WEBVTT mathematics/ap-calculus-ab/hovasapian
00:00:00.000 --> 00:00:04.200
Hello, welcome back to www.educator.com, and welcome back to AP Calculus.
00:00:04.200 --> 00:00:12.000
Today, we are going to finish off our AP practice exam by doing the no calculator portion of the free response question.
00:00:12.000 --> 00:00:16.200
Let us jump right on in.
00:00:16.200 --> 00:00:23.100
Once again, you can find this particular test, this particular section, all of them, at the following.
00:00:23.100 --> 00:00:24.600
The links are down below in the quick notes.
00:00:24.600 --> 00:00:27.300
I will go ahead and write it here for you anyway.
00:00:27.300 --> 00:00:51.900
They are at www.online.math.uh.edu/apcalculus/exams.
00:00:51.900 --> 00:01:02.400
This is going to be section 2, written, and we are using version 2.
00:01:02.400 --> 00:01:07.300
If you pull it up, leave it on the screen or print it out, however you want to do it.
00:01:07.300 --> 00:01:08.200
Let us get started.
00:01:08.200 --> 00:01:11.200
This is going to be question number 4.
00:01:11.200 --> 00:01:14.500
That is fine, I will stick with black, it is not a problem.
00:01:14.500 --> 00:01:17.500
Yes, that is fine, I will stick with black, at least for now.
00:01:17.500 --> 00:01:20.200
I should change colors in between.
00:01:20.200 --> 00:01:25.900
Number 4, it asked us to consider the graph of the following functions.
00:01:25.900 --> 00:01:34.900
We have f(x) = 2x⁴ - 4x².
00:01:34.900 --> 00:01:40.900
For part A, they want us to find the relative maxima and minima.
00:01:40.900 --> 00:01:45.200
They want us to find both the x and y coordinates.
00:01:45.200 --> 00:01:46.400
Let us go ahead and start.
00:01:46.400 --> 00:01:47.900
Basically, nice and straightforward.
00:01:47.900 --> 00:01:51.800
You just take, for relative max and min, local max and min.
00:01:51.800 --> 00:01:53.300
You take the derivative, you set it equal to 0.
00:01:53.300 --> 00:01:55.700
You solve for the x values.
00:01:55.700 --> 00:02:07.400
f’(x) is going to equal 8x³ - 8x and we are going to set that equal to 0.
00:02:07.400 --> 00:02:16.100
I’m going to factor out an 8x and we are going to have x² - 1 = 0.
00:02:16.100 --> 00:02:21.500
Therefore, we have each of those factors equal to 0.
00:02:21.500 --> 00:02:24.200
I’m just going to go ahead and factor the whole thing.
00:02:24.200 --> 00:02:29.300
It is going to be x - 1 × x + 1 equal to 0.
00:02:29.300 --> 00:02:39.900
We are going to have x = 0 is one point, x = 1, and x = -1.
00:02:39.900 --> 00:02:48.000
We go ahead and we draw ourselves a little line, 0, 1, -1.
00:02:48.000 --> 00:02:53.100
We are going to pick points over here, over here, over here, and over here.
00:02:53.100 --> 00:02:58.500
We are going to put them into here to see whether we get something that is greater than 0 or less than 0.
00:02:58.500 --> 00:03:00.000
Greater than 0 means the function is increasing.
00:03:00.000 --> 00:03:06.600
Less than 0 means the function is actually decreasing because the derivative is the slope.
00:03:06.600 --> 00:03:09.900
Let us go ahead and put some values in.
00:03:09.900 --> 00:03:19.600
When we try -2, we are going to get a negative number × a negative number × a negative number.
00:03:19.600 --> 00:03:25.300
That is going to give me a negative number, it is going to be negative, it is going to be decreasing.
00:03:25.300 --> 00:03:33.700
Over here, if I try, let us say -0.5, it is going to be a negative number × a positive number ×,
00:03:33.700 --> 00:03:34.900
Let me write this out.
00:03:34.900 --> 00:03:38.200
For -2, it is going to be negative × a negative × a negative.
