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Good day students AT part eight of the eighty Eight multiple-choice trivia questions off
for nineteen ninety-eight so is going to going over questions 86 to 92 in this uhm installment.
Right let's take a look at question number eighty-six so question eighty-six it says
the base of a solid is the region and the first quadrant bounded by the X axes y-axis
and the line X was to why equals eight is shown in the figure above. If cross-sections
of the solid perpendicular to the X axes are semi circles what is the volume of of the
solid okay so this is the basically assessing your understanding of solids with known cross-sections's
right so. The going to make a sketch of the situation, but if I do that for me labeled
this function right here so we looking for the salt the volume of the solid created by
of the base down device functions this function in a access. So let's call this the top function
since our segment is going to be perpendicular to the X axis and then look something like
this am that the cross-section of the semi circles so we have a this is the function
on the top the call this of F of X top right to going top-down and then at the bottom this
is this is G of X bottom right so for one of find the length of the segment which is
basically find the difference between these two functions and one thing one and notes
is that this function down here is basically the function why equals zero right and in
the top function is of why equals eight minus X over two ago ridiculous I got that later
but just know that this is the top in this is about a function so can you visualize the
solid that of is being created here to help you visualize it is go ahead and sketched
of the situation on going the rotate y-axis and just lay down flat. So this is my y-axis
and that's my X axis and that's the line of why equals X case but it like that okay so
what we have is of so this is why this is X the have semi circle sitting right on this
in such a way that you perpendicular to the X axis so this is one of the semi circles
limit to one of them. So this is one semi circle perpendicular to the X axis imagine
that this is ninety degrees right here okay that's ninety right so the semi circle look
something like this okay so that's one of them in a was going on is that you have an
infinite amount of semi circle just like that okay so the question is you have over to semi
circle just like that over this entire region what is the of of volume of that solid going
to be to that's what we doing in this problem okay right so of in order to do this we have
to find the of area the cross-sectional area of of our semi circle in the with and evaluate
an integral by the one and notes that for this problem with going from zero all the
way to eight so let's take a look at this the segment for example so the semi circle
right here how the damage are okay of the denominator is the and that D is basically
the difference between this function on the top and the function on the bottom right so
is basically F of X minus G of X where F of X is the one on top in G of X is about as
the bottom function okay so F of X is it is function right here of isolate is for Y of
X plus two why equals eighty by solid for why you have why equals eight minus X over
two so this is my top function by top function is eight minus two X answer is minus X eight
minus X over to into my bottom function is the x-axis which is why equals zero okay so
that's the denominator of our of semi circle okay eight minus X over to now of let's see
me write down the formula that were going to use to find the area of the resultant solid
so generally speaking is one of find the area of of solids is known cross-section a be do
it semi circle this is the formula right here so is putting green the formula if you going
along the X axis everything that in terms of X okay so the area is going to be the integral
from X to the left. The left was X to X to the right so in this case this is the leftmost
X zero and then one on the right is eight of of of with a between pie are square over
to basically the of the air of a circle divided by two so pirate of one half of pie times
the radius okay so what is the radius in this what is the radius in this case of the situation.
So the radius is going to be of the function F of X's right of his lower case F of X minus
G of X so F of X minus G of X represents the damage or right that we need to find the radius
how the find the radius if you noted that later is simply divided that by two okay F
of X minus G of X over to so this is our quantity squared the X so this is the formula for of
a semi circle okay, for semi circle right so is this from a look. So let's abstract
for you what we just did here is is we just evaluated the integral from eight to be of
one half pie are square the X that's what we did so this entire expression right here
is are square okay right so that's what were going to be applying to him this problem right
here, we need to find out what the radius is so what is our radius here of radius is
the diameter divided by two so our diameter is eight minus X over two so one half of that
one half of eight minus X over to will give us the radius of all semi circle okay so is
going to be eight minus X oval for so there goes the radius now the plug of this into
our formula so the plug-in to formula the area is going to be the integral of know that
X is going from left to right because our semi circle's are perpendicular to the X axis
is your program the with the Y axis everything we different this to be all why is why why
why why here okay the completely different and wheezing a different equation be using
X equals eight minus two write equation right so here the have with stuff from zero the
leftmost X to eight line right in then of this is a semi circle one half times pie are
square and we know what our is our is eight minus X over for the damning over to square
the X now this is a calculator of section so we going to waste our time trying integrate
this by hand going to use a calculator to do it okay so the reason the TI eighty-nine
titanium calculator does it occur the best calculator for the AP exam so is evaluate
this integral real quick so the integral of of one is one over to pie one divided by two
pie of parenthesis is of parenthesis eight minus X eight minus X close the numerator
divided by for the denominator close of this expression right here and raise it to the
second power and then the independent variable is X will limits of integration is zero "upper
limits of integration is a" enter were given the answer in the exact form I need and approximation
since my options here at of approximate value so diamond enter sixteen point seven five
five two so answer we can clearly see is of option letter C okay so answer is option C
area is approximately sixteen point seven five five right so the trick here is not forget
to divided dammit a by two okay. Him down the wrong answer is to go to the next problem.
