Related Rates in Calculus Part 1 Calculus 1 AB

Related Rates in Calculus Part 1 Calculus 1 AB

BAM!!! Mr. Tarrou. In the next couple of videos
we are going to be talking about in calculus a topic called related rates. Remember we
started studying calculus with the idea of limits we were able to bring in the idea of
motion into all the static measurements we had all the way up to precalculus. One of
the examples we are going to take a look at is going to involve, a couple of them actually,
but the one specifically is going to involve a right triangle. And what I mean by related
rates is this… You are on the side of the road, the road is right there perpendicular….at
least it will be in our diagram. You look down the road and you see a car coming. You
can look at that car and really pretty much not have to move you body or your eyes or
whatever, you are not going to have to move very much to keep that car in view until it
gets close. The closer that car gets, the more that distance from you is reduced, the
faster you are going to have to rotate to keep it within your field of view. Now with
your eyes you kind of… it kind of just happens. (the rotation) I am a big racing fan and you
will see photographers standing by the side of the track… I have taken some of these
pictures and I will share some right here in the corner of the screen to share my work.
uh… You are standing there and all of a sudden it like shoooooow. And then you go
back and slow slow slow, and then you have to pan really fast to keep the car in your
view, vision of view, as it comes closer and closer and closer. That distance is somehow
related to how fast you have to rotate to keep the car within your view and get the
shot. So that is, that is the kind of idea…of course there are many real world scenarios.
These are real world problems, so every problem could be different. Hopefully your textbook
does not make every single question different. Otherwise you will probably not be completely
grasping the concept. But any rate, I am rambling. Lets take a look at the four steps and do
our examples over the next couple of videos. First step. Guidelines for Solving Related
Rates Problems. Identify all given values and all values you need to find and sketch
a diagram. That sketch will help you place the variables in the right place. Once you
get a numerical answer if you make a reasonable sketch, you can use that to determine if the
number that you found is reasonable. Write an equation involving all variables of interest.
Remember the rate of change variables will not be seen until you use the chain rule to
implicitly differentiate both sides of the equation with the respect to t. Now why t?
T is time. If you are driving a car you either measuring your speed in miles per hour or
kilometers per hour. Your motion problems are going to be related to time. How fast
am I going? How fast is something spinning? Like in trigonometry we studied linear and
angular speed. That was distance over time or amount of radians over time. Finally, substitute
in known values of change and solve for the remaining variables of interest. So let’s
get this out of the way, these directions, and start looking at some examples. And these
examples will be in increasing difficulty as we go through these. So if you are struggling
with the initial concept of course maybe the first couple of examples will be fine. But
they will become more involved and difficult as we go. So let’s see what they look like!!!
BAM!!! Here is our first example. An object is moving along the given path of y equals
parenthesis x minus one squared plus three with a horizontal velocity of sixteen meters
per second. Find the vertical velocity at x equals zero, x equals one, x equals two,
and x equals three. So I gave you the equation of a parabola with a vertex is at (1,3). Now
when we are to the left of the parabola’s vertex as we read this left to right, the
graph is falling of course. And the closer it gets to the vertex of the parabola… You
can see that a parabola comes down very sharply, the velocity is very negative… very quickly
falling. And then it starts to slow down until it flattens out and the horizontal speed and
velocity is going to be zero at the vertex. And then it starts to accelerate back up the
other side of the parabola. So the vertical velocity is going to be changing, it is going
to be different at these values of x. We also have a vertex at the x value of one. It is
actually at one three. So we should be expecting this velocity to be negative, this vertical
speed to be zero, this vertical velocity to be positive, and because of the symmetry of
the parabola they should be the same velocity only with a different sign…just to kind
of let you see that. And then at the x value of three as we start to curve upward the vertical
speed should be accelerating. Now I have given you the values of, I have already sketched
it for you and give you the equation, and I have given you the values of uh… 1,2,
and 3 for x. I also gave you here where it says the horizontal velocity is sixteen meters
per second. Well that is a rate of change. Now it is rate of change over time. So this
is going to be the derivative of x with respect to time, or the horizontal distance because
the horizontal axis is a well.. that is how fast you are moving along the x axis with