00:03:38.200 --> 00:03:46.900
For 0.5, you are going to end up with negative, positive.
00:03:46.900 --> 00:03:55.300
I’m sorry, negative, -0.5 - 1 is negative and this is going to be positive.
00:03:55.300 --> 00:03:58.600
We are going to end up with positive, this is going to be increasing.
00:03:58.600 --> 00:04:07.800
If I try maybe 0.5, this is going to be positive, 0.5 – 1 this is going to be negative, 0.5 this is going to be positive.
00:04:07.800 --> 00:04:09.300
I will get a negative number here.
00:04:09.300 --> 00:04:10.800
This is going to be decreasing.
00:04:10.800 --> 00:04:17.400
If I try 2, it is going to be positive × positive × positive.
00:04:17.400 --> 00:04:19.500
It is going to be increasing.
00:04:19.500 --> 00:04:22.500
Decreasing, let me go to blue now.
00:04:22.500 --> 00:04:26.100
Decreasing, increasing, this is a local min.
00:04:26.100 --> 00:04:31.200
Decreasing, increasing, this is a local min.
00:04:31.200 --> 00:04:39.600
Our local mins happened at -1 and 1.
00:04:39.600 --> 00:04:45.300
Relative min at x = -1 and 1.
00:04:45.300 --> 00:04:46.800
We have relative max.
00:04:46.800 --> 00:04:51.100
Over here because it is increasing and decreasing, that is a relative max.
00:04:51.100 --> 00:04:55.900
Relative max at x = 0.
00:04:55.900 --> 00:05:00.400
Now let us go ahead and find the actual y values of this because that is what we wanted.
00:05:00.400 --> 00:05:03.400
Again, to find the y value, you put it back into f.
00:05:03.400 --> 00:05:06.100
You put these x values back into f not into f’.
00:05:06.100 --> 00:05:10.300
Just be very careful which function you are using.
00:05:10.300 --> 00:05:16.300
f(-1) it is going to end up equaling -2.
00:05:16.300 --> 00:05:19.600
Therefore, we have -1, – 2.
00:05:19.600 --> 00:05:23.000
f(1) is going to equal -2.
00:05:23.000 --> 00:05:27.500
Therefore, we have 1, –2.
00:05:27.500 --> 00:05:33.800
f(0), when I put it into the function, the original function, it is going to b 0.
00:05:33.800 --> 00:05:36.500
I end up with 0 and 0.
00:05:36.500 --> 00:05:48.000
My relative mins are going to be -1, -2 and 1,-2.
00:05:48.000 --> 00:05:53.100
My relative max is going to be 0,0.
00:05:53.100 --> 00:05:57.900
That is it, nice and straightforward.
00:05:57.900 --> 00:06:03.900
Now we take a look at part B, part B asks us to find the coordinates of the points of inflection.
00:06:03.900 --> 00:06:06.900
Again, the xy coordinates of the points of inflection.
00:06:06.900 --> 00:06:12.300
Our point inflection is where the concavity changes from positive to negative or from negative to positive,
00:06:12.300 --> 00:06:20.100
which means we are going to now look at the second derivative and set that equal to 0.
00:06:20.100 --> 00:06:28.600
We have f, the original f was 2x⁴ - 4x².
00:06:28.600 --> 00:06:36.000
f’ was equal to 8x³ - 8x.
00:06:36.000 --> 00:06:43.100
f” is equal to 24x² – 8.
00:06:43.100 --> 00:06:51.500
It is the second derivative that we set equal to 0, to find where the concavity changes.
00:06:51.500 --> 00:07:04.700
When I solve this, I end up with x² is equal to 1/3 which gives me x equal to + or -1/ √3.
00:07:04.700 --> 00:07:11.900
The points of inflection, the x values of the points of inflection are + and -1/ √3.
00:07:11.900 --> 00:07:17.400
We want the coordinates so let us go ahead and find the f values.
00:07:17.400 --> 00:07:19.100
Yes that is fine, I can do it over there.
00:07:19.100 --> 00:07:24.500
Let us go ahead and take f(1)/ √3.
