Up next in 10
How To Use Related Rates & Implicit Differentiation To solve Real -life problems
Feb 19, 2025
In calculus, we often come across word problems that ask us to find how fast something is moving. These types of questions are called "related rates" problems, and they can be solved using a technique called implicit differentiation. We'll take a look at how this works in this video
Show More Show Less View Video Transcript
0:00
okay so basically
0:03
this section has to do a lot with
0:05
implicit differentiation this is why we
0:08
covered at first okay so assume you have
0:11
water in this stack and the tank has a
0:14
shape of a cone okay now how do we use
0:17
related rate
0:18
let's say you want to see how quick the
0:21
water the height of the water is
0:23
decreasing in terms of time or how quick
0:26
the radius of the water is decreasing in
0:30
terms of time or how quick the volume is
0:33
decrease in terms of time so that's when
0:35
we use related rate now in this case the
0:38
volume
0:39
of this Chronicle uh shape there is pi
0:43
over 3 r squared times H now each one of
0:46
them is going to be expressed in terms
0:47
of the time because the rate rate of
0:50
change is something change in terms of
0:52
time so if I'm finding
0:55
um
0:55
DV
0:57
over DT this is basically finding a
1:01
quick the volume of this content is
1:05
changing in terms of time you get that
1:07
right and I'm going to do the same thing
1:09
with the radius the radius will be d d r
1:11
over DT and the height will be d h over
1:14
DT so basically you have water in this
1:18
Con in this tank
1:20
right suppose this is water
1:23
H2O right and this is changing in terms
1:26
of time so how quick if you go down here
1:29
the height is going to diminish because
1:30
if the water is coming out of this tank
1:33
something is happening to the height
1:35
okay the height is changing in terms of
1:38
time so you want to know what is the
1:40
rate of change of the water
1:43
you can find the rate of 10 or the
1:44
volume of the water we can find the rate
1:46
of change of the height of the water we
1:49
can also find the rate of change of the
1:51
water itself because when you have the
1:53
water inside the water looks like a
1:55
circle okay there's a radius to it and
1:57
that radius is going to change in terms
1:59
of time so how do we use implicit
2:02
differentiation to do this this is what
2:04
we're going to talk about so let me show
2:06
you how to work it out
2:08
so you guys know implicit it's not like
2:11
you don't know it so how would you find
2:13
the volume here
2:14
so what's the derivative of V
2:18
where would it be
2:20
what do you think
2:24
and
2:25
what's the derivative of the volume in
2:27
terms of time
2:29
so is it
2:32
in terms of what variable time right the
2:36
volume is changing in terms of time what
2:38
would it be wait what's available for
2:40
time every one of them every one of them
2:42
right every single one of them is
2:45
changing in terms of time you get that
2:47
in this case we are finding DV over DT
2:50
everything is changing in terms of time
2:54
so one of them is not expressed in terms
2:56
of the other it's not like I'm finding
2:58
the rate of change of the volume in
2:59
terms of the radius I'm finding each one
3:02
of them in terms of what time
3:04
so the derivative of the volume will
3:07
just be DV
3:09
over DT does that make sense
3:12
right and now pi over 3 is a constant so
3:15
that stays there since it's modified
3:17
right now I have R square H so what am I
3:21
going to use here
3:23
what am I going to use product food
3:25
right so I'm going to use a product tool
3:27
so I'm going to put it on the side
3:30
and I'll apply the product rule okay so
3:33
I have R square h
3:35
and I want to find the derivative in
3:37
terms of time okay time is my my
3:39
variable here so first I'm going to find
3:43
the derivative of r squared which is
3:44
what
3:46
where's the river for R square two R
3:49
times what
3:51
thanks
3:52
are right
3:54
T times H right now the derivative of H
4:02
the H over DT times what
4:06
are R square right
4:09
r squared that makes sense because we're
4:11
doing the first one F Prime
4:14
G
4:16
plus G Prime f
4:18
so the derivative
4:19
will be 2R
4:22
times the r over DT
4:26
times H and then Plus
4:28
the H of a DT
4:31
times r squared right now
4:35
this is the the rate of change of the
4:37
volume I can stop here I don't have to
4:39
do any more transformation now the
4:41
question comes how would you know the
4:44
actual
4:46
um rate of change of the volume if
4:48
certain one of these things are given
4:49
let's assume that the radius
4:52
is two centimeters
4:54
and the rate of change
4:57
of the radius let's say it's like 0.5
5:00
