I haven’t seen anyone put this yet, so I will. A simpler process can be used to ascertain the rate of the movement of the shadow tip:

Bc the length, L is the sum of x and s. dL/dt, the rate of change of the length with respect to time is the sum of the rates of s and x with respect to time.
L = s+x ->differentiate-> dL/dt = ds/dt + dx/dt (rule: derivative of sum is sum of derivatives)

This makes more sense to use bc after completing part A, we already have both components of this sum, dx/dt and ds/dt. This sum is 3 + 6/5 or 21/5.

It’s a faster method bc you simply compute one sum.

Lastly, I want to make it clear why the derivative of the entire length of the shape created tells us this (or at least my reasoning). The lamppost is a fixed point, so the combined length is dependent on the position of the shadow tip. Focusing just on the shadow tip and the lamppost, as the tip moves, the length expands.
Thus, finding the rate of change of the length gives the rate of movement of this tip.

Joseph Almaznaai

If you divide by 6s, you'll end up with

21/6 = x/s

Multiply by s from here and you get

21/6s = x
OR
(21/6)s^-1 = x

Differentiate

-(21/12)s^-2 * ds/dt = dx/dt

There isn't enough information to solve for ds/dt in this format, so the teacher chose a simpler method that's easier to use.

All comments from YouTube:

The Organic Chemistry Tutor

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Liezyl Davila

4y↑202

for those who are asking why 8ft from the street light is not included in part a. it turns out that as long as the man is moving at a constant rate, the rate of his shadow will also remain constant at 6/5 ft/sec. regardless of how
far away he is from the street light.

Nicholas Aguirre

3y↑6

Thank you for explaining that!

Vega

4y↑9

Thanks a lot....i wasn't able to solve the first part by myself, but after understanding it, i was able to solve the second part by myself :) Whenever you guys solve these related rates problems, your first step should be to make a diagram and find an equation which relates both the variables which are changing w.r.t time! Good Luck :)

Philip Y

2y↑9

for what it's worth.. the the rate at which the tip of the shadow is moving is equal to the Given velocity of the man walking PLUS the rate at which the shadow is moving. So, dx/dt + ds/dt is all you need. NO need to calculate using dL/dt. This was an extra step that was worthwhile to see. It's always good to see extra steps to learn as much about how a problem can be solved. GREAT VIDEO as USUAL.. You're the best!!! thanks for all the effort you've invested in all your videos...

SHR1MPy

3y↑92

Bruh I learn more from u in 10 minutes than I do from my teacher in 10 classes. TYSM

MarcusJ

3y↑10

Facts. Got a huge exam tmrw

AR

2y↑3

@MarcusJ how was ur exam lol

Muath Ahmed

2y↑8

This dilemma will never end (tha fact that teachers sometimes make it more complicated)

KPX_ Ebreezyy

facts.

Gabriel

4y↑53

How do you know math so well? And how are you able to teach is so well? I've watched several of your videos and found each one very helpful. Thank you! I also comment and like every video of yours I watched, you know, for the youtube algorithm, but your videos are always the first to pop up, so I guess I am doing it because I really appreciate what you do.

Bluefire64

Amazing insight. Randomely encountered a similar problem and had no clue what to do. Thank you so much.

Yal Lav 2.7

5y↑1

Thank you bro . You are so precise in explaining . Your way of explanation is way to good for us people who is not really good in direct solving. Thank you for explaining it step by step. 👍👆

Joe Crow

4y↑2

As a math tutor,I have been looking around for this concept. It is nice to see it here

Benjamin Beaudry

Thank you sir! Fantastic explanation.

Brett Martin

6y↑186

you r actually so good at explaining

TheMuffinMan

Thomas Mack Dont know if anyone gives a damn but i think its a hacking client that steals your cookies. Lose all of your account stuff enjoy!

TheMuffinMan

Royal Tanner Dont know if anyone gives a damn but i think its a hacking client that steals your cookies. Lose all of your account stuff enjoy!

