WEBVTT
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in this exercise we're asked to consider a situation of
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dialysis. And we are given a function. You
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represents the amount of your area in a patient's blood
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in milligrams at a given time t where time is
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minute. Sorry. It took me a lot to
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verify. And the equation or in the problem but
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it says that. Uh huh. The rate is
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milligrams per minute. So time is in minutes.
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R. Is the rate of flow of blood through
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the dialogue. This machine. Mhm. Yeah.
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Oh mm B. Is volume of the patient's blood
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in milliliters and Cease of zero is equal to the
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initial concentration of Yuria and blood. So looking at
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this R. Is the rate of flow divided by
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the volume. So that gives me the amount of
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blood that can how much of the blood flows through
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the ratio of the rate to the volume of the
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patient's blood. So what percentage of the patient's blood
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or what ratio of the patient's blood is flowing through
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the machine each minute times the initial concentration times he
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raised to the power of negative are times time divided
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by volume. So um the more time that passes
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, the smaller this exponential. Um Yes exponential expression
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is the less the higher percentage or the less of
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the initial concentration remains. And then we are asked
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to evaluate Yeah you can't a girl. Mhm.
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From zero. Yeah. Uh huh. Yeah.
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Yeah. 30. I believe this big away the
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integral from time equals 02 tiny those 30 minutes of
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this dialysis dialysis from shen with respect to T.
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Yeah and explain what I mean. So this integral
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written in terms of the actual equation actually gets simplified
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a little bit because our and the and see subzero
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which is the initial concentration are all constants in this
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situation. The only thing that is changing is time
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, time is my variable of integration. And so
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all of that can fall outside. Uh huh.
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The actual integral. And then I'm evaluating the furniture
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of 2 30. The definite it's a girl E
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. To the power of negative R. T over
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B. With respect to T. Yeah. Yeah
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. Well and when I do that that is going
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to require us to do some substitution. So everybody
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to switch to just being able to draw wrong.
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Yeah. I'm gonna change my color to black.
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It's so the problem is I've got this weird funky
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exponents in my exponential situation. I know how to
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find the integral of E. To the X power
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dx is just itself. Right? The exponent here
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has more than just the variables. I am going
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to do a substitution. I'm going to call function
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you that whole exponent negative. RT divided by V
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. And in order to evaluate an integral I need
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to know what the derivative of that is. So
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do you is just going to be negative are over
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being DT to make the substitution a little easier because
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I have a D. T. Here. I
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am going to find out what D. T.
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Equals by rewriting that. Mhm. So if I
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divide both sides by negative are over B. I
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get the DT equals negative B. Over our.
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Do you? And now I am going to substitute
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negative Rt over be for you. And I am
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going to substitute. No. Yeah T. T
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. With negative V. Over R. D.
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Use put the DT here and change that. I'm
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going to put to you here. I'm gonna rewrite
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all of this. Mm. Yeah. So I
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still have the R. I still have to be
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I still need the initial concentration. I am still
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evaluating an integral. Yeah I'm going to come back
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to the limits of integration in just a second and
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I still have but instead of raising it to the
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power of negative RT over mean I'm going to raise
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it to the U. Power. And instead of
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D. T. I now have negative. This
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is multiplied by negative being over R. D.
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T. Yeah simplifying that a little bit negative.
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The over our is still a constant negative B.
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Over our times are over. B cancels each other
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out. So. Uh huh. Look what happens
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to all of that. So this are going to
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cancel how this are because our divided by R.
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Is one. This baby is going to cancel how
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the Mississippi because a divided by B. Is one
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the negative is a constant. So I can go
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outside and can rewrite all this as. Uh huh
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negative. I still have my initial concentration of the
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definite and to grow you know where that came from
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. Get rid of it. The definite integral of
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he to new power. Okay that should be D
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. You because I rewrote D. T. As
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negative V. Over R. D. Use this
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was negatively over hard to do instead of D.
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T. That makes this. Yeah. But then
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I've got these limits. I have changed my variable
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of integration from T. T. You. So
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I need to change the limits from T. T
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. You as well. If T equals zero you
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is going to equal negative. Are times T divided
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by a Z. Are divided by the negative are
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times zero is zero divided by via zero. So
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when T equals zero U. Equals zero. When
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T equals 30 you is going to equal negative are
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times 30 divided by the so my upper limit of
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integration is going negative. 30? Mm hmm.
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Over. Okay. Got Okay give us me definitely
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integral that I can that I can find the anti
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derivative for I know that the anti derivative of each
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of the U. D. U. Is just
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each of the U. Power. So this all
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becomes negative. Initial concentration keeps Sorry about that initial
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concentration times. Yeah. Uh huh. Yeah.
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To the youth power Evaluated over the integral from zero
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To-34. Right? Yes. And then evaluating
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that I still have initial consultation tom. Uh huh
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. Okay. Very quickly being as one else today
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, times Plugging the negative 30 are over be into
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. Each of the power gives me me to the
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negative. Uh huh. There are over B.
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Yeah Plugging zero into the eat the U. power
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anything to visit all. Power is one. I
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can multiply the initial concentration through. And given the
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initial concentration of syria and the blood minus. These
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are the power of the amount of time. 30
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minutes times are divided by the total volume of blood
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. And this will give me the total milligrams of
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syria that remain in the patient's blood. The initial
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amount of area and the patient's blood minus the amount
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of your area that has been removed. Which is
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based on the rate Of the patient's blood flowing through
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. As a ratio to the total amount of patients
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blood. Multiplied by the amount of time that has
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passed 30 minutes. Oh okay. Uh huh.