respect to time. So dx over dt is going to be sixteen meters per second. Right now I
have got a parabola and its equation. I don’t have variables related to change. I don’t
have any derivatives so we are going to take this equation and find the derivative of both
sides implicitly in respect to time. The derivative of y, well if it was the derivative of y with
respect to y it would just be one. If it was the derivative of y with the respect to x
like we have been doing recently it would be dx over dy. But now it is relation to time
or respect to time so this is going to be the chain rule. We are going to take the power
of one and drop it down front, this exponent comes to be zero, so this just going to be
dy over dt. This equals…now here we can see more like a natural chain rule it is an
inside function and a power of two. It will be the two out front, the general power rule,
two times x minus one. This power reduces to a power of one which we don’t need to write.
So I am not going to write because it will look like a prime symbol. Then we need to
find the derivative of the inside function. Well if you were finding the derivative with
respect to x this would just be one. That is a constant so it will just be zero. The
derivative of x with the respect to x is dx over dx which is one. But we are taking the
derivative with respect to time, so this is also….the x is not just going to go away
like it has with our other implicit differentiation. This is going to be the derivative of x with
respect to time. Well now this is again like I said we are going to go from easy to hard.
So this is it. This is what I have got. We are looking for the vertical speed, so I am
looking for dy over dt. Change in y, that is the vertical change…that is vertical
velocity/speed. dx over dt, that is the 16 so all we really have left to do is…. Let
me write this final answer. Dy over dt is equal to two times x minus one times 16. Or
that is going to be two time sixteen which is 32 times x minus one. So that is dy over
dt. Now that I have got my equation solved for my variable of interest. I will just plug
in my four values of x and see what we get. So, let’s say x equals zero. Then dy over
dt is equal to thirty two times zero minus one, or negative thirty two meters per second.
Let me make sure I am not saying something silly. And this should be negative because
the graph is falling to the right. Lets let x equal one. That is where the vertex is so
we should be getting an answer of zero. dy over dt is equal to 32 times one minus one
and indeed this does come out to be zero which it should of. We have symmetry in this parabola
so when we go to use the value of two we should have that thirty two show up again but only
be positive this time. We are rising up to the right. So x is equal to two. So dy over
dt is equal to thirty two times two minus one and indeed that is thirty two meters per
second. Finally when we let x equal three, we have dy over dt is equal to thirty two
times three minus one, or that is thirty two time two which is sixty four meters per second.
And that is the end of our first example. NEXT!!! For our next example we have a problem
that sounds very similar. Find the rate of change of the the distance between the origin
and a moving point on the graph y equals one half x squared plus two if dx over dt, the
horizontal velocity… the rate of change horizontally is equal to four units per seconds
or whatever. Ok, well first of all I have not given us any specific values of x. We
are not going to find the rate of change at any specific values of x. We are just going
to find an equation that describes the rate of change. So that will be a little bit different.
But the biggest deal here that the previous example the rate of vertical velocity. It
asked for how quickly the point is falling or rising with respect to the y axis, dy over
dt. Here I want the rate of change between a point and the origin. I don’t want to find
a vertical distance. So I don’t need or want to see dy over dt because that is just the
vertical rate of change. I want the rate of change from the point to the origin, it is
a slanted line. So, I can’t just take this parabola and find the derivative of both sides
implicitly because I am going to have dx over dt on one side and dy over dt on the other.
I don’t want the vertical rate of change. I want the rate of change of the distance
between the point and the origin. Now I keep saying the distance right so that means I
need to use the distance formula. I am going to use the equation of the parabola but I
want the rate of change between the point and the origin. That slanted line requires
that to find that distance we need to use the distance formula. D is equal to the square
root of x sub two, or the x from the point, minus the x sub one from the origin. So we
have x minus zero squared plus y minus zero squared. That is going to simplify to be the
square root of x squared plus y squared. This describes the distance between any point,
excuse me this point, and the origin. Ok. I can write this as d equals the square root
of x squared plus y squared and I don’t like finding derivative with the square root symbol.
So I can take this and write it with a set of parenthesis with all this raised to the
