BAM Mr. Tarrou. In this math lesson we're going to be taking a look at question
six of the 2018 calculus a B test it's going to involve some slope field
sketching it's going to involve some linear approximation and then finally
solving a differential equation a separable differential equation so let's
see what we got going on here consider the differential equation of
dy/dx is equal to one-third x times y minus 2 squared Part A says sketch the
solution curve that passes through the point zero two and the one that passes
through the point 1 0 and I've done my best to approximate what that slope
field looks like by the way in the description you will find timestamps to
let you jump her head to each of the three parts you'll also find a link to
the AP calculus AP calculus try and get that five on our AP test when you look
at the slope field you want to make sure that you sketch a solution curve that
goes all the way through the slope field passes through the 1 or in this case two
points that they're indicating and you kind of just look through here and get
the general gist of what that graph looks like in other words you know we
don't want to look at this point of 1 0 and say oh well the slope is positive
maybe a slope of positive 1 and then just draw a straight line going through
that point because if you were to draw a straight line at some point you'll get
down over here and you're like man my graph is going down but the slope field
is showing or when my graph is going up but my slope field is showing that the
slopes are negative you kind of want to to the best of your ability just kind of
go with the flow of that slope field like here at 0 a 2 through this point
that slope is horizontal as I extend horizontally I end up hitting more
blue marks which represent the slope of the solution curve at that particular
point and we kept we just keep hitting slopes that well appear to stay
horizontal so you can put arrows on this which is never a bad idea or just let
that curve run right off to the right off the edge of the slope field if you
stop this short and make it look like a line segment you will lose that point at
this point of one zero we see a slope that appears to maybe have a slope of
one and as we go up and up and up well you're not going to go up with that same
constant slope forever because it seems like this sort of curves down and even
to the best of my ability here so I've trying to draw the slope field even as I
get to this next column with tick marks if I were to make a linear function I'm
already looking like I have a steeper slope than what's indicated by the slope
field so you want to kind of just curve that over go to the end and or include
an arrow at the end of that as we cross through the x value zero
all of these slopes regardless of what the Y button the y coordinate is equal
to R zero or these are all horizontal tick marks so I'm going to come through
this point and level off and kind of come through horizontally through the y
axis where X is equal to zero and as we come to the other side we see that the
slopes now I'm kind of drawn this from right to left which isn't really how the
flow of the graph works right these slopes over here seem to start off being
either nearly zero or quite negative depending where you are vertically along
your soap field but if I match over to this other side it kind of seems like
the slopes or maybe then a little bit positive here but they start to fall
down I think that's probably just my poor drawing they start to fall down
come to the y axis and then rise back up again so something like this would be
your two approximations of the slope field of the solution curve going
through the slope field make sure you of course pass through the point of zero to
make sure you pass through the point of one zero or whatever they happen to give
you for this particular test or any question with the slope field and
just follow along as best you can if you're a little bit off we're not going
to be concerned about that just not wildly off you can't again have
something coming down here with a positive slope or clearly in this entire
quadrant three all those slopes are negative and you'll be good now Part B
says let y equal f of X be the particular solution to the given
differential equation with the initial condition of F of 1 is equal to 0 so
we're talking about a solution that passes to the point of 1 0 basically
maybe we're looking at this curve down here write an equation for the tangent
line to the graph of y is equal to f of X so if I have done a reasonable job of
drawing my solution curve then we're talking about a tangent a line tangent
to this graph y is equal to f of X at x equals 1 so we're gonna find the
equation of a tangent line that looks something maybe something like this you
know my sketch through the slope field is just that a rough sketch but
somewhere along that and we're going to use the equation of the linear tangent
line to approximate the functions value at 0.7 so we know you know given this
initial condition of the solution curve goes through the point of 1 0 we know
that this is exact the rest of this is just a sketch but we're going to
estimate the value of this orange solution curve at the value of point O 7
and the tangent line can actually be a good approximation for any curve as long
as you're close to that point of tangency let's so let's go ahead and
point O 7 or excuse me 0.7 is close to the next value of 7 so
let's take this off so I have some room to work and find the equation of that
tangent line or the line tangent to f of X at
x equals one okay so we're looking at first we need slope right what do you
need to write the equation of line a point and a slope well we have the point
it's 1 0 we need the slope so dy/dx at the point because we end this derivative
is asking for both the X&Y coordinate so at the point of 1 0 is equal to 1 third
our x value of 1 times our y value of 0 minus 2 squared so okay so negative 2
squared is 4 4 times 1/4 times 1/3 so we have a slope of 4/3 so the equation of
our tangent line is in point-slope form y minus y sub 1 is equal to M 4/3 times
X minus X sub 1 now don't please be too quick to put
this equation of a tangent line into slope-intercept form which is actually
the form that's most commonly requested you don't if you give an equation of a
tangent line as a final answer if a ap question is asking for an equation of a
tangent line they are going to want that in slope-intercept form but you really
want to leave this in point-slope form to make it a little bit clearer as to
how you're going to find that approximation at the x value of 0.07 at
least the way I like to look at this so we are looking at Y is equal to 4/3
times point zero point 7 minus 1 and at least in terms of doing linear
approximations I like my tangent line to be in point-slope form as opposed to
