begins flowing into the empty trough at the rate of 2 cubic feet per minute. At what rate is the height h feet of the water in the trough changing 2 minutes after the water begins to flow?

Answer 1

Answer:

1/6

Step-by-step explanation:

Given:

• Length of the trough: 9 ft

=> The volume of the trough: V =$\frac{9}{2}$ * (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 = $\frac{9}{2}$  *(2h *h) = 9$h^{2}$  

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$h^{2}$    

<=> 9$h^{2}$     = 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