Joseph Byron

@TheMuffinMan wtf

TheMuffinMan

@Joseph Byron comments got deleted or something

George Sadler

MR. Organic Chemistry Tutor, this is another solid explanation of Shadow Problems in Related Rates. Calculus has some great Related Rates problem in its library.

Kashyap Maddala

First of all, thank you so much for this video. I really needed it.

And second, couldn't you just implicitly differentiate L = x + s and then substitute your values?

Nathan Heide

awesome explanation! really appreciate it!!

Michael Smith

4y↑14

Great video and explanation. After this video, setting up and working out this type of problem made a lot more sense. The only question I have, which is more for a deeper understanding is why are the two "distances from the light" (8 ft for part A and 10 ft for part B) irrelevant?

systematic

3y↑6

ikr, i guess the rate is constant no matter how far away

Jacob Lee

35w

he is walking at a constant speed

Hanson Ji

4y↑4

everytime I face a difficult problem in calculus, your always here for me. Your the real MVP

Jacky Tang

Same

S B

Thank you for being the one video that actually explained it

Kasler

5y↑4

Beautiful, you’re so good at explaining

Ripsnort McGee

3y↑3

I am sure this has been explored, but if 3d space is a "shadow" of a higher dimension, is this a way to approach an explanation of the accelerating expansion of the universe?

Matematika

2y↑2

Hello.
I can't understand the difference between the "tip" and "length" of the shadow. Don't they have to be the same things?

Samuel Kuran

1y↑1

Neil Lawrence

A little late to this video. When working with similar triangles my preferred method is to use proportional sides. In this case I say to myself 21 is to (x+s) as 6 is to s. 21 is to x+s is written 21/x+s, as means equals to, 6 is to s is written 6/s: 21/(x+s)=6/s. All part geometry ends up based on triangles (CNC programmer for 58 years).

Dulan F

amazing video, helped a lot!!!

Dark Matter

4y↑2

The explanation of why the 3 ft/sec was meant to be dx/dt was amazing, but I'm still wondering why it can't be dS/dt. If I recall how shadows work correctly, the farther you move from a light source, the longer your shadow will be, so S increases too. I need a precise method to determine which rate belongs to what variable.

a.a

3y↑11

I'm a year late, but maybe this will help someone else. It's because 3 ft/sec refers only to the rate at which the man is walking away from the light source, per the question. The shadow will indeed increase too, as you said, but not necessarily at the same rate that the man is walking at.

The length "x" between the man and the pole depends directly on the rate at which he is walking. The length "s" of the shadow depends directly on the angle at which the light is hitting the man, which is a fundamentally different thing, so the two will not necessarily change at the same rate. ("S" will change as the angle changes, and this angle will change as the length "x" changes, which is what makes this a related rates problem)

Von

40w↑1

@a.a and a year later this helped at least one other person. Thanks for comment! Very insightful.

Anurag

Helped alot , thank you for this video !

Adrian Gibbs

Outstanding. I have a better intuitive understanding now.

Athol Erskine

4y↑1

You are a genius. Keep up the good work. God bless

Matt10y

Oh my goodness thank you so much! Such a simple solution lol, this problem fried my brain.

Jack Jeffries

Thank you so much seriously this makes so much sense

Victor Nnah

3y↑1

This man is GOATED. I like the formula he used here, it just makes everything look easy. 🙏🔝

Eric Renner

I love how this guy is the sole reason why people graduate.

eixn

Thank you for the video! I was able to finish my homework problem cause of this so thank you very much. ^-^

Adam Baxter

5y↑2

Thanks so much dude you made it so clear

Random Stuff

I just thought of something crazy, since L in this problem is just x + s, dL/dt = dx/dt+ds/dt. And since we solved for ds/dt in part a, dL/dt = 3+1.2 = 4.2! You don't need to do all of that work if you just use your answer from part a I believe

The sakura princess

Any alternative solutions with those values didn't used?

Joshua Yoo

You teach these things like a GOD, you really deserve ur subscribers. In fact, I'm gonna sub right now!

