BAM! Mr Tarrou In this video we're going to be doing an example where we find the equation of
a tangent line to the function f of theta is equal to the sine of theta at or when theta is equal to
5 pi over 6. we're going to find the equation of the tangent line we're going to find the equation
of the normal line and then we're going to discuss you know why is it or how are we going to
you know use this line equations of lines that we've been you know finding
since algebra 1 tie the old in with the new how is it going to get used in our course of calculus
and give you some links in the description below if you want to jump ahead in your calculus course
and see maybe potentially how you're going to be using these tangent lines you're also going
to find a link to the home page of my channel and website where you can find any one of the 700 plus
math videos i've done so far from algebra 1 to calc 2 and statistics all righty then
we're going to find the line again that is tangent to this function at theta is equal to 5 pi over 6.
what do you need to write the equation of a line that's not going to change just because we are in
calculus you need a point and you need a slope so that you then can plug it into whatever form you
like i generally start off with point slope form well let's find the point of tangency basically
so we have that theta we're going to use the value of theta which
is 5 pi over 6. so let's plug 5 pi over 6 into our function and see what we have hopefully remember
your unit circle this won't be too terribly difficult we even started talking about maybe
product rule or quotient rule or especially chain rule so we are going to be thinking about
f of 5 pi over 6 and that is going to be equal to the sine of 5 pi over 6
off the unit circle that is going to give us an answer of one half so step one
if not you know if you're not given a full point of tangency find
that point so our point is going to be the value the coordinate of 5 pi over 6
one half okay so we have a point that our tangent line is going to go through
now we need a slope how are we going to do that what did we just learn in calculus that
tells us slope didn't we just learn how to find derivatives using the definition of a derivative
or using our basic trig rules or power rule yes our second step let's pick a different
color here orange for our second step we're going to work on finding the slope of our tangent line
meaning what we need to do is we need to find what f prime of theta is we need to take the derivative
with respect to theta of the left hand side i'm writing this really for teaching purposes
normally i just write f prime and maybe even just write the answer to the right hand side but we are
taking the derivative of both sides of this so we have the derivative with respect to theta of the
sine of theta so the derivative with respect to theta of f of theta is going to be just simply
f prime and the derivative of the sine of theta with respect to theta and as long as the variable
with which we're taking the derivative with respect to is the same as our independent
variable there's no chain rule we haven't even learned that yet maybe but we have the cosine of
theta so we have now a function a derivative function which is going to tell us the slope
of our original function you know anywhere that we desire as long as we remember that you cannot
evaluate or find a derivative at any type of discontinuity a sharp bend where the original
function maybe goes vertical or the end point of a closed interval so we're going to take this theta
is equal to five pi over six just plug it in we have f prime of 5 pi over 6 is equal to the cosine
of 5 pi over 6 which again off of the unit circle is going to be negative square root of 3 over 2.
yes we're in advanced calculus but what you need to write the equation of line is not
going to change from pre-algebra or algebra 1 wherever you first
learned how to find the equation of a line i'm sure that was algebra 1. so let's take step three
and i like to grab it in color but the green and blue sometimes is a little bit hard to see so step
three we're going to take that point of 5 pi over 6 and one-half take that slope that we just found
of square root of 3 negative square root of 3 over 2 and just go back to our point-slope
form the point-slope form of a line of course is y minus y sub 1. so we have y minus one half
is equal to m we just found that the instantaneous rate of change for the
equation or function f of theta is equal to the sine
of theta that instantaneous rate of change is negative square root of three over two
the slope of the sine function when theta is equal to five pi over six then we have x minus x sub 1.
so we have x minus 5 pi over 6. we're going to go ahead and keep all this in exact format
so distributing that negative square root of 3 over 2 through we're going to have y minus one
half is equal to negative square root of three over two x we have negative times negative which
is positive so plus uh five pi times square root of three of course just five pi square root of 3
multiplying our denominators in here we have 2 times 6 which is equal to 12.
