Related Rates - The Shadow Problem

This calculus video tutorial explains how to solve the shadow problem in related rates. A 6ft man walks away from a streetlight that is 21 feet above the ground at a rate of 3ft/s. At what rate is the length of the shadow changing?
Related Rates - Free Formula Sheet:
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Calculus 1 Final Exam Review:
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Introduction to Limits:
• Calculus 1 - Introduct...
Derivatives - Fast Review:
• Calculus 1 - Derivatives
Introduction to Related Rates:
• Introduction to Relate...
Derivative Notations:
• dy/dx, d/dx, and dy/dt...
Related Rates - The Cube:
• Related Rate Problems ...
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Inflated Balloon & Melting Snowball:
• Related Rates - Inflat...
Gravel Dumped into Conical Tank:
• Related Rates - Gravel...
Related Rates - Area of a Triangle:
• Related Rates - Area o...
Related Rates - The Ladder Problem:
• Related Rates - The La...
Related Rates - The Distance Problem:
• Related Rates - Distan...
____________________________________
Related Rates - Airplane Problems:
• Related Rates - Airpla...
Related Rates - The Shadow Problem:
• Related Rates - The Sh...
Related Rates - The Baseball Diamond Problem:
• Related Rates - The Ba...
Related Rates - The Angle of Elevation Problem:
• Related Rates - Angle ...
Related Rates - More Practice Problems:
• Related Rates - Conica...
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[@TheOrganicChemistryTutor]

Related Rates - Free Formula Sheet: bit.ly/48hJymz Final Exams and Video Playlists: www.video-tutor.net/ Calculus 1 Final Exam Review: kzbin.info/www/bejne/jZ6lq6B-p9pqbtk Next Video: kzbin.info/www/bejne/sKrXqHmJrq1jlZI

[@liezyldavila306]

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.

[@nicholasaguirre8764]

Thank you for explaining that!

[@vardaankola2185]

I think you meant part b. Thanks!

[@danielshamlian2800]

I hope this makes sense to explain where the 8 would go mathematically -- if you're thinking in terms of derivatives, the 8 ft would be the constant in the equation, so the derivative of the 8 would be "0," which is why it wouldn't affect the rate of change.

[@Diego_Cabrera]

Thank you so much!

[@wakuwaku3190]

THANK YOU’

[@shr1mpy194]

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

[@marcusj3287]

Facts. Got a huge exam tmrw

[@AR-fh4qu]

@@marcusj3287 how was ur exam lol

[@thebainfiles]

facts.

[@popkpoplyrics8366]

You are having point

[@toastycrystaleclxpse3423]

@@AR-fh4qu We used to be in the same class. It went poorly

[@merkov8715]

I can't even believe I am now capable of solving these type of problems on my own. Thanks a lot man.

[@hansonji9578]

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

[@jackytang5668]

Same

[@ptyptypty3]

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...

[@BrentArtuch]

I'm proud of myself for seeing this before reading your comment.

[@ta_ogboy9998]

Yeah no idea why he set up L = x + s but not dL/dt = dx/dt + ds/dt

[@JonathanBrown-emboldened]

5 years later and you are still saving lives!

[@brettmartin536]

you r actually so good at explaining

[@MuffinManWasTaken]

@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!

[@MuffinManWasTaken]

@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!

[@josephbyron7004]

@@MuffinManWasTaken wtf

[@MuffinManWasTaken]

@@josephbyron7004 comments got deleted or something

[@HKS-Digital]

Since my assignment question was asking for only part b, I realized I could write s = L - x -> 21/L = 6/(L-x) -> 15L = 21x , differentiate d/dt ->15(dL/dt) = 21(dx/dt) and solve for dL/dt by plugging in dx/dt that was given in the question. Thanks for video!

[@georgesadler7830]

MR. Organic Chemistry Tutor, this is another solid explanation of Shadow Problems in the Related Rates section of Calculus One. Calculus has some great Related Rates problem in its library. This is an error free video/lecture on KZbin TV with the Organic Chemistry Tutor.

[@sabrinabong9320]

saved me. i owe you my firstborn child.

[@Pushed2InsanityYT]

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 :)

[@gabriel9668]

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.

[@joecrow2847]

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

[@sin_2]

For those wondering, dL/dt is enough to calculate everything. If we find how fast the tip of the shadow is changing, we can just extract 3 ft/s (because shadow gets smaller from that side) to find ds/dt.

[@Nathanator]

Got an exam in 15 minutes, this video was more helpful than my teacher’s lessons, thank you

[@bluefire6470]

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

[@baisfaqirzai1265]

I wanted to drop my calculus class because of these, look how beautiful and easy you just made it for me. Thanks alot

[@darthTwin6]

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.

[@joshuayoo1306]

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

[@victornnah3920]

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

[@ThatTechyStoat]

Genius. God bless you sir. Thank you for existing 🎉

[@s--b]

Thank you for being the one video that actually explained it

[@danielbyun]

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.

[@keithhowen9493]

Daniel Byun yeah i was thinking the same thing too

[@evanpage6569]

@@sportsarelife2211 doesnt mean he is right

[@LavenderReview]

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

[@JasonVaysberg]

The key words in the problem are "constant rate"

[@vrichaprentado5989]

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

[@Fyreshield]

We never went over this kind of problem when doing related rates in my calc 1 class so when I saw it on a test I had no idea how to even set it up

[@Say_Tin]

This is more helpful than my calculus course right now lmao

[@grandpa-30]

You save me every time I go and do math homework

[@stevennotthe2997]

I have my unit test tomorrow and this was a big help as on the homework i could not understand this kind of problem(the calculus was easy i just struggled with how to set it up. It is common in my class to understand calculus but forget basic algebra)

[@LiveGoodPhilippines888]

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. 👍👆

[@mister_allmond]

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.