Michael Myers

3y↑1

So when the problem asks you at which rate is the shadow moving when its 8ft away, you dont have to worry about the 8ft?

Isaiah lobert

This man is doing Gods work

Esra

6y↑16

Do you make your videos while you study? You sound very prepared.

JasonV12

Matin Nawabi adults can't study?

Grandpa -

3y↑1

You save me every time I go and do math homework

Mar0nat0r

Absolutely GOATED, thank you so much for the help🙏

Porphyrin Hamburger

5y↑3

After solving for ds/dt, couldn't you just add dx/dt to it to get dL/dt since both of the rates are independent of where the man is?

TheEpicPineapple56

Yeah, just tried it. That works too

Charlie Haberstock

No it doesn’t you get 3.2

Joyful Piano跳舞的手指

5y↑13

For question b, it was asking for the rate when he is 10 ft from the light, but the solution for part b did not include that 10 ft. Is it not related?

Queenis Mymiddlename

5y↑4

Yes, those datas are not necessary in solving shadow problems

Justin Santos

5y↑6

The rate is the same no matter the distance

Austin v123

@Justin Santos then wouldnt ds/dt be the same as dl/dt? im just confused

Justin Santos

@Austin v123 No because S is not the same as L. They are different variables. So their rate of change is not the same. What I said was that the rate ds/dt will be the same no matter the distance of the man is from the light. The rate of change of the shadow's length is constant and does not depend on any variables such as the man's distance from the light. At any distance of the man from the light, the rate of change of the shadow's length will always be 6/5 ft/s.

Austin v123

@Justin Santos yeah thats why I'm confused how is the answer different if the derivative is a constant if the only thing you're changing is distance from the pole? Sorry I'm honestly just confused not trying to be rude!

1 More Replies...

Daniel Byun

5y↑24

Can you assume ds/dt is 6/5 from part (a)? I thought it was only 6/5 since the length of his shadow changing was 8 ft. from the light not when he is 10 ft. from the light.

Keith Howen

Daniel Byun yeah i was thinking the same thing too

Evan Page

4y↑1

@robby mcdonnell doesnt mean he is right

Chloe Hall

4y↑6

Since he is moving away from the light at a constant rate, ds/dt will be the same at every length.

JasonV12

The key words in the problem are "constant rate"

Vrich Aprentado

For question (a), shouldnt 8 be used for the length of x?

BlackDragonGaming

5y↑2

omg you just saved me so much! thank you why can't you be my ap calc teacher. Is there any way I can dm you for more questions about related rates. I have a test next week and am lost aff

EmpyreanLightASMR

4:50 i figured everything out up to this point. but how on earth are you supposed to know that, at this point, you use implicit differentiation? I would've kept solving for s and then not known what to do afterwards 😂

AC

If L is s plus x, then dL could also just be ds +dx right?

Dev Pratap

39w

This same question came in IIT JEE Paper of India in 😮2023 😮

Richard Villanueva

So great !!

Karo K.

Will the answers ever be negative because of decreasing? or is it always positive

Alex An

3y↑1

Yes, it is possible for a rate of change to be negative. It just means that the value is decreasing. Let's say a ball is falling to the ground. The rate of change of the height of the ball would be negative since the height is decreasing.

David Stevensom

Can you please explain part B better? I don't understand why the rate of the tip of his shadow moving is L... Please. I need more help with related rates and with implicit differentiation.

John Price

2y↑3

This from a year ago, so not sure if this helps, but it's a good reference anyways. X is the distance the man is from the pole, and s is the distance the tip of the shadow is from the man. These things together are the distance from the pole to the tip of the shadow, or L. Thus, the rate of change for the tip of the shadow is going to be how fast it is moving away from the pole, which is dL/dt. I hope that helps who ever needs it.

S M

@John Price thank you. That actually clarified one of the confusion I had with these shadow problems

mastershrew82

3y↑4

I love how he just ends with, "Yeah these other values are irrelevant". Thank you god for sending this angel to explain work made by assholes

Amanda Liss

thank you for this!

