In Exercises 43 and 44, let S represent the pounds of salt in ...

Question

In Exercises 43 and 44, let S represent the pounds of salt in a tank at time t minutes. Set up a differential equation representing the given information and the rate at which S changes. Then solve for S and answer the particular questions.

Pure water flows into a tank at the rate of 4 gal/min, and the well-stirred mixture flows out of the tank at the rate of 5 gal/min. The tank initially holds 200 gal of solution containing 50 pounds of salt.

b. How many pounds of salt are in the tank after 1 minute? after 30 minutes?

Accepted Answer

Hello there. Today we are 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 reservoir initially contains 300 gallons of a saline solution with 75 pounds of dissolved salt. Clean water flows into the reservoir at a rate of 6 gallons per minute. While the mixed solution exits the reservoir at a rate of 8 gallons per minute through an outlet pipe. Let q of T be the pounds of salt in the reservoir at time t in minutes. How many pounds of salt are in the reservoir after 2 minutes and after 20 minutes? Round your answers to 2 decimal places. OK, so it appears for this particular prong we're asked to ultimately solve for 2 separate answers. Our first answer is we're asked to determine how many pounds of salt will be in the reservoir after 2 minutes. That's our first answer. And then we're asked to determine how many pounds of salt will be in the reservoir after 20 minutes. And we're then asked to round both of our final answers to 2 decimal places. OK. So now that we know what we're ultimately trying to solve for, our first step that we need to take in order to solve this problem is we need to define what our variables are. So once again, Q oft or function Q of T will be the pounds of salt in the reservoir at a time t, where t denotes once again our time in units of minutes. Fantastic. And we need to consider our initial salt amount in our in our reservoir, which the initial amount will be Q 0, and Q 0 will be equal to 75 pounds, which will denote as LB for short for pounds, and we need to note our initial volume, which we'll call V0, and V0, our initial volume, once again is 300 gallons. Fantastic. So, we need to consider that clean water flows in, and when clean water flows in, that means that no salt is added. And considering the outflow, the mixed solution will exit, which I'll put a capital E in quotes, so exit at. 8 gallons per minute. Fantastic. So, and I've denoted gallon as GAL for short. OK, so once again, our outflow for our mixed solution will be exiting at 8 gallons per minute. So, our next step now, step two, is we need to write the differential equation. So the volume at time T will be V of T and via T will be equal to 300+ entheses of 6 minus 8 multiplied by T, which will be equal to 300 minus 2 multiplied by T. So, we also need to note, based on what the prom tells us, the salt will leave our reservoir via the outflow at any time, the concentration of salt in our reservoir will be denoted as QT divided by V of T. So that will mean that the rate of the salt leaving our reservoir will be 8 multiplied by QT divided by VT in units of pounds per minute. Fantastic. So that will mean, therefore, the rate of change of the salt in our reservoir will be DQ by DT will be equal to the rate in minus the rate out, which will be equal to 0 minus 8 multiplied by QT divided by VT. Which will be equal to -8 multiplied by Q divided by 300 minus 2 multiplied by T. Fantastic. So our third step is we now need to solve for our differential equation, and we need to be able to recognize that this is a separable equation. So we can write that DQ divided by Q is equal to -8 multiplied by DT divided by 300 minus 2 multiplied by T. So now, at this point, we need to integrate both sides. So we need to take the integral of DQ divided by Q will be equal to -8 multiplied by the integral of DT divided by 300 minus 2 multiplied by T. So we need to use a substitution method. So we need to in order to solve this interval, so we need to state that u will be equal to 300 minus 2 multiplied by t, which will mean that duu will be equal to -2 dt, which will mean that dt will be equal to du divided by 2. So that will mean that we could write that the integral of DQ divided by Q will be equal to -8 multiplied by the integral of duu divided by 2 divided by u, which will be equal to 4 multiplied by the interval of duu divided by u. Which will be equal to 4 multiplied by the natural log of you, specifically 4 multiplied by the natural log of the absolute value of you. Which will be equal to 4 multiplied by the natural log of the absolute value of 300 minus 2 multiplied by T. And don't forget that 300 minus 2 multiplied by t is an absolute values signs. So that will mean therefore we could write that the natural log of the absolute value of q will be equal to 4 multiplied by the natural log of the absolute value of 300 minus 2 multiplied by t plus c where c denotes some constant variable. So now we need to exponentiate, so we need to exponentiate both sides. So we need to note that Q will be equal to a multiplied by 300 minus 2 multiplied by T, so the power of 4. And we need to note where A will be equal to E to the power of C. And we need to note at this point that we can safely drop the absolute value signs because due to the fact that senseq represents the amount of salt, and this must be a non-negative value. So, that will mean that also, since the reservoir volume is 300 minus 2 multiplied by T, the volume must remain positive for the problem to be physically meaningful. So that means for when T is less than 150, that will mean that 300 minus 2 multiplied by T will be positive. So I put a little positive symbol in quotes. So that will mean for our fourth step is we now need to apply our initial condition. So at T is equal to 0, that will mean that Q of 0 is equal to 75. So that will mean that we could write that 75 is equal to a multiplied by 300 to the power of 4. And when we rearrange this expression to isolate and so solve for a using a little bit of algebra, we will find that A is equal to 75 divided by 300 to the power of 4. Therefore, we could write that q of T. Is equal to 75 divided by 300 to the power of 4 multiplied by parentheses of 300 minus 2 multiplied by t in parentheses to the power of 4. We are almost done, I promise. So our fifth and final step is we need to compute the values at T is equal to 2 minutes and at T is equal to 20 minutes. So that means we need to evaluate Q2 and Q of 20. So q of 2, so once again q of 2, meaning we're plugging in a value of 2 n for t, will be equal to 75 divided by 300 to the power of 4 multiplied by parentheses of 300 minus 2 multiplied by 2 to the power of 4. And when we round to two decimal places, we will find that it's approximately equal to 71.08 pounds, which we could write as LB for short and similarly to save some time, I won't rewrite the whole expression out, but when you plug in a value of 20 in fort, we will find that q of 20 will be approximately equal to. 42.31 pounds when we round to two decimal places. And that's it. Those are our final two answers. Hooray, we did it. So going back to look at the prom itself, we could safely say that our final two answers without a doubt will be once again, so for our values for q of 2 and q of 20 will be once again q of 2. Is approximately equal to 71.08 pounds and Q of 20 will be approximately equal to 42.31 pounds. And that's it, those are our final two answers. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

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