Right let's take a look at eighty-seven it as which of the following an equation of the
line tangent to the graph of F of X equals X the forklifts X square at the points where
F prime of X is equal to four coming equals to what right so basically we're being told
that the slope is one in the asked to write down the equation of the tangent line right
to is right I equation the tangent line just a write down the point the points the form
of the equation of the line one is why one equals and X minus X one so we need is and
which Rodino is of one remember M is your slope F prime we need X one and we need why
one okay so we know that the slope at X one why one is one so what is the exit value of
that with that gives us and out the one for the slope so we have to do is we have to solve
this equation F prime of X equals one so what is F prime of X how do we find F prime of
X will simply differentiate this function right so that F of X is X to the four plus
two X square and the follows that F prime of X using the power rule for differentiation
is for X to the third plus four X okay since F prime of X is equal to one we have to solve
this equation that to solve four X to the third plus four X equals one is to other the
X value what the slope has a value of positive one okay so how do we solve this equation
the different ways we can do it of one easy way to this we can use the Later here is by
simply graph in the two functions are looking for the intersection right so is go and do
it. So we have that into the first function for X to the third plus four X enter that
the first function the second function is one enter the the graph both of them that
one F-3 to graph the go the cubic function in the linear function you become of the second
there goes the linear function okay so what we looking for is the intersection so F5 is
you the axis all the calculation and music and on the graph so option five is intersection
first first curve enter second curve enter lower bound any points to the left of the
intersection will suffice upper bound any point to the right of intersection enter any
X value points to three six of order but with the run it up to three decimal places okay
so one two three seven so the solution here is going to be X equals points to three seven
not using the TI eighty-nine titanium is a different ways you can do this is much quicker
so ensure you from the home screen. Or you simply do is go out to graph base select solve
F2, one DV the axis is so function into the function four X so the equation 4 X the third
plus four X equals one in the what they with IE solving for is solving for X enter the
okay so you have a TI eighty-nine titanium have of can be our system here can take advantage
of two so quadratic equations which you have it you eighty-nine titanium you have eighty-three
or eighty-four is go to the graph menu and then just to this is that you showed you okay
so of this is the X value what the slope of the tangent line is positive one so we have
X one now not only have X one of we of we can skip steps and use a calculator to find
equation of the tangent line all weekend find X two and of right on the equation of the
line okay so don't to be lazy distance I can do of I can graph the original function in
in the original function is X to the fourth plus two X square enter rights he graphic
right so what you looking for the is the slope of the tangent line at X equals one two three
seven okay so of when is go to F5 and option six is derivatives the why the X option one
now what does does is the calculate the slope and also using the equation of the tangent
lines so it says the why the X at what X value let's enter it points to three seven enter
and why is us of the equation okay I tried is the run one is telling the slope and equation
of the tangent lines so is go back to F5 I A1 accuses option right here scroll down to
tangent okay this is the one to tangents select that so says tangent at what's points here
points to three the state again points to three seven enter so draw the tangent line
for you any gives you the equation of the tangent line okay so that's basically why
equals the approximated why one X minus one points to one and run it up it's your answer
is going to be present calculators Cup so why equals one X minus one points to two okay
isolate is point one two two two point one two two write so that secant see clearly see
that answer is option be that is what we of the next to find the equation of the tangent
line at any point on a and our graph within known function so another we can do it you
want to do this by hand you take a look at longer but Tom divide this by hand we just
have to figure out what of why one is to why one is basically F of X one okay which is
F of points to three seven right so what is F of twenty-three seven that what you get
the street draw the graph of that is what you gets when inputs point two three seven
into your function right so the graph is coming there goes the graph in then unit of use the
trace menu traced in any answer points to three seven for the X and Y need is the why
so my wife point one one five okay so my why one is point one one five point one one five
now I have my and that's the way have my am which is one my X one is points to three seven
in my why one is point one one five now put all of this into my point slope equation of
the line the have why minus point one one five equals one times X minus points to three
seven eighty so this for why is basically at this to both sides you have why is equal
to X of minus zero point one two two okay subtract this from both sides committee at
this to both sides Dalby a final answer so of the can use the shortcut we can do it use