King T Animations

1y↑1

math is only hard in college because they dont fully explain solutions. Saw a "solution" video for the same question provided by Web assign and they skip over some crucial steps/ dont explain how they got certain values

Sourav Bairagya

Thank you so much!!

Yobany Tix

6y↑1

What program are you using to make your videos?

PrabhatKC 70

Watch
MorgueOfficial
Anonymous Official
The crow House

Zakka Wakka

4y↑2

this man saves my grade repeatedly

daisy pawar

Was panting to get the solution for this question......😅 thanks for the explanation 😃

Lisa Liz

GOD BLESS THIS MAN

Charles Bautista

Is time rates and related rates the same??

Pong Lee

Is him being 8 feet away insignificant for part a?

deleteaman

5y↑18

lol I decided to go with
L'=x'+s'
or
dL/dt=dx/dt + ds/dt
got the same answer. Neat. two ways to solve it.

Mister Allmond

5y↑15

At 4:00 do we need to introduce tangent? we know those 2 ratios are equal because the two triangles are similar by AA similarity.

Louis Hennick

5y↑1

Doesn’t matter you can do either I’m pretty sure

18lucky17

4y↑3

No, but it is helpful for some people. By that logic, do we need this video?

Advik Puri

why don't we use 8ft and 10 ft? is it because it has a constant rate of change?

LadderFromMGS3

Thank you!

JohnT

Getting from a to b, why not just add? dL/dt = dx/dt + ds/dt

Samriddha Malla

5y↑33

i have a feeling i will be getting this problem in the final

RJ Arcayna

5y↑2

yeah me too

vide0gameCaster

5y↑8

It is indeed a classic problem of Related Rates

Rania Safira

bro same

Thomas Eubank

I wish you were my Calculus Professor.

Asher James Daniel

Majority of the comments of this video tells about how he saved them.

KRISHNENDU C P

Thanku sir

Waout

Thank you :)))

Random Ones

so we're not using the value of x at any point?

Kevin L

in 2021 this ques is still in my syllabus😶

Form Melon

Thanks!

Ishiita Pal

Thankyou so much

Sabbir Ahmed

34w

What is the source of this math?
I mean where you had found this math?plz tell me the source 😢

Amir Osman

5y↑1

Very very nice thank you veryvery match

surfing da web

5y↑1

thank you so much

Elisabeth Tausig

THANK YOU

Karl Baker

49w

thanks!

Israel Sumagang

Omg i love you so much you just saved me

naidely

Related Rates is so hard for no reason

Sae Rin

Help! why it is subtracted by 6s instead of dividing it, at 4:54?
Thanks!!!

Joseph Almaznaai

@_sf_editz

Usefulbro

Math with nour

You are my first choice ... creative

jessie garcia

so the 8ft and the 10ft have no purpose

elevation

4y↑3

So the fact that he’s 8ft away doesn’t matter in part a???

menandro aguinaldo

That's also may question 😅

Ahmed Ghanoum

Thank you but why we you didnot use the informastion that he is 8 fet from the light in part (a)?

seoulpeach

thank you

Deagon

I got my calc final tomorrow wish me luck

Brandon B

im crying thank you

moosegoose12

5y↑10

whats the point of 10 if we never use it?

Justin Manangan

5y↑8

its kind of used as distract information

D3ATH G4MING

Depends on the problem like the cylinder doesnt need a "3m deep"
But a circular cone needs that.

Dank Dale420

If I had a desire to learn this math I would come here, very nice. but as soon as you got to ds/dt I noped out.

Ivan Chen

hey why did u decide to use tangent, why not cosine or sine?

S M

Because to use sine or cosine you need to know the hypotenuse and the hypotenuse of these triangles are not defined

Osama Malik

Fucking genius

Caitlin Villanueva

5y↑2

not sure why im getting the wrong answer following this way?

Hassan Saeed

You are a god

Conner Oh

who else have a geometry test tomorrow.

Liam Kandel

legend

harkrish ahuja

Why is tip of shadow represented by L

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