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?

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thaovtp1407 thaovtp1407 · Answer 1 · 2020-04-25T06:13:41+02:00

Answer:

1/6

Step-by-step explanation:

Given:

=> The volume of the trough: V =[tex]\frac{9}{2}[/tex] * (b * h) (1)

=> the other two sides of the triangle is:

= tan(45 degrees) = h/(b/2)

<=> b = 2h substitute in (1), we have:

V = [tex]\frac{9}{2}[/tex] *(2h *h) = 9[tex]h^{2}[/tex]

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[tex]h^{2}[/tex]

<=> 9[tex]h^{2}[/tex] = 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