the using the log argument to option number question number eight eighty-eight so far
eighty eighty says of that F of X the anti-anti-derivative of the natural of them of X the third of X
is F of one the anti-derivative at one is equal to zero then what is the anti-derivative
at nine so we can easily use FTC part two here to of to solve this problem so remember
that the FTC part two that of the anti-derivative at that's misuse start from the anti-derivative
evaluated at the minus anti-derivative evaluated at a is equal to the definite integral from
eight to be of the function F of X the X in this problem F of X is equal to the natural
of them of X to the third over X okay and the is the upper bound which is nine and a
is the lower bound which is one the argument of our anti-derivative functions so let's
set of the integral so using FTC two we can see that the anti-derivative at nine minus
the anti-derivative at one is equal to the integral from one to nine of the function
natural log name of X to the third over X right of so let's see we looking for what
we looking for the looking for of F nine so we can just had F one to both places it is
implies that anti-derivative at nine is equal to the integral from one two nine of the natural
log of them of X to the third over X plus the anti-derivative evaluated at one right
so let go to plug is in our calculator we know that F of one the anti-derivative at
one is zero so can just the range of that of the zero so just for this in a calculator
and that will give us our final answer right so calculators out go back home we looking
for definite integral here of second function seven integral of parenthesis is the natural
log of them of X equals the of logarithmic function of the of the numerator area this
is third power and divide that by X case studies calculate again of have is a problem here
right divided by X of independent variable is X lower limits of integration is one "all
the way to nine" we need an approximation diamond enter five point eight two seven okay so F of minus
five point eight two seven plus write that down is approximately 5 point eight two 7
to 3 decimal places our answer is option C right let's take a look at the next problem
problem number eighty-nine so says energy is a differentiable function such that G of
X is less to zero all the numbers to G of X is always negative and is F prime of X is
X square minus four times G of X which of the following is true to have situations here
about Max and mean of the function them of the determine relative Max and mean what you
do is you find the critical points of and then you set up you number line in look at
the science right so of the critical points are where you have the derivative having a
Valley of zero Re: have are nonexistent derivative that's out corner cost a vertical tangent
or no tangent at all okay of discontinuity right so let's look at this in this problem
we are focusing mainly on the critical points associated with this function right here so
of if you set this component of the derivative X square minus four two zero a solvate you
can have X is equal to plus a minus two okay this can generate some critical points for
the really needed of to answer this problem so what we going to do is make in number line
in his graph are critical points of we have of negative to into so these are the points
where the derivative of the is the of is equal to zero okay now in order to determine it
is increasing to decreasing look at the break it up into three integral's integral one integral
to and interval three we need test numbers for each integral so for this integral we
can use X equals of negative three your keys X equals zero in here can use X equals of
positive three right so want to look for the sign of F prime on this interval there can
determine the behavior this extrema here okay now of we can plug-in these the values into
this function to determine the sign a look at this look at of the the behavior of the
graph okay so are we looking for here is the of sine of sine of F prime interval number
one X is equal to negative three so when X is equal to negative three of we have we looking
at X square minus four times G of X now let's look at the graph of you sketch the graph
are your real quick that's a easily can so let's if you sketch the graph of X for minus
four it's a problem that shifted for units down so as to this is negative for here in
the Y intercept other roots is got negative to into umm thirty X intercept so look something
like this right is in the construct so something like this right so what this shows us is that
when you are and interval one this quantity is negative is positive okay only between
negative to two it's negative and then when you are the given to is positive right so
we can use that here so of for X square minus four in his about one this is minus our G
of X was to sign of G of X when Amanda problem were told earlier G of X is always less than
zero so regardless of where you're at G of X to be negative. So he is going to be negative
negative times negative and is positive of so can determine mistake of its X square minus
four is positive here and then G of X is negative right so negative times positive is negative
of integral two we use X equals zero you can look at the graph we can plug it in here if
you plugging here to see have negative for if you look at the graph your the beneath
the X axis is negative. So X for minus four is negative here and G of X is always negative
we multiply these two of you going to have minus times minus which is positive okay now
of lastly we have integral three X is equal to three of this is positive on the interval