[@louishennick12]

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

[@18lucky17]

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

[@joyfulpianomt]

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?

[@queenismymiddlename2115]

Yes, those datas are not necessary in solving shadow problems

[@justinsantos5751]

The rate is the same no matter the distance

[@austinV321]

@@justinsantos5751 then wouldnt ds/dt be the same as dl/dt? im just confused

[@justinsantos5751]

@@austinV321 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.

[@austinV321]

@@justinsantos5751 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!

[@Kasleryoutube]

Beautiful, you’re so good at explaining

[@samriddhamalla6046]

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

[@rjarcayna5740]

yeah me too

[@vide0gameCaster]

It is indeed a classic problem of Related Rates

[@raniasafira5255]

bro same

[@deleteaman]

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.

[@atholerskine719]

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

[@Nerdyplaneresident]

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?

[@boxmeister3059]

Indeed you can! You end up with the same answer :)

[@MichaelSmith-nc9iy]

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?

[@systematic3550]

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

[@peelysl]

he is walking at a constant speed

[@flour5135]

This was my question. Calculus is hard enough we done need excess info 😭😭😭

[@ripsnortmcgee8830]

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?

[@isaiahlobert7843]

This man is doing Gods work

[@codeintherough]

In what class do we learn that length of shadow is determined by the hypotenus?

[@ahmedghanoum8136]

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

[@matt10y27]

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

[@mar0nat0r31]

Absolutely GOATED, thank you so much for the help🙏

[@cellmany8160]

What is tge use of 8ft on the the first part

[@michaelmyers6098]

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?

[@ZakkaWakka002]

this man saves my grade repeatedly

[@adriangibbs6349]

Outstanding. I have a better intuitive understanding now.

[@porphyrinhamburger6996]

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

[@charliehaberstock9878]

No it doesn’t you get 3.2

[@neillawrence4198]

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).

[@anurag02075]

Helped alot , thank you for this video !

[@bryanclark5805]

this can't be correct. why didn't you use the 8 from part a or the 10 from part b.

[@ericrenner8761]

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

[@deagon5535]

I got my calc final tomorrow wish me luck

[@AC-tn4it]

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

[@saerin9181]

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

[@anjokii]

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.

[@eixn728]

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

[@blackdragongaming3004]

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

[@Shovon-17-02]

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

[@thomaseubank1503]

I wish you were my Calculus Professor.

[@jackjeffries1369]

Thank you so much seriously this makes so much sense

[@advikpuri2385]

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

[@daisypawar5625]

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

[@thesakuraprincess8393]

Any alternative solutions with those values didn't used?

[@esra423]

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

[@JasonVaysberg]

@Matin Nawabi adults can't study?

[@mastershrew82]

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

[@benjaminbeaudry8642]

Thank you sir! Fantastic explanation.

[@moosegoose1282]

whats the point of 10 if we never use it?

[@justinmanangan9968]

its kind of used as distract information

[@d3athg4ming20]

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

[@ivanchen2170]

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

[@SM-qn7sr]

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

[@ETHANMARCOARREZA]

Thanks man

[@jt3539]

Why does this dude sound just like Mark Wahlberg?

[@bbyskittles91]

He sounds black to me.

[@DP-nf8fk]

Not even close

[@sanjurotsubaki3683]

All because of good vibrations

[@Raaaahhhhbbbie]

"Hey streetlight, howsit goin'? how's ya mother?"

[@BrotherofWord]

Wahlberg is to Boston accent what Organic Chemist is to...c) Brooklyn accent

[@mathwithnour9780]

You are my first choice ... creative

[@boredomgotmehere]

Hi, at 5:20 , how did you get 15ds/dt = 6dx/dt ?

[@dakotaburwell]

Take the derivative of both sides

[@fredericksquirlly]

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.

[@lisaliz6564]

GOD BLESS THIS MAN

[@Mymelodyismy123]

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

[@DrArtorias]

awesome explanation! really appreciate it!!

[@DarkMatterr]

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.a9021]

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)

[@von970]

@@a.a9021 and a year later this helped at least one other person. Thanks for comment! Very insightful.

[@Sungka101]

your drawing is very human

[@water875]

😂

[@roybautista160]

thank you gob bless you idol❤

[@davidstevensom4330]

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.

[@johnprice4420]

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.

[@SM-qn7sr]

@@johnprice4420 thank you. That actually clarified one of the confusion I had with these shadow problems

[@souravbairagya3073]

Thank you so much!!

[@dulan_fernando]

amazing video, helped a lot!!!

[@israelsumagang4801]

Omg i love you so much you just saved me

[@adambaxter8865]

Thanks so much dude you made it so clear

[@gabbydavis3034]

can anyone explain why he was able to multiply 1/6 to each side

[@amieyang809]

since both sides have a 6 or is being multiplied by a 6, he is essentially dividing both sides to simplify the equation to the simplest form.

[@randomstuff940]

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

[@dedede5586]

THANK YOU 🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏

[@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 😂

[@amirosman6342]

Very very nice thank you veryvery match

[@adeeshastha2479]

Thankyou mr man

[@i_sometimes_leave_comments]

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

[@karok.3626]

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

[@alexan406]

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.

[@richardvillanueva1782]

So great !!

[@AssasinSpike]

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

[@jhomarivillanueva7818]

I hope i pass my exam today

[@surfingdaweb1975]

thank you so much

[@amandaliss]

thank you for this!

[@elisabethtausig7293]

THANK YOU

[@g4m3rb0y123]

Is him being 8 feet away insignificant for part a?

[@LadderFromMGS3]

Thank you!