Best lesson I've ever seen on Related Rates. Very very helpful the way Mr T set-up and solved this Rocket problem . Great teachers are hard to find . He is surely one !
OMG why cant school teachers be like this, set up the question and give us general steps to solving related rates, instead of just solving a problem once on the board and expecting us to understand that, you are a hero Mr. S Math thank you for the crystal explanation.
This was the Pythagorean Theorem. a^2+b^2=c^2. From the given information I knew that a=5000 and b=10000. To solve the Pythagorean Theorem for c I took the square root of both sides of the equation. Here, the square root of a^2+b^2 is the square root of 5000^2+10000^2, which simplifies to 5000 root 5.
Great question! That last part is a bit tricky because the numbers are so large. To simplify 5000^2+10000^2, first realize that 5,000 squared is 25 million and 10,000 squared is 100 million. Adding these gives a sum of 125 million. This is equal to c^2, so to find c you take the square root of 125 million. Simplifying the square root of 125 million is difficult because it is so large. Since 125 million is equal to 1 million times 125, I began by rewriting the square root of 125 million as the square root of 1 million times the square root of 125. The square root of 1 million is 1000, so it becomes 1000 root 125. Then, I rewrote the square root of 125 as the square root of 25 times the square root of 5. The square root of 25 is exactly 5, so now I have 1000 times 5 times the square root of 5. Multiplying 1000 and 5 gave me the final simplification which is 5000 root 5.
He didnt multiply each term, he multiplied it by d(term)/dt. You do this because you are derrivative with respect to something else. So if you are derivating with respect to x in an equation, whenever you derive any other term you add d(term)/dx(x in this case since you are derivating with respect to it). If you were wondering about the reason he put a' and b' and c', it's just another way of putting da/dt, db/bt and dc/dt. That's why
The given information describes a situation that can be represented as a right triangle. Only two sides of the triangle are given in the problem. At 7:45 I used the Pythagorean Theorem to find the length of the third side. This theorem is a^2+b^2=c^2 where a and b are the legs of a right triangle and c is the hypotenuse. We are told what side a and side b are in the problem, at 7:45 I used this theorem to find side c.
In the video, I explain that side c is the square root of 5000 squared plus 10,000 squared. 5000^2 + 10000^2 is 125,000,000. The square root of 125 million simplifies to 5000 sqrt 5. I think the easiest way to simplify this is to find the square root of 1,000,000 (which is 1,000) times the square root of 125 (which is 5 sqrt 5). Then multiply these to get 5000 sqrt 5.
![How to Solve ANY Related Rates Problem [Calc 1]](http://i.ytimg.com/vi/gBHIZlF0TX8/mqdefault.jpg)
![How to Solve ANY Related Rates Problem [Calc 1]](/img/tr.png)
Best lesson I've ever seen on Related Rates. Very very helpful the way Mr T set-up and solved
this Rocket problem . Great teachers are hard to find . He is surely one !
From the integral of my bottomless heart, thank you, Sir.
You're very welcome!
This is the best explaination on related rates,hope theres more video from mr S Math
This really saved me! Cannot tell you how thankful I am for this video. Fantastic explanation. Wish me luck on my test tomorrow!
How it go
WOW. There are so many ways to solve and show the work. This way is the best and very organized. Thank you for the video.
You are very welcome!
OMG why cant school teachers be like this, set up the question and give us general steps to solving related rates, instead of just solving a problem once on the board and expecting us to understand that, you are a hero Mr. S Math thank you for the crystal explanation.
Simple explanation. Love it! Awesome video!
This is explained so well. I'd love to see you cover some conical tank and shadow problems using related rates.
Nice clear method and explanation. Thank you, Mr. S!
Glad it was helpful!
Mr. Smith you made it so easy to understand thank you so much may god bless you Thanks again !!
You are very welcome! Glad it helped!
May you forever be granted the best gifts. Thank yo so much!!!!!
You are welcome!!
Great video sir.You really broke down the concept for easy understanding.
Glad it was helpful!
i was kind of feeliing stressed about my test later this week but after this video i think im gonna do great.
Good luck on the test - hope this helped
W mans, underrated bro
Thank you so much! I actually feel like I understand the concept now!
Thank you so much for this video you single handedly saved me from failing my calc test tmrw.
You’re welcome - good luck on the test!
@@MrSMath Thanks I'm taking it in 1 hr
I have a clarifying question at 8:00, how did you simplify the a^2 and b^2 to get 5000 root 5??
This was the Pythagorean Theorem. a^2+b^2=c^2. From the given information I knew that a=5000 and b=10000. To solve the Pythagorean Theorem for c I took the square root of both sides of the equation. Here, the square root of a^2+b^2 is the square root of 5000^2+10000^2, which simplifies to 5000 root 5.
@@MrSMath Yea I know that part but I’m talking about the last part. How does it simplify to 5,000^5?
Great question! That last part is a bit tricky because the numbers are so large. To simplify 5000^2+10000^2, first realize that 5,000 squared is 25 million and 10,000 squared is 100 million. Adding these gives a sum of 125 million. This is equal to c^2, so to find c you take the square root of 125 million.
Simplifying the square root of 125 million is difficult because it is so large. Since 125 million is equal to 1 million times 125, I began by rewriting the square root of 125 million as the square root of 1 million times the square root of 125. The square root of 1 million is 1000, so it becomes 1000 root 125. Then, I rewrote the square root of 125 as the square root of 25 times the square root of 5. The square root of 25 is exactly 5, so now I have 1000 times 5 times the square root of 5. Multiplying 1000 and 5 gave me the final simplification which is 5000 root 5.
@@bobjoe2538 Thank god, someone questioned this. Cause I had no literal idea how to solve this part. T-T
very helpful. thanks
Well done.
That was really fun to do.
Can you PLEASE DO A VIDEO ON introduction to related rates with TWO EQUATIONS!? In calculus
good idea
9:50 this is the part that still confuses me. Why do you differentiate and multiply each term?
He didnt multiply each term, he multiplied it by d(term)/dt. You do this because you are derrivative with respect to something else. So if you are derivating with respect to x in an equation, whenever you derive any other term you add d(term)/dx(x in this case since you are derivating with respect to it). If you were wondering about the reason he put a' and b' and c', it's just another way of putting da/dt, db/bt and dc/dt. That's why
very well done
Thanks ❤
Welcome!
life saver
7:45 i just need a clarification im hella confused
The given information describes a situation that can be represented as a right triangle. Only two sides of the triangle are given in the problem. At 7:45 I used the Pythagorean Theorem to find the length of the third side. This theorem is a^2+b^2=c^2 where a and b are the legs of a right triangle and c is the hypotenuse. We are told what side a and side b are in the problem, at 7:45 I used this theorem to find side c.
@@MrSMath no not that gow do you get the 5000 sqrt5?
@@MrSMath and also whrre did you pull out the 5 bruhh
In the video, I explain that side c is the square root of 5000 squared plus 10,000 squared. 5000^2 + 10000^2 is 125,000,000. The square root of 125 million simplifies to 5000 sqrt 5. I think the easiest way to simplify this is to find the square root of 1,000,000 (which is 1,000) times the square root of 125 (which is 5 sqrt 5). Then multiply these to get 5000 sqrt 5.
Thank you
thank you!
You're welcome!
tysm!!
How do you do Step 4 on V = πr^2h
In that situation, the Product Rule is required when differentiating the right side of the equation. The result would be V' = π (r^2*h'+2rh*r')
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W in the chat thx man
yw - glad it helped
Refering to Mr S math ..not Mr t