in then G of X is negative always negative multiply see have minus right so these are
the signs whichever we want to do it of so for example look at this right here its positive
negative positive so says G of X is always negative what it does is the just let's this
patterning to negative positive negative that's another shortcut you can use right us so let's
see we have of integral one is of what is negative negative and positive to the negative
right so what does what is the signs tell me about the behavior of F now that we have
the sign of F prime what is the sign of F prime tell us about the behavior of F behavior
of F we know that when F prime is of negative the function is decreasing what is positive
function is increasing is negative is decreasing right so we decreasing and increasing what
can of extrema the have here at negative to you have a minimum right to relative minimum
in here is it increasing and decreasing asked to into the have a relative Max right so let's
see which one of is the correct answer relative Max at no relative mean at X equals negative
of the method X equals two excellent our answer is option letter b right let's take a look
at the next question is ninety percent is the base be of a travel is increasing at a
rate of change inches per minutes was height is six this area three just a minute which
of the following must be the truth muzzy true about the area of the triangle so this is
the of related rates problems on the area formula of a triangle okay so of the area
of a triangle area of a triangle is one half base times height so this is both variables
are of change with the differentiate implicitly okay to the derivative of a is just one the
ADT in a we can factor out the one half these two it's variables but of them a constant
so we have to differentiate then to them is variables so the two variable functions we
have to use the product rule here UV prime we recall of is the you prime plus you the
prime because the UV prime position you prime doesn't matter because we know that to you
think of addition commutes okay so is go ahead and do that so the this H the derivative of
the is just one the DDT plus of you press right is elsewhere get confused. On this is
you that this is you right here that be is you and H is V okay right so of now we have
you which is H that you is some the anti-the derivative of be is one the using implicit
differentiation be H DT right now we're told that of that of the base is increasing at
three inches per minutes so this Valley right here is the T DT the change in the base per
unit time and a height this one right here it's so this one is increases is positive
and then this was decrease this is getting negative in in this is the H DT the change
in the height right) does values into equation so we have the ADT which is the rate at which
the area is increasing or decreasing is one half times H the DDT is positive three plus
the base be times the HCT says is decreasing is negative three okay so have this equation
here of so the question is which of the following must be true about the area of the triangle
right so we need to find out we need to take what we want to find out the one of find out
the conditions for where the ADT is increasing or first decreasing now what you and note
is that of if it's looking at increasing with a look for where the ADT is rated and zero
and decreasing look for the ADT is less than zero okay so the have always the decreasing
since I have decreasing here decreasing here and decreasing so of a is decreasing when
the derivative the ADT is less than zero remember we talked about this before of the connection
between the sign of the derivative and the function the derivative is negative them is
a function is decreasing okay so of in our to solve for when the area is decreasing we
just can us that this expression to less than zero's is go ahead and do that simplified
a little bit so will have of one of find where one half of of three H minus three be when
is this less than zero and a rate less than zero know that the ADT is less than zero so
the ADT is equal to this expression right here okay so multiply what sites by two we
have three H minus three B is less than zero of at three be to both sides three H is less
than B be in in the Bible sites by three H is less than B so in our if you less than
B that implies that the a DT is less than zero which implies that a is decreasing right
so of when H is less than B a to B is equivalent to saying B is greater than H right to and
is greater than H of the area is decreasing the clearly see that the answer is option
letter D okay so that's that right us take a look at question number ninety-one that
says that F the function that is differentiable on the open interval one tan if F of two is
equal to negative five F of five is equal to five and F of nine is equal to negative five which of the following
must be true now we have, differentiability and is open interval I'll that means that
within this integral of the function is going to be continuous of a here now going to be
any corner costs or vertical tangents right so we have these points right here let's make
also that the charts to organize this information in the with that's easy to of understand so
we have a chart we have three inputs to five and nine other X is in our F of X our outputs
negative five five and negative five now let's give ourselves a possible sketch this is not
the only of this is the exact sketches is one of the possible sketch is that satisfies
of this description here so let's sketch the graph of the situation X goes from to all
the way to nine and then okay from negative five two five spread and center so here in
Right so let's put graph the points so to negative five one two three four five to negative