one half power. I can find the derivative of this equation implicitly. But there is
a couple of problems with there. One, I will have the change in distance with respect to
time, or d over dt. I will have through the chain rule dx over dt, which I am given. But
I am going to dy over dt and no where in my problem am I given a rate of vertical change.
Secondly if I go and try to find the derivative of this, what am I doing just completely ignoring
the equation of the parabola? I am completely ignoring the actual path that point is following?
I have to put these together. I have to put the distance formula together with formula
of the parabola so they are actually related to each other. I need to keep in mind I am
only given dx over dt, I am only given the horizontal velocity. So I don’t want to see
a dy over dt after I am done finding the derivative because I don’t know what that is. So we need
to do some kind of substitution with this equation of the parabola, after all this is
the path the object is following, with the distance formula. I need to do this in such
a way that I only have x’s left so that after I find the derivative I will only have dx
over dt. That is what I am given and that is what I have to use. So that means that
we are going to before we find the derivative, take the y out and replace it with what it
is equal to from the function. This is going to be equal to the square root of x squared
plus one half x squared plus two squared. Ok, let me make sure I am not writing something
silly here and I do the rest of the problem with a mistake in it. I have to work this
out a little bit and then simplify it. D is equal to the square root of x squared plus
hum…we are going to need to foil this binomial out. Do really really need to do that now?
We don’t have to do that now but I my notes have it where I have foiled this out. Let
go ahead and foil that. We have one half x squared plus two, distribute it out, one half
x squared plus two. That is going to be one half x squared times one half x squared which
is one fourth x to the fourth. One half of two is one so we have plus one x squared.
And another one x squared… and then two times two is equal to four. This is going
to be one fourth x to the fourth power plus two x squared plus four. Lets simplify this
up some more. This is one and here is two so adding like terms we have d is equal to
the square root of one fourth x to the fourth power plus three x squared plus four. We are
going to find the derivative. I don’t know why I keep writing this radical sign because
it does not make finding the derivative that friendly so let me erase that and write that
as the one half power. Now we are going to find the derivative of both sides with the
respect to t. So we get d over dt is equal to…this is the general power rule… so
we are going take the power of 1/2 and move it out front. Now 1/2 times one fourth x to
the fourth power plus three x squared plus four, reducing this by one we have the negative
one half power. Then inside we have got one fourth times four is one. So we have x to
the third. Now we are now finding the derivative of the inside function and remember we are
finding the derivative with respect to t not x. This is not going to just be x to the third.
This is going to be x to the third times dx/dt. Plus bring down this power of two, three times
two is six x dx/dt… the chain rule there. This is a constant so the derivative is zero.
Now we have got… Lets see here. We have got the derivative with respect to t of…
this is going to come down to the bottom… so we have one over two times the square root
of one fourth x to the fourth plus three x squared plus four times x cubed dx/dt plus
six x dx/dt. Sorry for the interruption there. So now we have one over two times the square
root of all this stuff. Over here we have x cubed dx/dt plus six x dx/dt. We have a
lot of variables going on here but we know what dx/dt is, it is equal to four. I can
take these values out… I am going to erase this because I am running out of room. We
are going to take out the dx/dt and replace it with what it is equal to which is four.
We plug in those values of four. Now we can write our final answer which is going to be
d over dt. The change in the distance between this point and the origin with respect to
time is going to be equal to four x cubed plus twenty four x over two times the square
root of one fourth x to the fourth plus three x squared plus four. Now this radical sign
is grouping all that stuff together in the denominator and the factor of two. Both of
the terms in the numerator are even so we can cancel a two out of here. Our final reduced
answer… four divided by two is two… two x to the third plus twelve x over the square
root of one fourth x to the fourth plus three x squared plus four. We now have d over dt,
the change in the distance between that point on the parabola and the origin given to us
by this expression. Now we can take any value of x we like, plug it into that expression….we
can plug in x value from the curves domain we like… and it will tell us how quickly
that point is…what the rate of change is.. how quickly that point is approaching or leaving
the origin with that horizontal speed of being four units of measure per what ever time you
are using in the problem.