00:07:24.500 --> 00:07:27.800
When I put 1/ √3 into this,
00:07:27.800 --> 00:07:32.400
I will just write it all out, that is not a problem.
00:07:32.400 --> 00:07:45.400
1/ √3⁴ - 4 × 1/ √3².
00:07:45.400 --> 00:07:48.700
Let me see, what do I get.
00:07:48.700 --> 00:07:51.400
I think I may have done the calculation wrong on my paper but that does not matter
00:07:51.400 --> 00:07:55.900
because this is the answer, the rest is just arithmetic.
00:07:55.900 --> 00:08:00.100
1/ √3 × 1/ √3 is 1/3.
00:08:00.100 --> 00:08:02.800
1/3 × 1/3 is 1/9.
00:08:02.800 --> 00:08:06.700
It looks like I did not do make arithmetic mistakes.
00:08:06.700 --> 00:08:10.600
2/9 - 4/3 = -10/9.
00:08:10.600 --> 00:08:16.000
That is the y valve for 1/ √3.
00:08:16.000 --> 00:08:26.600
When I do f(-1/ √3), I end up with - 10/9.
00:08:26.600 --> 00:08:47.300
The points of inflection are 1/ √3 – 10/9 and -1/ √3 – 10/9.
00:08:47.300 --> 00:08:51.300
That is it, nice and straightforward.
00:08:51.300 --> 00:08:56.200
Part C, they want us to find the intervals of increase and decrease for the function.
00:08:56.200 --> 00:09:02.800
We already did that when we took care of Part A, when we found the critical values, the max's and min’s.
00:09:02.800 --> 00:09:05.500
Let us recall what it is that we did.
00:09:05.500 --> 00:09:09.700
We had 0, we had 1, and we had -1.
00:09:09.700 --> 00:09:19.300
We ended up with a decreasing, increasing, decreasing, increasing.
00:09:19.300 --> 00:09:25.300
The intervals of increase turns the intervals on which the function is increasing.
00:09:25.300 --> 00:09:34.300
It is going to be from -1 to 0 union 1 to positive infinity.
00:09:34.300 --> 00:09:39.700
That is it, nothing strange, just read it straight off.
00:09:39.700 --> 00:09:42.400
The interval of the points of concavity.
00:09:42.400 --> 00:09:50.000
Part D, I do not need to write it down.
00:09:50.000 --> 00:09:58.400
I’m sorry, not the point of concavity, intervals of positive concavity.
00:09:58.400 --> 00:10:10.400
Positive concavity, we said that the points of inflection are -1/ √3 and positive.
00:10:10.400 --> 00:10:20.900
We said that we had -1/ √3 and 1/ √3.
00:10:20.900 --> 00:10:26.000
Those are going to be in this interval right there.
00:10:26.000 --> 00:10:30.500
Basically, we already know what the graph looks like.
00:10:30.500 --> 00:10:39.500
The graph looks like this.
00:10:39.500 --> 00:10:45.800
We know that it is positively concave here and we know that it is positively concave here.
00:10:45.800 --> 00:10:53.900
For positive concavity, we really do not need to do any more work.
00:10:53.900 --> 00:10:59.300
The interval is -infinity to -1/ √3.
00:10:59.300 --> 00:11:06.800
From here all the way to this point, wherever it actually changes from concave up to concave down.
00:11:06.800 --> 00:11:16.200
And then, from here to positive infinity to 1/ √3 to positive infinity.
00:11:16.200 --> 00:11:18.000
We do not need to do any more work.
00:11:18.000 --> 00:11:23.400
We have already done all the work necessary.
00:11:23.400 --> 00:11:29.400
Let us move on to question number 5 here, let us see what we have got.
00:11:29.400 --> 00:11:42.900
Question number 5, we are going to be dealing with the equation x² - 2y² + 3xy = -4.
00:11:42.900 --> 00:11:50.400
Part A wants us to find an expression for the slope of this implicitly defined function
00:11:50.400 --> 00:11:53.700
anywhere along the curve, at any point xy.
00:11:53.700 --> 00:11:59.100
Basically, they are just asking for the derivative dy dx, y’, however you want to think about it.