five plus one point and then we have five five one two three four five one two three
four five and then the last one that satisfies to six seven eight nine nine goes take the
right back to five negative five okay right so this is the sketch can help is really understand
what's going on here so this function is differentiable from one all the way to ten a much bigger
interval I'll we know that of this is good we smote so the have any costly with is happening
here of so it we can assume that a look something like this is the possible sketch goes like
that and when hits that maximum the better have a significantly smooth the rough it's
not differentiable social right there and then make a look something like this okay
so this is a possible interpretation the looking at his graph let's see of what conclusions
the control using these of options right here is this is F of F has at least two zero's
is that possible if you look here we guaranteed to have a zero somewhere between of two and
five and another zero between of five and nine we can conclude that this is accurate
using the intermediate Valley there okay since this function is continuous on this interval
of from 2 to 5 any attains of values of negative five and five the function was attain every
Valley in between five and negative five. So sin zero is the in between five and negative
five we guaranteed to have a zero between two and five same argument here is an intermediate
Valley terms of the function is continuous on this interval and he goes from five to
negative five it must also attain the value of every Valley between five and negative
five which includes zero and that's another zero right there so we have to zeros of on
this interval now the next of option a says that F has at least one horizontal tangent
now let's think about the extreme points that we have two and nine we can have a closed
interval there what do we have in this situation we have of a horizontal tangent line okay
so the endpoint has a horizontal tangent line what can we guaranteed to have somewhere in
between here you remember the mean value theorem semi about their tells is that if you old
draw the secant line on any function that's continuous on. Of an open interval and differentiable
on the closed interval is draw secant line the slope of the secant line is going to be
up points in with you the interval where the slope of the tangent line is equal to the
slope of the secant line right so what is the slope of the secant line on from to all
the way to nine we can clearly see that the slope of the secant line is zero so am here
the slope of the secant line is zero so of am is equal to zero so F of be minus F of
a remember that in the can set for the mean value theorem over B minus a this of of secant
line in this case is equal to zero so there has to exist the see of it has to be a C in
a be where F prime of see is equal to zero this is by the mean value theorem so this
is true vitamin A Valley care to be specific this is a special case of the mean value theorem
known as arose there okay to this is the slope of secant line have the value zero there must
be another value in between two and nine what the slope is also zero so that guarantees
that we have at least one horizontal tangent line on this interval right and then lastly
it says between two and five F of see is equal to three so let's look at two and five two
five has values are five to negative from negative five two five so by this is an is
Valley Therm we guaranteed that this function at attains value to about that of at these
every value in between negative five and five so of this three in between negative five
and five absolutely so by the intermediate Valley Therm we can also conclude that the
statement is true so all these three statements the correct about answer is option letter
D right let's move on to the last question so this is says that if case which is zero
and pi over two and the area under the curve of why equals cosine X from case two of X
equals pi over two the zero point one and what is the value of K right so we know how
the find is the underneath the curve we use the definite integral so what we have here
is a situation this can go ahead and set it up for you so we have of the integral from
so Valley can all the way to pie over to of the function cosine X the X equals zero point
one so the question is what is the Valley of can't so to set of this equation this the
integral set up of it was so that will equation. We just going to apply FTC two to the left
side okay so is the Valley the anti-derivative of this function if you am evaluate this is
can become are the anti-derivative of cosine a sign of negative sign X cosine X positive
sine X so if you integrate some sign to make integrate cosine you have sign so we evaluate
in this anti-derivative from K all the way to pirate to and that is equal to zero point
one now apply FTC part two to this expression the have sign evaluated at the upper limits
pi over two of - evaluated at the lower limits which is K and that is equal to zero point
one and then of sine of of pirate to know that sine pirate to is one so we can have
one - K is equal to zero point one the can of subtract one from both sides you have negative
sign K is equal to of negative zero point nine so sign K is equal to zero point nine
and K is the inverse sine of zero point nine seven a plug that in your calculator of K
is approximately 1 point one two zero right so that's that the thanks a much for taking
the time to watch this presentation because the free to subscribe to my channel and please
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exiting for watching and have a wonderful day