45 thoughts on “Related Rates in Calculus Part 1 Calculus 1 AB

  • August 8, 2012 at 4:39 am

    i like the pictures you took of the race cars they would make good wallpapers

  • August 8, 2012 at 10:22 am

    Thank you very much:):):)

  • November 5, 2012 at 2:48 pm

    I am sorry I missed your comment. Do you still need help? I am extremely busy at work and… 🙁

  • November 11, 2012 at 3:48 am

    THANK YOU SO MUCH!!!!!!!!!!

  • November 11, 2012 at 9:24 pm

    YOU ARE SOOOOO WELCOME! Thanks for watching and please share with your friends and fellow students! Thanks:D

  • November 12, 2012 at 4:02 pm

    Thank GOD you do a video on related rates. Favoriteeeee teacher ever.

  • November 12, 2012 at 4:58 pm

    Thank you, I love being a favoriteeeeee teacher! I am slowly building my calc library, so I hope you will find more useful videos to watch and share:)

  • January 4, 2013 at 4:09 am

    Thank you. Photography is a pretty major hobby of mine:) Thank you for watching as well!

  • January 27, 2013 at 2:34 pm

    That is a good question…which I am not sure how to answer. There are no questions in the textbook I use like that…and I have not been looking at any Calculus for many months as I have building my Geometry playlist for a new class I am teaching. I don't teach Calc to a live class, so I don't live and breath it everyday. I look forward to getting back to calc in a couple of months. Of hand, you can probably have the rate of water flow be negative and keep the cone in the same orientation.

  • February 19, 2013 at 6:43 am

    Hello! Biochemistry major form Dallas here. I made my first "perfect" score on a calculus exam last week and I have you to thank Mr. Tarrou!!! Please keep it up!

  • February 21, 2013 at 11:15 am

    WOW, "great" job Biochemistry major from Dallas! Sorry I only have the first 4 chapters in Calc done…but after my Geometry library is complete I will be moving on to Calc (that won't be in time to help you but I sure do appreciate the nice comment and your viewership!

  • October 18, 2013 at 8:03 pm

    As well you should be confused, you are absolutely correct!!! I went off on a tangent there and apparently went out of my head as well. I am going to trim the last example off this video and remake a video for that example. I am very sorry for the error.

  • October 20, 2013 at 1:18 pm

    Your videos got me through Trig and Precalc, and they're the only reason I have an A in calc now! Thank you so much! Derivatives were so hard for me in the beginning, and now they make a ton of sense.

  • October 20, 2013 at 3:32 pm

    You are welcome…and thank you for learning from my videos. You played a big part in those grades…after all, you were the one who reached out for extra help…I was just lucky enough for you to choose my channel to learn from!!! Thanks for your continued support:)

  • October 31, 2013 at 3:56 am

    What does your shirt say?

  • October 31, 2013 at 6:05 pm

    Try not to let your mind wander, it's much too small to be outside by itself.

  • January 10, 2014 at 12:31 am

    Thank God for this video

  • January 24, 2014 at 7:25 am

    Thank you. This video was a good refreshment for when I go back to school, and take calc. 2. Thanx. your an awsome teacher.

  • January 24, 2014 at 7:26 am

    lol I just realized I will not come across related rates problems hahahaha but non the less a good refreshement.

  • February 12, 2014 at 5:04 pm

    Thank you, you have helped me a lot, you are my favorite one and you deserve more likes * I love you * ☹❤️❤️

  • March 4, 2014 at 12:38 am

    Awesome! My biggest passion is building and racing cars! The entire reason i am working to my mechanical engineering degree is to build and race cars! Nice pictures!

  • March 6, 2014 at 1:58 am

    Sir do you have solid geometry/mensuration videos? thanks in advance… these calculus videos has helped me a lot by the way. yeah i completely agree on your comment on 2:03, its as if the textbooks does not want to share the information at all.

  • March 15, 2014 at 5:16 am

    Thanks sir this is awesome! i have told all my cousins in middle east about ure channel 🙂 and they love it.

  • March 25, 2014 at 6:55 am

    Prof Rob hands down to you, I go to school but the real learning is in your channel. pretty much you teaching me calculus instead my teachers. Greetings from Mexico! 

  • April 16, 2014 at 6:40 am

    Man you're awesome. You're trig videos got me an A and now getting me an A in calc 1. Where do you teach?

