{Use of Tech} Spring runoff The flow of a small stream is mo...

Question

{Use of Tech} Spring runoff The flow of a small stream is monitored for 90 days between May 1 and August 1. The total water that flows past a gauging station is given by v(t) = <matrix 2x2> where V is measured in cubic feet and t is measured in days, with t=0 corresponding to May 1.
c. Describe the flow of the stream over the 3-month period. Specifically, when is the flow rate a maximum?

c. Describe the flow of the stream over the 3-month period. Specifically, when is the flow rate a maximum?

Accepted Answer

Below there today we're going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A farmer is monitoring the growth of a crop over a 120 day period. The height of the crop in centimeters is given by the function H of T is equal to curly bracket of 9 divided by 5 T2 if 0 is less than or equal to T and T is less than 60. 9 divided by 5 multiplied by the power of 2 minus 240 T plus 120, if 60 is less than or equal to T and T is less than 120, where T is the time and days since initially planting their crops. When is the growth rate of the crop at a maximum? Awesome. So it appears for this particular problem, the final answer that we're ultimately trying to solve for is we're trying to figure out when the growth rate of this particular crop is at a maximum. So that is our final answer. So we're trying to figure out in what the amount of days. Is the crop growth rate going to be a maximum value? So with that said, first off, what we need to do in order to help us solve this problem is we need to first determine the derivative of the given growth function, meaning we need to find the derivative of H of T. So using our expression for the function that's given to us in the prim itself for H of T, we'll find that H of T. is going to be equal to curly bracket of, so when we take the derivative of 9 divided by 5 T 2, you'll get 18 divided by 5 T. if 0 is greater than or equal to or actually 0 is less than or equal to T and T is less than drum roll, if T is less than 60. Fantastic. So once again, We're taking the derivative of 9 divided by 5 T squared, and we do take the derivative, we'll get 18 divided by 5 T 0 is less than or equal to T and T is less than 60. And similarly, we will Find that are derivative of negative 95 or 9 divided by 5 multiplied by T2 minus 240 T plus 120. That will give us an answer of when we take the derivative, -9 divided by 5, multiplied by 2 T. -240, and don't forget that. We're gonna be dealing with the range of 60 is less than or equal to T and T is less than 120. Fantastic. So, now that we've found the derivative of the given growth function, our next step that we need to take is we need to draw the graph of the derivative of the growth function. So we need to note that for H. Of T, we need to know that this is going to be equal to 18 divided by 5 T. And the graph of the function will be a line, as this is a linear function. So now we need to find two points on the interval, 0 is less than or equal to T and T is less than 60. So when we do that, we will find that H of 0 is going to be equal to 18 divided by 5 multiplied by 0, because we're plugging in a value of 0 in for T, and when we do that, we will find that it's equal to 0. And for H of 30, it's going to be equal to 18 divided by 5, multiplied by 30 is going to be equal to 108. So now we need to mark our points, 0.0 and 30,108 on our graph, and then we need to join these two points to form a linear curve. And that will help us get the graph of H T is equal to 18 divided by 5 T. And we also need to make sure we keep the domain as the graph as the closed interval, which is square bracket 0.60 open interval, which is a parenthesis. Note once again, as we should recall, a parenthesis indicates an open variable, whereas a square bracket indicates a closed interval. So now we need to shift our attention to H of T. So H T is equal to -9 divided by 5, multiplied by 2 T minus 240. So now we need to graph this function as A line since it's due like we did for our previous derivative, this is also a linear function. So now we need to find two points on the interval, where 60 is less than or equal to T and T is less than 120. So now, let us say that H of 90 is going to be equal to -9 divided by 5 multiplied by 2 multiplied by 90 minus 240, which is going to be equal to 108. And let's say that H of 110. So once again, we're plugging in different values for T. So H of 110 is going to be equal to -9 divided by 5, multiplied by 2 multiplied by 110, minus 240 is going to be equal to 36. So now we need to mark the points 90,108 and 11036. On our graph, and you need to join these two points by connecting them with a like a linear curve. And that will help us produce our graph of the derivative H T is equal to -9 divided by 5 multiplied by 2 T minus 240, and we need to also make sure that we're keeping the domain of closed interval 60, 120 open variable. Fantastic. So with that said, when we officially graph all this information that we just found together, we will receive the following graph. So I've gone ahead and plugged it into some graphing software, our different derivative values, and as you can see, since we were given an empty graph earlier, we have our Y axis which denotes our H of T, and our X-axis denotes the time, T and days. And as you can see, we have our two linear curves, which are diagonal lines, and as you can see, they connect in a V shape, almost, almost like a V graph, so it's like a mountain peak. And it's an inverted V graph shape. And as you can see, we have our different points. We have our point at 0.0. We have our point at 30.108, and then we have our point at 90.108, and then we have our fourth and final point at 110.36. And as you can see, looking at our graph, We can see that the growth rate where the crop is a maximum, which the maximum will be the peak on our graph, which when we look at our peak, which I'll circle it in red, we can see that our peak, which will be our when our growth rate of our crop is a maximum, is going to be at T equals 60, it appears. So that's our final answer. So that means from our graph, the growth rate of the crop will be a maximum when T is equal to 60 days. And that is our final answer. Hooray, we did it. We found our final answer. So now, Just to quickly recap, so this problem is very easy to solve as long as we understand how to take derivatives, and then we also need to note how to focus our attention on the proper domains, because if it doesn't fit within the domain, then we won't get the correct answer. So once you have a strong grasp of those two concepts, this problem will be very easy to solve. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Key Concepts

Flow Rate

Integration

Critical Points

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