ap is going to want the equation of your tangent line in slope intercept form i also teach ib
international baccalaureate program and a lot of times they're happy with the equation of their
lines in point-slope form so just put the equation of your tangent line in whatever form your teacher
is requesting or requiring we're going to undo the minus one-half over here with the y
of course by adding both sides by one half we're going to then need to find a common denominator
so we're going to multiply one half by six in the numerator and six in the denominator a form of one
and we have y is equal to negative square root of 3 over 2 x plus we've got 5 pi square root of
3 plus 6 over 12 and this is the equation of the line which is tangent to the function f
of theta is equal to sine of theta at or when theta is equal to 5 pi over 6. now i'm going
to write this over here on the left hand side and i will graph the or we'll be looking at the graph
of sine and the graph of the tangent line and the normal line before we go on and again discuss when
you may be using these tangent lines later on in this course and how you're going to be using them
so let me get this out of the way sometimes we're not going to be asked for the equation
of a tangent line we're going to be asked for the equation of a normal line now the difference
between a normal line and a tangent line is at the point of tangency the equation of the tangent line
well it's going to have the same slope as the slope of the function
or effectively we're you know kind of if you think about that definition of a derivative
and how we have two points that are becoming ever closer together as we let delta x approach
zero we're using the slope between two points that are infinitely close we're kind of using the slope
of the tangent line to estimate the slope of the function at some given value a normal line
is just simply perpendicular so instead of a line which is parallel to the function at a particular
domain value we are looking for the equation of a line which is perpendicular
oh ok so for the normal line we have i will do this in orange for our normal line
we're still working with that same point of tangency which for some reason i erased
five pi over six and one half
but for the normal line and understanding now that it's perpendicular our perpendicular
slope is the opposite reciprocal of the well slope of the tangent line or the
slope of that function at some particular value for which of domain 5 pi over 6. so
the opposite reciprocal of negative square root of 3 over 2 is positive 2 over square
root of 3 of course going to rationalize that denominator or at least i'm going to
so we have a point that we're going to use to find the equation of the normal
line which again is 5 pi over 6 and one-half and the perpendicular slope
is again two square root of three over three
point slope form of a line y minus y sub one so y minus one half is equal to m two square
root of three over three times x minus x sub one x minus five pi over 6 we're going to distribute
that slope which is now positive 2 square root of 3 over 3 through the parentheses yielding us y
minus one half is equal to 2 square root of 3 over 3 x minus because of course
it's positive times negative 2 times 5 is 10 so 10 pi square root of 3 over we've got 3 times 6
is 18 and we need to find a common denominator as we go through the process of undoing the
minus one-half that's over here with the y we're going to add both sides by one-half
we need a common denominator between this term that doesn't have the variable of x in it and
our one-half so we're going to multiply the numerator and denominator by nine
and that is going to give us a normal line which is y is equal to 2 square root of 3 over 3 x
and i'm going to write plus and bring these two terms together we have a
we have a negative 10 pi square root of 3 and a positive 9. and i want to write
this as an addition so i'm going to write that 9 which is positive and then minus 10
pi square root of 3 over a common denominator of 18.
if you
if you write like a minus here and then 10 pi square root of 3
plus 9 you would actually need to take that negative and distribute it through the two terms
that's in the numerator and have a little bit of a mistake so we have plus the 9 is positive here
and it's still here we have a negative 10 pi square root of 3 and it's still negative 10 pi
square root of 3 and that is the equation of our normal line
so for a visualization there we have our sine wave coming through the sine of negative pi
is zero the the sine of negative pi over two is negative one the sine of zero is uh
zero the sine of pi over two is one and so on and so on and if you enter this into your
calculator say is y sub 1 and then enter this into your calculator as y sub 2 and y sub 3
well and maybe i have of course the sine function in blue we are going to see that
sine wave coming through and then here comes the equation of the tangent line coming through and
my drawing's a little bit messed up but we have a point of tangency at of course theta
is equal to pi over 5 pi over 6 or the full coordinate of that point
which is the point of tangency which is 5 pi over 6 and one-half and then you have that normal line
which is intersecting the sine wave but with a slope which is an opposite reciprocal of the
instantaneous rate of change of sine when theta is equal to 5 pi over 6. now it's kind of doing
like a i really tried to make this a pretty decent graph without the aid of technology which i have
desmos on my la my ipad and this picture already set up but you might look at something here
actually the amount of separation that i have between the sine wave and the tangent line is
actually slightly exaggerated here so we have an area of the sine wave that may be almost
indistinguishable between the tangent line and the sine wave and indeed when you're close to
and if you take that calculator and start zooming in closer and closer and closer to
that point of tangency you might zoom in so much that you can't tell the difference
for a bit of the domain a little bit to the left and right of the point of tangency you
won't be able to tell the difference between the slope of the line and the slope of the end of the
function at that value indeed that's exactly kind of what's going on as we go through the derivative
process but you can also then estimate the values of y so there's this idea of called linearization
where we can use a linear functional line to estimate not just the slope of a function
near a point of tangency but actually its value as well that's one way you use tangent lines in
calculus another way that you use tangent lines is something called newton's method
where you're approximating zeros so we have like say we have an x-intercept here
and maybe it's such that we can't actually find the x-intercept because the function is too
complicated well we can take this idea of newton methods with some small points and come up with
some tangent lines that will working with the tangent line we can start you know
hitting around that x-axis and estimate the x-intercept of our function using
tangent lines and it's probably going to be a series of small steps of building new tangent
lines on top of other ones it's called newton's method there's also the euler's method spelled
e-u-l-e-r-s euler's method i always want to say it's euler's euler's method which is a way of
estimating the solution to a differential equation maybe if you and this is way beyond where we're at
right now in calculus but we're eventually going to learn how to solve a few types of differential
equations well differential equations can be quite difficult and may be impossible to solve
so we can use a series of taking small steps and creating tangent lines along the way or just lines
to estimate the shapes we'll be able to estimate the shape of a solution or estimate the function
with something called euler's method for a differential equation so
even though it's just the equation of a line something we've known how to do
and and build or create since algebra one it still has some really powerful applications in calculus
in the description below i'll put links to those lessons so you can get ahead maybe and just kind
of see how all this is going to play out but for now i'm mr true bam go do your homework
you