00:11:59.100 --> 00:12:01.500
We are just going to differentiate implicitly.
00:12:01.500 --> 00:12:15.000
This is going to be 2x - 4yy’ + 3 ×, again, I’m somebody who likes to separate out my constant and just deal with my functions.
00:12:15.000 --> 00:12:26.700
It is going to be this × the derivative of that, xy’ + this × the derivative of that y × 1 = 0.
00:12:26.700 --> 00:12:39.600
We get 2x - 4yy’ + 3xy’ + 3y = 0.
00:12:39.600 --> 00:12:44.300
2x + 3y = y’.
00:12:44.300 --> 00:12:54.800
I’m going to move the y, in terms of y’ over to the right, factor out the y’, 4y - 3x.
00:12:54.800 --> 00:13:06.200
I get y’ is equal to 2x + 3y/ 4y - 3x and that is the expression we were looking for.
00:13:06.200 --> 00:13:11.700
That is dy dx, the slope along the curve at any given point xy.
00:13:11.700 --> 00:13:14.700
Nice and straightforward, just a nice application.
00:13:14.700 --> 00:13:35.500
Part B wants us to find the equation of the normal line to the curve at the point 1 - 1.
00:13:35.500 --> 00:13:42.700
In other words, if that is the curve, that is the tangent line, that is the normal line.
00:13:42.700 --> 00:13:46.600
The normal line is going to be, this is the tangent, this is the normal.
00:13:46.600 --> 00:13:50.800
We are going to find the equation of the tangent line and we are going to basically take the reciprocal of the slope,
00:13:50.800 --> 00:13:56.000
to get the slope of the normal line passes through the same point 1 - 1.
00:13:56.000 --> 00:13:58.700
We have the equation for it, nice and straightforward.
00:13:58.700 --> 00:14:02.600
The first thing we want to do is find the slope of the tangent line.
00:14:02.600 --> 00:14:07.400
We are going to do y’ at 1 - 1.
00:14:07.400 --> 00:14:12.800
y’ is this thing, we just put in 1 - 1 into x and y respectively.
00:14:12.800 --> 00:14:24.500
It is going to equal 2 × 1 + 3 × -1/ 4 × -1 - 3 × 1.
00:14:24.500 --> 00:14:28.100
It is going to be 2 – 3, it is going to be -1.
00:14:28.100 --> 00:14:33.800
This is going to be -4 - 3 – 7, we end up with 1/7.
00:14:33.800 --> 00:14:45.500
When we take the reciprocal of that, the slope of the normal line is -7.
00:14:45.500 --> 00:14:54.200
We get y - y1 which is -1 = -7 × x - 1.
00:14:54.200 --> 00:14:57.600
There you go, that is your answer, and you are welcome to leave it like that, not a problem.
00:14:57.600 --> 00:15:05.200
You do not necessarily have to simplify it, unless it asks you to do so.
00:15:05.200 --> 00:15:09.400
Part C, what is part C asking of us?
00:15:09.400 --> 00:15:25.300
Part C, they want to know what the second derivative d² y dx² is at 1 – 1.
00:15:25.300 --> 00:15:40.100
y’ we know is 2x + 3y/ 4y - 3x.
00:15:40.100 --> 00:15:43.000
Basically, we are just going to find y”, that is what we want to find.
00:15:43.000 --> 00:15:48.100
We want to find y”, what is that?
00:15:48.100 --> 00:15:50.200
Let us just use the quotient rule, that is fine.
00:15:50.200 --> 00:15:53.800
I mean, there are other ways we can do it but you go ahead and use the quotient rule.
00:15:53.800 --> 00:16:02.200
y” = this × the derivative of denominator × the derivative of numerator – numerator
00:16:02.200 --> 00:16:06.400
× the derivative of the denominator divided by the denominator².
00:16:06.400 --> 00:16:07.600
This × the derivative of that.
00:16:07.600 --> 00:16:36.200
We have got 4y - 3x × the derivative of this which is 2 + 3y’ - 2x + 3y × the derivative of this, 4y’ - 3/ 4y -3x².