centimeters per second
5:02
and then the height
5:04
is about two let's say about five
5:07
centimeters
5:09
and the rate of change of the height the
5:11
H of a DT
5:12
is about
5:15
0.8 centimeters
5:17
per second if I give you this which they
5:20
will give those will be given to you
5:22
they're not just going to leave you
5:24
hanging okay so if these are known how
5:27
do you find the derivative they change
5:28
the rate of change of the volume in
5:30
terms of time all you have to do now is
5:32
just plug it in does that make sense
5:35
right so if I want to know the rate of
5:37
change of the volume of the water in
5:40
terms of time I'm just going to replace
5:41
it given these so DV over DT
5:45
will equal 2R which is 2 times 5 right
5:49
times the r over DT which is 0.5
5:54
and then
5:56
I'll put this in brightness because I'm
5:57
multiply pi over three I don't want to
5:59
forget that right times the height the
6:01
height is five right and then Plus
6:05
DH over DT which is 0.8
6:10
times
6:11
r squared which is basically uh
6:15
also if two squared okay and then the
6:19
whole thing times five or three
6:22
now you can plug this in your calculator
6:23
and that's going to give you the rate of
6:25
change of the volume in terms of time
6:28
any questions
6:30
that make sense
6:31
so basically you can use the related
6:34
rate to solve all types of problems so
6:36
if I want to know how quick is um
6:41
I don't have anything on top of my hair
6:43
right now but we'll find some examples
6:45
in the book but this is really practical
6:46
okay and it's all based on what implicit
6:50
Deliverance yes I'm confused with the Dr
6:55
DT and DH or TT in the derivative yeah
6:59
all right
7:04
over DT not times h plus the H of TT
7:09
well we use a product called let's do
7:10
that again okay let me erase this and
7:12
let me do that let's do that again
7:14
so
7:16
let's just remember we are finding the
7:18
derivative in terms of what
7:20
DT right all right so if I have so in my
7:24
formula here right I have R square times
7:28
H now what is that what what does the
7:31
protocol say
7:32
the derivative of this if this is f
7:34
right and this is G we're going to do
7:37
what the derivative will be F Prime
7:39
Times G right plus G Prime times 1 F now
7:45
if this is f
7:47
all square what's the derivative of this
7:51
2R times what
7:57
right and if the derivative of H is
8:00
what's the derivative of H
8:03
one times
8:06
DH over DT right now if I were to use
8:09
the formula now right
8:10
so what would I equations
8:13
will be the derivative of this right
8:15
which is 2R
8:17
times d r over DT
8:20
times what times h
8:24
okay ah okay right and then Plus
8:29
the H of a DT right average continues to
8:33
Y there's three numbers on one side and
8:34
two numbers oh this is true because this
8:37
one was just H right times r squared
8:39
does that make sense Yeah so basically
8:41
this is how you use the related rate
8:43
okay now I'm gonna give you an example
8:45
and I'm gonna have you guys work on it
8:47
so let's assume that
8:49
you have an equation
9:06
y square equals
9:10
to
9:11
now y equals
9:13
x squared plus three
9:21
and then what you'll find
9:26
d y over DT
9:29
when
9:31
X is equal to one okay they want you to
9:34
find d y or DT
9:37
well X is equal to one when
9:43
the X over DT is also equal to two right
9:46
so basically what is this equation
9:49
represent here
9:52
what do I represent who knows
9:54
what kind of ship do we have here
9:57
yeah the parabola right opens upward
10:00
because this is positive positive now
10:03
they want to know how quick this
10:05
Parabola is changing in terms of time
10:06
when the derivative of x is 2 and when
10:10
the x is equal to one so how are we
10:12
going to solve this we need to find d y
10:14
over DT right
10:17
so
10:20
um
10:23
so what is d y of DT what do you think
10:27
what's the derivative of x squared
10:31
what's the derivative of x squared
10:35
2X and then one
10:38
times
10:41
DX over DT right what's the derivative
10:45
of three
10:48
zero so that's gone right so now how are
10:51
we going to find U over DT
10:53
all we have to do is just plug in the
10:56
values right we know what x is X is what
10:59
positive one okay
11:02
and what's DX over DT
11:05
2. so I'm just solving
11:08
so this is going to give you four
11:11
does that make sense
11:14
any questions so basically
11:17
once they give you the equation they
11:18
will give you some of the values and
11:20
then they want you to find
11:21
one of the variable in terms of the
11:23
other using the implicit differentiation
11:27
so let's work on one word problem here
11:29
and we can
11:31
we can continue on
11:37
so if you have your books
11:39
go to page 150.
