Question

SCALCET8 3.9.026.MI. A trough is 10 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 13 ft3/min, how fast is the water level rising when the water is 5 inches deep

Answer 1

Answer:

The answer is "0.624"

Step-by-step explanation:

$b = 5x\\\\h = x\\\\ l = 10$

Using formula:

$V = (\frac{1}{2})(b)(h)(l)$

   $= \frac{1}{2} \times (5x)\times (x) \times (10)\\\\= 5x\times x \times 5\\\\= 25x^2$

$\frac{dV}{dt} = 50x \ \frac{dx}{dt} \\\\\text{(where x represents the height in feet and}\ \frac{dv}{dt} = 13\ \frac{ft^3}{min})\\\\\frac{dx}{dt} = (\frac{1}{50})(\frac{1}{x}) \ \frac{dV}{dt}$

When the water is 5 inches deep then:

$x = (\frac{5}{12})\ ft\\\\13 =50(\frac{5}{12}) \ \frac{dx}{dt}\\\\13 \times 12 =50(5) \ \frac{dx}{dt}\\\\\frac{13 \times 12}{50 \times 5} = \frac{dx}{dt}\\\\\frac{156}{250} = \frac{dx}{dt}\\\\ \frac{dx}{dt}= 0.624$