00:16:36.200 --> 00:16:39.300
Let us go ahead and multiply everything out.
00:16:39.300 --> 00:16:58.000
This is going to equal 8y + 12yy’ - 6x – 9xy’ - 8xy’
00:16:58.000 --> 00:17:28.300
+ 6x – 12yy’ + 9y/ 4y – 3x².
00:17:28.300 --> 00:17:38.900
Let us see, what have I got, 12yy’ – 12yy’ – 6x + 6x.
00:17:38.900 --> 00:17:47.900
I have got 8y 9y, that left and that left.
00:17:47.900 --> 00:18:14.300
y” is going to equal 17y - 17xy’/ 4y - 3x².
00:18:14.300 --> 00:18:28.100
Find y’ of 1 - 1 first, to put into here and then we will find the y”.
00:18:28.100 --> 00:18:40.700
We already know what y’ is at 1 – 1, it = 1/7, that was the slope of the tangent line.
00:18:40.700 --> 00:18:58.100
Once again, we have got y”, let me write it again, our expressions 17y - 17xy’/ 4y - 3x².
00:18:58.100 --> 00:19:03.500
When I put in the 1 - 1 including this which is that,
00:19:03.500 --> 00:19:35.900
I get y” at 1 - 1 is equal to 17 × -1 - 17 × 1/7/ 4 × - 1 - 3 × 1² = -17 – 17/7/ 49.
00:19:35.900 --> 00:19:44.300
You end up with -1/36/ 343.
00:19:44.300 --> 00:19:52.200
There you go, nice and straightforward.
00:19:52.200 --> 00:20:01.200
Let us see, what does part D asks us?
00:20:01.200 --> 00:20:13.900
At the point 3a – 2a, what is the nature of the tangent line to the curve for any point
00:20:13.900 --> 00:20:17.700
that has the coordinates 3a and -2a.
00:20:17.700 --> 00:20:32.700
Let us say if a = 3, then it would be the point 9 – 6, what is the nature of the tangent line at 9 - 6 or at 3 – 2, or at 6 – 4.
00:20:32.700 --> 00:20:36.000
For different values of a, it is going to be different points.
00:20:36.000 --> 00:20:37.600
We literally just plug it in.
00:20:37.600 --> 00:20:40.700
They want to know what the nature of the line tangent.
00:20:40.700 --> 00:20:42.500
It is going to be y’.
00:20:42.500 --> 00:20:51.500
We have an expression for y’, that was 2x + 3y/ 4y - 3x.
00:20:51.500 --> 00:20:57.500
Let us basically put in this and this into x and y respectively.
00:20:57.500 --> 00:21:23.000
y’ at the points 3a and -2a, it is equal to 2 × 3a + 3 × - 2a/ 4 × - 2a - 3 × 3a.
00:21:23.000 --> 00:21:30.800
You are going to get 0/-17a = 0.
00:21:30.800 --> 00:21:35.500
Whenever you have anything that is of a nature of 3a – 2a,
00:21:35.500 --> 00:21:48.300
the tangent line through any point whose coordinate is given by 3a - 2a is horizontal.
00:21:48.300 --> 00:21:59.700
That is it, straight application of basic differential calculus, nothing strange about it.
00:21:59.700 --> 00:22:05.200
Number 6, number 6 is a related rates problem.
00:22:05.200 --> 00:22:10.300
Let us go ahead, as you can see it here, you have water being poured into a tank.
00:22:10.300 --> 00:22:12.700
They gave you the height of the tank.
00:22:12.700 --> 00:22:14.500
They gave you the radius.
00:22:14.500 --> 00:22:22.900
They gave you the water level of the tank is increasing at a constant rate of 0.5 ft/s.
00:22:22.900 --> 00:22:24.700
Find different things.
00:22:24.700 --> 00:22:25.700
Let us go ahead and get started.
00:22:25.700 --> 00:22:29.600
Let us draw a picture before anything else.
00:22:29.600 --> 00:22:31.700
Let us go ahead, we have something like this.
00:22:31.700 --> 00:22:36.200
We have an inverted conical tank.