11:42
page 150 and an example
11:45
two
11:54
if you have a DJ 150 page
11:57
page 150 example two
12:00
so ripples in a pond right so have you
12:04
gone to the river and through you throw
12:06
a rock in the water what happens what do
12:08
you see
12:12
I don't know if you guys have done it I
12:14
like I'd like to just sometimes keep a
12:16
rock or throw it in the water just to
12:18
see the concentric circles right have
12:21
you ever done it right now
12:23
maybe you're not much of an outdoor
12:25
person I'm not much but I like doing
12:27
that
12:28
right so the problem says that it's
12:31
purple huh no I can't
12:34
okay
12:47
oh you did seven yeah that's crazy
12:51
anyway so the question says the pebble
12:54
is dropped into a corn Pond right
12:55
causing reports in a form of concentric
12:58
circles as shown in figure 2.34 and the
13:02
radius R of the outer record is
13:04
increasing at a constant rate of one
13:06
foot per second
13:08
and when the radius is four feet at what
13:10
rate is a total area of the Disturbed
13:13
water is changing
13:15
okay
13:17
so what kind of figures do we have here
13:19
we just need to know first thing what
13:21
kind of figure do we have concentric
13:23
circles right it's a circle right so it
13:26
makes our life a lot easier
13:28
so we know that we're going to have
13:32
so let me kind of give you a quick
13:33
picture here so this is what you have
13:37
right
13:41
so this is like concentric Circle so you
13:44
drop the water in there you drop a a
13:47
pebble a rock right
13:49
and then causing because he's causing
13:51
ripples now they are saying that the
13:53
radius of the outer is increasing and a
13:56
rate of one foot per second
13:58
so we know that the variables uh R and A
14:02
are going to be related by the equation
14:04
a equals what pi times what r squared
14:08
right this is how you find the area of a
14:11
circle
14:12
okay and the question is saying how
14:15
quick
14:16
the water is changing the rate is
14:19
changing uh when the radius is four feet
14:22
so what is the rate of change of the
14:24
radius how much is it what do we know
14:28
so there you go so d r over DT is what
14:33
one right one foot per second
14:36
and what's the radius
14:39
four right they want us to know how
14:42
quick this area is changing so we are
14:45
finding what d a over DT does that make
14:48
sense
14:49
so the a of the DT
14:54
equals pi times what two R times d r
14:59
over DT and all you have to do now just
15:02
replace all by what four and D over DT
15:06
by one and just solve it that makes
15:09
sense
15:10
any question on that
15:12
the hardest part in this section is
15:14
really figuring out what's going on
15:16
understanding how to interpret the given
15:19
question okay in this case we have
15:21
concentric circles in that case we have
15:23
water that's coming out of a tank and in
15:26
some cases you may have whatever they
15:28
give you so this is basically all you
15:30
need to know so we're gonna work on some
15:33
problems in the book