  • July 11, 2014 at 5:43 am

    Mr. Tarrou do you have videos for Fourier Series? im really dumb at it and your teaching skill are legend! BAM:)

  • September 29, 2014 at 3:23 am

    thank you for helping me in my ap calculus class. you dont know how much of great help you are.

  • October 24, 2014 at 3:06 am

    Thank you for the videos! I had to sadly drop my Calculus 1 one class because I lacked the basics required. I love math, but it takes longer for me to learn it (at least I believe it does). My prof wasn't very approachable either lol 

  • November 5, 2014 at 9:35 pm

    Can you do a video on related rates where the cone is not inverted?

  • November 14, 2014 at 3:54 am

    What calculus textbook do you prescribe for your students? The class that I am in is using the Stewart's book and I really can't stand it.

    Love your videos, they are what has gotten me this far.  Thank you!

  • December 6, 2014 at 10:08 pm

    Your videos are so helpful ! Thank you I wish my teacher was as enthusiastic and helpful as you lol 🙂

  • December 9, 2014 at 3:28 am

    I love you videos and are my go to for help. So please keep it up! Question though, you say you enjoy watching car racing, where did you go to take those pics? And I also have to ask, do you have a fun car to drive? 🙂

  • January 20, 2015 at 8:20 pm

    Related Rates in Calculus now with Closed Captions.  WOOHOO!!!

  • April 15, 2015 at 5:52 pm

    Oh man! THANKS AGAIN :D.  It's easier than i thought, but my text book makes it seem like rocket science.

  • July 23, 2015 at 9:31 pm

    Another great lesson! i'm just loving math. One quick thing though: In the first example when the object reaches the vertex you said: "the horizontal speed (velocity) is going to be 0". But that conclusion would contradict the given data of the horizontal velocity being 16 m/sec, which implies that is staying constant along the entire path. So, correct me if i'm wrong but my logic says that the vertical velocity is actually reaching 0 not the horizontal velocity. And also in the second example, i'm deducing, if i understand correctly, that since the horizontal velocity is positive and constant, that would imply that the moving point moves only in one direction which is from left to right – and in that case, the sign (positive or negative) of [d(distance)/dt] when we evaluate it for an arbitrary position would tells us with certainty if the distance between the point and the origin is actually increasing or decreasing. Hope you can help me with that Mr.Tarrou, and as always, loving the videos!

  • October 2, 2015 at 9:04 pm

    car lover! thats awesome!

  • October 15, 2015 at 3:51 am

    i love you and your new haircut.. my savior <3 lol
    thank you so much

  • May 29, 2016 at 8:35 pm

    I am confused at 19:50 I thought you cannot simply terms in both the numerator and denominator if the terms on either are being added/subtracted. For example, you cannot simplify (4+5x)/2. You can't simplify the two and the four.

    Can someone explain to me how Rob was able to do it in this case?

  • February 10, 2017 at 3:34 pm

    Mr Tarrou,
    Im taking my first calculus class at 37. I started refreshing my math skills a year ago on some of you algebra 1 series. Many of your videos have helped me reiterate the lessons from classroom and books. My sincere gratitude to you. Your effort and love of math are legendary!

  • February 24, 2017 at 7:15 am

    In question 2 @ 10:18 . Your lecture on how the question worked is a little shakey, but tell me if I understood it wrong because it seems I took a slightly different approach to get the same result

    You are given the dx/dt and with that and the f(x), you can find dy/dt. Since you have both dx/dt and dy/dt , it essentially acts as 'a' and 'b' in a pythagorean triangle. With a^2 and b^2 you can find c^2 which is the slanted line on the graph. Im missing something in my understanding, im not sure what. Im not sure if its right either. (Love the vids btw, been subbed to this channel since precalc 11)

  • May 9, 2017 at 2:40 am

    I have my calc exam tomorrow and this really helped

  • March 22, 2018 at 3:01 am

    you must be on the council of math gods. I think I'm praying to you before and after I take my exams.

  • August 25, 2018 at 9:58 am

    Really helpful! Thanks

  • December 26, 2018 at 8:36 pm


  • March 22, 2019 at 3:44 am

    And you're a photographer?? How many things can one guy be great at?? What's your favorite car?


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