00:22:36.200 --> 00:22:41.900
Of course, we have the water level.
00:22:41.900 --> 00:22:52.400
This is going to be our h, they tell me that the radius is 24.
00:22:52.400 --> 00:22:56.900
They tell me that the height of the tank is 120.
00:22:56.900 --> 00:23:01.200
Water is being poured in here and it is filling up.
00:23:01.200 --> 00:23:05.700
Let me just go ahead and draw little something like this.
00:23:05.700 --> 00:23:13.200
They tell me that the water level in the tank is increasing at a constant rate of 0.5 ft/s, dh dt.
00:23:13.200 --> 00:23:29.400
Dh dt, the height is changing at 0.5 ft/s.
00:23:29.400 --> 00:23:37.800
Part A says, find an expression for the volume in the tank in terms of the height.
00:23:37.800 --> 00:23:49.200
We know that volume for a conical tank is going to be 1/3 π r² × h.
00:23:49.200 --> 00:23:51.600
This is a function of two variables, r² and h.
00:23:51.600 --> 00:23:55.800
We want it to be only in terms of height.
00:23:55.800 --> 00:24:03.700
We need to find an expression relating the radius and height that I can substitute in here.
00:24:03.700 --> 00:24:11.200
Once again, I’m going to use similar triangles here.
00:24:11.200 --> 00:24:15.100
I’m going to take half.
00:24:15.100 --> 00:24:18.100
This is a cross section of this thing right here.
00:24:18.100 --> 00:24:24.700
This is 24 and this is 120.
00:24:24.700 --> 00:24:30.400
This is h and this is r.
00:24:30.400 --> 00:24:31.900
There is a relationship.
00:24:31.900 --> 00:24:42.400
h is to 120, as r is to 24, similar triangles.
00:24:42.400 --> 00:24:47.600
I get 24h = 120r.
00:24:47.600 --> 00:24:53.000
I'm going to get r is equal to 1/5h.
00:24:53.000 --> 00:24:58.100
I’m going to take this 1/5h, I’m going to plug it in here.
00:24:58.100 --> 00:25:08.400
I get volume = 1/3 π × h/ 5² × h.
00:25:08.400 --> 00:25:23.700
The volume = π h³/ 75, that is my answer for part A.
00:25:23.700 --> 00:25:34.500
Find an expression for the volume, in terms of the height.
00:25:34.500 --> 00:25:37.800
Part B, let us see what part B is asking us to do.
00:25:37.800 --> 00:25:48.600
It wants us to know how fast is the water being poured in at the instant, that the depth of the water is 20 ft?
00:25:48.600 --> 00:26:04.800
When h is equal to 20 ft, how fast is the water being poured in?
00:26:04.800 --> 00:26:09.300
How fast water being poured in, there is a little bit of typo here.
00:26:09.300 --> 00:26:16.800
It says feet per second, it should be cubic feet per second.
00:26:16.800 --> 00:26:19.600
It should be cubic feet per second.
00:26:19.600 --> 00:26:27.700
If for any reason there is not a typo, I will explain that after I finish this particular, the part B of the problem.
00:26:27.700 --> 00:26:31.900
When we say how fast, they are asking for how fast does the volume changing?
00:26:31.900 --> 00:26:33.400
How fast does water being poured in?
00:26:33.400 --> 00:26:36.700
Volume is cubic feet per second, not feet per second.
00:26:36.700 --> 00:26:43.800
What they are asking us basically do is to find dv dt, how fast does the volume changing?
00:26:43.800 --> 00:26:52.200
Dv dt = what, when h = 20?
00:26:52.200 --> 00:27:00.300
We have an expression volume = π h³/ 75.
00:27:00.300 --> 00:27:04.800
Let us differentiate both sides with respect to t.
00:27:04.800 --> 00:27:23.400
dv dt = 3π h²/ 75 dh dt, which is equal to π h².
00:27:23.400 --> 00:27:29.400
I’m going to put the dh dt on top and I’m going to leave this down here, 25.
00:27:29.400 --> 00:27:39.100
We already have dh dt, the rate at which the height is changing was 0.5 ft/s.
