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A water trough is $ 10 m $ long and a cross-section has the shape of an isosceles trapezoid that is $ 30 cm $ wide at the bottom, $ 80 cm $ wide at the top, and has height $ 50 cm, $ If the trough is being filled with water at the rate $ 0.2 m^3/min, $ how fast is the water level rising when the water is $ 30 cm $ deep?

$\frac{1}{30} \mathrm{m} / \mathrm{min}$ or $\frac{10}{3} \mathrm{cm} / \mathrm{min}$

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Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

So what we see here is we have a water trough and our ultimate goal is to find the volume. Um Oh, are ultimately to find how fast the water level is rising. So we want to, um, refer to these values with different equations so we can take the derivative of them. So we know that the volume is going to be equal to the area of the cross section times the length of the trough. So if we have our trough right here has depth like this. And if this is the cross section right here because this goes all the way through and then we have the cross section times the length that will give us the volume. So we know the length of the trough is given as a constant of 10 m so we can differentiate both sides. Um, with respect to t, we get TV. D t is equal to l times. D a d t. So l is 10. So we could just write that is 10 times g eight et. Then we want to find the area of the cross section, which we know is a It's a nice sauces trapezoid. So if we look for the area of I saw Seles trapezoid. What we end up getting is the length of the upper base, plus the lower base divided by two times the height. So we know that the length of the upper base at any time is two K plus 30. Um and that's going to be also we know that the using triangles within our trapezoid. So if this is our trapezoid right here using this diagram where we have these small triangles with links K, this is 30 30 30. This is gonna be 25 25 thing we have right here the height and then right here is going to be 50. So based on all of this, we know that 25 over K is equal to 50 over H. So knowing that we see that H is equal to two K, um, and therefore, if we have two k 30 that's the same thing. A two K plus 30. We could rewrite that as a judge, plus 30. Then we know that the lower bases length 30. So we add that, and then that's gonna be over to times the height. So another way we could write this now is, um, going to be HCGH squared over to plus 30 h and this is helpful because we now have our area which we can take the derivative of and end up getting that d a d t is equal Thio h plus 30 times d h d t. Thankfully, though we consult itude this expression in for the previous equation because our goal is to find d h d t. So we have the DVD t is equal thio l times h plus 30 times d h d t. We know the values that we've been given our the volume So 0.2, uh, meters cubits per minute And then we also have, um since we're dealing with centimeters the way that we're gonna one right, this is 0.2 times 10 to the sixth centimeters, cubes per minute. And then we know that since the length is 10, the height is 30. So we have a length of 10 right here and a height of 30. So based on that, we now have that two times 10 to the fifth is equal to six times 10 to the fourth D h DT. We divide these and we end up getting that D h d t is going to give us, um, 13th centimeters per minute, so that will be our final result. And that's the rate at which the water level is increasing.

California Baptist University