Related Rates - The Shadow Problem
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- čas přidán 28. 02. 2018
- This calculus video tutorial explains how to solve the shadow problem in related rates. A 6ft man walks away from a street light that is 21 feet above the ground at a rate of 3ft/s. At what rate is the length of the shadow changing?
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:
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Gravel Dumped Into Conical Tank:
• Related Rates - Gravel...
Related Rates - Area of a Triangle:
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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:
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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.
Thank you for explaining that!
I think you meant part b. Thanks!
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.
Thank you so much!
THANK YOU’
Bruh I learn more from u in 10 minutes than I do from my teacher in 10 classes. TYSM
Facts. Got a huge exam tmrw
@@marcusj3287 how was ur exam lol
facts.
You are having point
@@AR-fh4qu We used to be in the same class. It went poorly
5 years later and you are still saving lives!
I can't even believe I am now capable of solving these type of problems on my own. Thanks a lot man.
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...
I'm proud of myself for seeing this before reading your comment.
you r actually so good at explaining
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@@josephbyron7004 comments got deleted or something
everytime I face a difficult problem in calculus, your always here for me. Your the real MVP
Same
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.
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 CZcams TV with the Organic Chemistry Tutor.
As a math tutor,I have been looking around for this concept. It is nice to see it here
Amazing insight. Randomely encountered a similar problem and had no clue what to do. Thank you so much.
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 :)
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. 👍👆
Thank you for being the one video that actually explained it
Beautiful, you’re so good at explaining
Genius. God bless you sir. Thank you for existing 🎉
This man is GOATED. I like the formula he used here, it just makes everything look easy. 🙏🔝
You teach these things like a GOD, you really deserve ur subscribers. In fact, I'm gonna sub right now!
Oh my goodness thank you so much! Such a simple solution lol, this problem fried my brain.
I wanted to drop my calculus class because of these, look how beautiful and easy you just made it for me. Thanks alot
Thank you sir! Fantastic explanation.
i have a feeling i will be getting this problem in the final
yeah me too
It is indeed a classic problem of Related Rates
bro same
awesome explanation! really appreciate it!!
You are a genius. Keep up the good work. God bless
Outstanding. I have a better intuitive understanding now.
Thank you so much seriously this makes so much sense
Helped alot , thank you for this video !
Absolutely GOATED, thank you so much for the help🙏
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?
ikr, i guess the rate is constant no matter how far away
he is walking at a constant speed
This was my question. Calculus is hard enough we done need excess info 😭😭😭
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
Got an exam in 15 minutes, this video was more helpful than my teacher’s lessons, thank you
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.
Thank you for the video! I was able to finish my homework problem cause of this so thank you very much. ^-^
This man is doing Gods work
amazing video, helped a lot!!!
saved me. i owe you my firstborn child.
You save me every time I go and do math homework
thank you for this!
Thank you so much!!
Was panting to get the solution for this question......😅 thanks for the explanation 😃
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?
thank you so much
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.
This is more helpful than my calculus course right now lmao
Very very nice thank you veryvery match
Omg i love you so much you just saved me
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.
Daniel Byun yeah i was thinking the same thing too
@@sportsarelife2211 doesnt mean he is right
Since he is moving away from the light at a constant rate, ds/dt will be the same at every length.
The key words in the problem are "constant rate"
For question (a), shouldnt 8 be used for the length of x?
Thank you!
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?
Indeed you can! You end up with the same answer :)
Thanks so much dude you made it so clear
So great !!
Thanks man
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.
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GOD BLESS THIS MAN
Thank u soooo much
THANK YOU
Thankyou so much
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.
Doesn’t matter you can do either I’m pretty sure
No, but it is helpful for some people. By that logic, do we need this video?
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?
Yes, those datas are not necessary in solving shadow problems
The rate is the same no matter the distance
@@justinsantos5751 then wouldnt ds/dt be the same as dl/dt? im just confused
@@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.
@@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!
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?
Any alternative solutions with those values didn't used?
Thank youuuu
Do you make your videos while you study? You sound very prepared.
@Matin Nawabi adults can't study?
thanks!
Thank you but why we you didnot use the informastion that he is 8 fet from the light in part (a)?
thank you
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
Thank you :)))
Is time rates and related rates the same??
In what class do we learn that length of shadow is determined by the hypotenus?
What program are you using to make your videos?
Watch
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The crow House
What is tge use of 8ft on the the first part
I love you. Simple
I wish you were my Calculus Professor.
Usefulbro
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?
Yeah, just tried it. That works too
No it doesn’t you get 3.2
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).
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this man saves my grade repeatedly
If L is s plus x, then dL could also just be ds +dx right?
why don't we use 8ft and 10 ft? is it because it has a constant rate of change?
You are my first choice ... creative
so we're not using the value of x at any point?
erm actually there is an easy way to solve part b. I love the video, but here is how I solved part b. Simply identify that L=x+3. Than take the derivative of both sides of the equation in respect to t. We get dL/dt=3ft/s + (6/5ft)/s. Now just add the fractions and you get dL/dt = 4.2ft/s.
Hi, at 5:20 , how did you get 15ds/dt = 6dx/dt ?
Is him being 8 feet away insignificant for part a?
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.
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)
@@a.a9021 and a year later this helped at least one other person. Thanks for comment! Very insightful.
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
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.
not sure why im getting the wrong answer following this way?
What is the source of this math?
I mean where you had found this math?plz tell me the source 😢
Thanku sir
Why does this dude sound just like Mark Wahlberg?
He sounds black to me.
Not even close
All because of good vibrations
"Hey streetlight, howsit goin'? how's ya mother?"
Wahlberg is to Boston accent what Organic Chemist is to...c) Brooklyn accent
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
I got my calc final tomorrow wish me luck
You are a god
I love you bro
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 😂