The number of gallons of water in a tank t minutes after the t...

Question

The number of gallons of water in a tank t minutes after the tank has started to drain is Q(t) = 200(30 - t)². How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min?

Accepted Answer

Welcome back, everyone. In this problem, a large container initially holds 500 kg of a chemical solution. The mass of the solution is described by the function M of T equals 500 multiplied by 1 minus 0.05 T cubed for a T greater than or equal to 0 but less than or equal to 20, where MRT is the mass of the solution in kilograms and T is the time in hours. Determine the rate at which the mass of the solution is decreasing at T equals 6 hours and compute the average rate of change of the mass of the solution during the 1st 4 hours. Now this problem has two parts, so let's tackle them one by one. First, let's determine the rate at which the mass of the solution is decreasing at T equals 6 hours. To find the rate of change, we need to differentiate M with respect to T and then figure out what that rate is when T equals 6. Now, notice our function MFT is a composite function, so to differentiate it, we can use the chain rule. So now by the chain rule of differentiation. OK. Then the derivative of MFT. is going to be equal to first, we write the constant 500 multiplied by our power, we bring down our power multiplied by 1 minus 0.05 T squared, OK, and we subtract one from the power. In other words, we've differentiated the alter function. Multiplied by the derivative of the inner function, the derivative of 1 minus 0.05 T is negative 0.05. Now, when we simplify here, 500 multiplied by 3 multiplied by negative 0.05 is -75. So the derivative of M of T is -75 multiplied by 1 minus 0.05 T squared. Now that we have the derivative, we can figure it out at T equals 6 because now. To find it then we're gonna substitute 6 into our derivative. So it's M 6 equals -75 multiplied by 1 minus 0.05 multiplied by 6, OK. All squared Now, 0.05 multiplied by 6 equals 0.3, OK. 1 minus 0.3 equals 0.7, and negative 75 multiplied by 0.7 squared is going to be equal to essentially -147 divided by 4 or -36.75, OK? So this is the value of, this is our rate of change at T equals 60s, OK? In other words. That's right. I thought here, therefore. OK, at T equals 6 hours. The mass of the solution. The mass of the solution. I decreasing, OK. It's decreasing at a rate. Of 36.75 kg per hour, because remember it's negative, so that tells us our rate is decreasing, OK. Now let's move on to the second part of our problem. No, we want to compute the average rate of change of the mass of the solution during the 1st 4 hours. No, how are we going to find the average rate of change? Well, we know that the average rate of change, OK, so the average rate of change. We're looking at it for the 1st 4 hours, so we're gonna find the difference between our starting and ending point. So that's M 4 minus M of 0. OK. That is the value of our, of our mass, OK, divided by or change over time. So in this case, that's gonna be 40 minus 0. So it's gonna be M 4 minus M of 0 divided by 4. Before we start, we need to find MF 4 and MO 0. And to find out, all we need to do is to plug 4 and 0 into our function M. Now we know that M of 0. It's gonna be equal to 500. Multiplied by 1 minus 0.05, multiplied by 0 cubed, OK? And that's gonna be 500 multiplied by 1 cubed, which equals 500 kg, OK? Next, we know that Emma Ford. It's gonna be equal to 500 multiplied by 1 minus 0.05 multiplied by 4, OK, cubed. 0.05 multiplied by 4 equals 0.02 0.2 and 1 minus 0.2 0.8. So this is gonna be 500 multiplied by 0.8 cubed, which is gonna be equal to 256 kg. Know that we have M of 0 and MF 4, then that means our average rate of change. It's gonna be equal to 5 256 kg. Minus 500 kg divided by 4 hours, OK. I know when we were that all, that's -244 divided by 4 or -61 kg per hour, OK? Thus, our average, to summarize, sorry. Or Let's let me put this together. The mass of the solution is decreasing at a rate of 36.75 kg per hour at T equals 6 hours, and its average rate of change during the 1st 4 hours is -61 kg per hour. Thanks a lot for watching everyone. I hope this video helped.

The number of gallons of water in a tank t minutes after the tank has started to drain is Q(t) = 200(30 - t)². How fast is the water running out at the end of 10 min? What is the average rate at which the water flows out during the first 10 min?

Key Concepts

Derivative

Average Rate of Change

Quadratic Function

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