00:27:39.100 --> 00:27:43.000
They gave us the h, now we just solve for dv dt.
00:27:43.000 --> 00:27:49.700
Dv dt, when h = 20, is nothing more than π.
00:27:49.700 --> 00:27:59.300
Put the 20 in, 20² × dh dt which is 0.5/ 25.
00:27:59.300 --> 00:28:05.800
That is going to equal 200π/ 25.
00:28:05.800 --> 00:28:14.200
You answer is 8π ft³/ s, that is your answer.
00:28:14.200 --> 00:28:15.700
I know that there is a typo.
00:28:15.700 --> 00:28:19.900
If for any reason there was not a typo and this is a trick question.
00:28:19.900 --> 00:28:23.800
If they say how fast in feet per second is the water being poured in,
00:28:23.800 --> 00:28:29.800
at the instant that the depth of the water is 20 ft, feet per second is the change in the height.
00:28:29.800 --> 00:28:32.200
The change in the height is constant.
00:28:32.200 --> 00:28:34.400
The answer would just be 0.5.
00:28:34.400 --> 00:28:40.100
If you run across a situation like this, you definitely want to call your proctor and ask them is this a typo,
00:28:40.100 --> 00:28:42.600
how can I interpret this, or do both.
00:28:42.600 --> 00:28:45.000
Just do the problem and then write down.
00:28:45.000 --> 00:28:49.500
If this is not a typo, the answer is this.
00:28:49.500 --> 00:28:53.100
All your bases are covered.
00:28:53.100 --> 00:29:07.200
Let see with c says, they want to know how fast the area of the surface of the water is changing, when the depth is 20?
00:29:07.200 --> 00:29:12.300
Now area is π r².
00:29:12.300 --> 00:29:17.700
We know that r is 1/5 × h.
00:29:17.700 --> 00:29:30.400
Therefore area, in terms of the height of the water is π × h/ 5² which is just π h²/ 25.
00:29:30.400 --> 00:29:36.400
Now that we have that expression, we differentiate with respect to t.
00:29:36.400 --> 00:29:50.200
How was the area changing, da dt = 2 π h/ 25 dh dt.
00:29:50.200 --> 00:29:55.900
Because h is a function of t, we are taking the derivative.
00:29:55.900 --> 00:30:21.700
Da dt, when h = 20 = 2 × π × 20 × 0.5/25 is going to equal 20 π/ 25 = 4π/ 5, this is ft²/ s area.
00:30:21.700 --> 00:30:24.400
That is it, straightforward application.
00:30:24.400 --> 00:30:26.200
All related rates problems are the same.
00:30:26.200 --> 00:30:30.100
There is a rate that they give you, at least one, maybe more.
00:30:30.100 --> 00:30:33.100
There is a rate that they want.
00:30:33.100 --> 00:30:38.200
There is a rate that they give you, let us say dm dt.
00:30:38.200 --> 00:30:41.500
There is a rate that they want, let us say dp dt.
00:30:41.500 --> 00:30:47.400
Your job is to find a relationship, an equation, that expresses one in terms of the other.
00:30:47.400 --> 00:30:48.900
It could be this in terms of that or that in terms of that.
00:30:48.900 --> 00:30:50.400
It does not really matter.
00:30:50.400 --> 00:30:54.300
Once you have that expression, you differentiate with respect to t.
00:30:54.300 --> 00:31:01.500
And then, you rearrange the equation as you see fit, plugging in the values that are given to you.
00:31:01.500 --> 00:31:05.400
My friends, we have come to the end of our AP Calculus course.
00:31:05.400 --> 00:31:11.400
It has been an absolute pleasure being able to present AP Calculus to you like this.
00:31:11.400 --> 00:31:13.000
I wish you the best of luck in all that you do.
00:31:13.000 --> 00:31:16.000
I wish you the best of luck on the AP Calculus exam.
00:31:16.000 --> 00:31:19.000
I thank you for joining us here at www.educator.com.
00:31:19.000 --> 00:31:20.000
We hope to see you soon, take care.