slope-intercept form that you would be giving for a final form of a tangent
line now this 7/10 is best worked with in
terms of a fraction so we're looking at 7/10
- ten over ten now we're looking at 4/3 times negative 3 over 10 and the
negative the threes are going to cancel out the 4 and the 10 are going to cancel
out to be 2 & 5 so our final answer here is that our approximation is at negative
2/5 so now this is the value that we're getting from our tangent line it's just
an approximation for that f of X function and I think it's probably a
good idea since we're being asked to estimate the value of f of 0.7 that
therefore f of 0.7 is approximately equal to negative 2/5 and if you want to
see how the points are divvied up again find that link to the AP College Board
website where you can download the scoring guideline and see where the
points are divvied up you can also look at the master scores comments to find
out where students most commonly made their mistakes as they were working out
these particular in all the particular problems on the AP test part see coming
up right now and for the last part C we're going to find a particular
solution to the differential equation the separable differential equation
using that initial condition of F of 1 is equal to 0 so this is our solving
separable differential equations my students would have seen this in chapter
6 of our Larson book so it's the only type of differential equation we need to
worry about solving in calculus a B or BC for that matter how are we going to
separate these variables well we're going to move things around with
multiplication and division we're going to bring the DX over with an apparent
multiplication and bring the Y minus 2 squared over to the left-hand side
giving us 1 over Y minus two squared dy is equal to one-third DX
now that we have these variables separated everything in terms of Y on
the Left terms of X on the right we're going to
set up an integral on both sides I'm also going to bring this y -2 up to have
a negative power so we're going to be integrating y -2 squared dy I'm also
going to bring this coefficient of 1/3 out front this constant multiplication
of 1/3 out front and we're gonna be integrating X with respect to X ok well
the right-hand side is quite straightforward actually both of them
are but on the right-hand side which is simply X we're going to raise that power
by 1 and divide by that raised power so we have 1/3 times x squared times 1/2
because we're divided by that increased power we're gonna have a plus C on the
right hand side no need to do a plus C on both sides of differential equations
as you're solving them on the left you have two choices you can take this y -2
and square it make it three terms integrate each of those three terms with
the basically the power rule individually or you can complete the
chain rule pattern or maybe that chain rule pattern is already completed that's
going to allow you to have fewer terms it's going to allow you to have on wine
only one term instead of two you have a choice here not saving maybe necessarily
a tremendous amount of time oh that should be negative 2 but I'm starting to
get comfortable and realizing how even just a little bit of time save them some
part of an AP question a time to test how beneficial that can be and if you
look in here and you say well let's let u equal y -2 well then u prime would be
just simply 1 and we already have a multiplication of 1 or a factor of 1 out
here but if you want to see that as a addition not in addition but in
introducing that code that factor of one you can divide by that factor of one if
you like nothing's really going to happen here but if that was like a 2y
then you would need to introduce a multiplication of two to complete that
chain rule where you have au and au Prime and if you introduce a
multiplication of two you need to balance it with the division of two so
we're going to integrate y -2 to the negative 2 power if now if this had no
power on if it was a power of negative 1 then the result of our integration would
look like a natural log would be a natural log function some of my students
that maybe some struggling a little bit might make that common mistake but we're
going to raise this power by 1 and then divide by that increase power and at
this point we're going to remember that you're going to save yourself the most
amount of time with these problems usually if right after you finish the
integration you go ahead and plug in that x and y coordinate to help you
solve for C don't clean this necessarily up for Y first and then find out what C
is equal to so our Y value is 0 our x value is 1 and so we're looking at what
we have here 1 divided by negative 2 so we have 1/2 x and we're going to this x
1 so we're just going to take this this positive 1/6 and subtract it over to the
other side find a common denominator and find out that C is equal to 2 6 or that
C is equal to 1/3 now that means that we're ready to go ahead and finish
writing up a cleaned up version making a cleaned up version of this differential
equation so we have negative 1 over Y minus 2 is equal to 1/6 x squared
is equal to one-third not equal to but plus one-third and they're not generally
happy until you get an equation that is solved for y so we're going to let's see
here what might be the best way to do this how about we multiplying both sides
by six so if negative six over y minus two is
equal to x squared plus two that gets distributed we're going to multiply both
sides by Y minus two and actually at the same time we're going to divide those
sides of the equation by x squared plus two and then finally add both sides by
two they are not generally happy with your final answer when you're solving
separable differential equations until you get Y by itself sometimes it's okay
to get like a Y squared or Y cube by itself and you know the math is getting
a little bit tough course this is an AP calculus test we should be expected to
do some fairly complex algebra but don't doubt yourself as you're doing this
because maybe you think oh wait what's going on here with this Y minus two in
the denominator that seems a little bit funny and the right-hand side is a bit
complex just have confidence in what you've done the skills that you have as
I check my answer but we are good and just keep going to that problem and then
when you have time if you have time at the end of your test go back and look
over your work for the questions that you're maybe a little bit unsure of but
don't start doubting yourself because the algebra looks a little bit maybe
complicated or funny or unexpected they're trying to really make sure you
know your concepts and you're confident in your work so that is number six of
2018 calculus a B I'm mr. Tarrou BAM go to your homework and get a five on that
AP test
you