Your tutorial on applying the fundamental principles of implicit differentiation to solve real-world problems is truly excellent. Your ability to convey complex concepts in a practical context is commendable, making it easier for learners to grasp and apply these essential mathematical techniques. Keep up the great work!
I just found your channel. I am literally watching your videos in my holiday and i love it 😅 If only i had a teacher like you, math would be so much more interesting ☺️
Excellent solution This guy is a genious in Maths I dare to compare him to Euler, Gauss and all the other great mathematician through centuries Thats it!
For Shift 8 Academy, students. Brillant explanation in simplicity of how the earth time, been 2 years, ( 4 min and 4 sec off time count ). Thank you, instructor Newton, he will be great for the school. Also, understanding , why and how, the Islamic calendar is off, and not accurate, their going off Saturn , but do not equate the sun knowledge, as the natives calendar, equate Sun and Saturn, the 6/8 duo to help Venus balance time.
Fantastic method! Thank you, Prime Newtons! Another way I found would be to first differentiate the surface area (A = 6a^2), which would give us dA/dt = 12x(dx/dt), and then find what dx/dt is by differentiating the original volume formula (V = x^3), and getting dV/dt = 3x^2 (dx/dt). Now, we know that dV/dt = 4 cm^3/min, so just substitute that into the equation and we will get 4/27 after some algebraic manipulation. So after this, we can come back to dA/dt and substitute 4/27 into the equation giving us: dA/dt = 12x * (4/27) = 144/27 = 16/3 cm^2/min :) Have fun learning! Related rates can be fun after you have done enough problems!
Really a Nice Video on application of Derivatives as a rate measurer. Your smile is infectious❤. Your Informative Videos on Tetration, Pentation, Hexation and Graham's Number was excellent. From India.😊❤
The entire left side of the board is unnecessary: The mantle of a cube is 6 times the area of a side, which is the side length cubed. And since the side length (typically called a) of a cube is ³√V, you get M = 6 * ³√V² = 6V^(2/3). It doesn't matter, if you know any value or if it is all variable. This is just the relationship between mantle and volume of any cube - it is universal. Once you got that, you calculate the derivative, so M' = 4/³√V (6 * 2/3 * V^(2/3 - 1) = 12/3 * V^(-1/3) = 4 * 1/V^(1/3) = 4/³√V) This is the rate of change between the volume and the mantle. But ³√V, as we said earlier, is just the side length, so we can write it as M' = 4/a. This means however, that the rate of expansion of the Area is dependent on the original side length of the cube. Unfortunately this is what makes this task so ridiculous: Since we are given a change rate over time, this rate is literally only valid for less than a femto second, as the moment the cube grows, the side length of the cube changes and the entire calculation would have to be done again. In other words, it is completely useless to calculate the rate of change of the mantle to the cube over time, as it changes constantly. It is not a fixed value!
You Can Also Solve It Like This If I'm Wrong Then Sorry 1. Assume H & B In Cube 2. In volume get The value of x 3. Put This x value in Ar. Equation (If Getting Simplified The Simplify) 4. Now Differentiate It 5. Put Differentiate equal to 0 6. Now Solve It U Shall Get Answer.
What about your series of algebra videos I dont know what should be next (I am not a teacher) but I would like to see video about similar matrices , characteristic polynomial ,eigenvalues and eigenvectors , Cayley-Hamilton theorem , some matrix decomposition like diagonalization , Jordan form and maybe SVD
Good job 😍 but there is a way too much easier than yours since that V=L³ , L=3 & dV/dt =4 cm³/min Therefore : dv/dt = 3L² dl/dt =4 Therefore dl/dt = 4 /27 And since Area of cube = 6L² Therefore dA/dt = 12 L dl/dt = (12)(3)(4/27) dA/dt=16/3 cm²/min
A(x)=6x^2 so: dA(x)/dt= 12*x*dx=4*(3*x*dx/dt) (1) V(x)=x^3 so: dV/dt=3x^2*dx/dt (we will need 3*x*dx/dt) 3*x*dx/dt=(1/(x))*dv/dt (2) after substitution (2) to (1) we get: dA(x)/dt= 4*(1/x)*dv/dx for x=3 and dV/dt= 4 dA/dt=16/3
I'm sure that if this guy were a professor at every high school, 90% of students would choose mathematics in college.
The way everything came together at the end made me smile. Maths is beautiful
55 years Oldman, love maths, and love when I discover a teacher like you seem to be !
Thanks
Your tutorial on applying the fundamental principles of implicit differentiation to solve real-world problems is truly excellent. Your ability to convey complex concepts in a practical context is commendable, making it easier for learners to grasp and apply these essential mathematical techniques. Keep up the great work!
Glad it was helpful!
I just found your channel. I am literally watching your videos in my holiday and i love it 😅 If only i had a teacher like you, math would be so much more interesting ☺️
Excellent solution This guy is a genious in Maths I dare to compare him to Euler, Gauss and all the other great mathematician through centuries Thats it!
For Shift 8 Academy, students. Brillant explanation in simplicity of how the earth time, been 2 years, ( 4 min and 4 sec off time count ). Thank you, instructor Newton, he will be great for the school. Also, understanding , why and how, the Islamic calendar is off, and not accurate, their going off Saturn , but do not equate the sun knowledge, as the natives calendar, equate Sun and Saturn, the 6/8 duo to help Venus balance time.
Fantastic method! Thank you, Prime Newtons!
Another way I found would be to first differentiate the surface area (A = 6a^2), which would give us dA/dt = 12x(dx/dt), and then find what dx/dt is by differentiating the original volume formula (V = x^3), and getting dV/dt = 3x^2 (dx/dt). Now, we know that dV/dt = 4 cm^3/min, so just substitute that into the equation and we will get 4/27 after some algebraic manipulation. So after this, we can come back to dA/dt and substitute 4/27 into the equation giving us: dA/dt = 12x * (4/27) = 144/27 = 16/3 cm^2/min :)
Have fun learning! Related rates can be fun after you have done enough problems!
Helps a LOT. This is so well explained
Really a Nice Video on application of Derivatives as a rate measurer. Your smile is infectious❤. Your Informative Videos on Tetration, Pentation, Hexation and Graham's Number was excellent. From India.😊❤
Brilliant! 🎉😊
Bro l like your teaching
excellent expalnantion sir. Thank you so much
Great video 😮😮🎉
The entire left side of the board is unnecessary:
The mantle of a cube is 6 times the area of a side, which is the side length cubed.
And since the side length (typically called a) of a cube is ³√V, you get M = 6 * ³√V² = 6V^(2/3).
It doesn't matter, if you know any value or if it is all variable. This is just the relationship between mantle and volume of any cube - it is universal.
Once you got that, you calculate the derivative, so M' = 4/³√V
(6 * 2/3 * V^(2/3 - 1) = 12/3 * V^(-1/3) = 4 * 1/V^(1/3) = 4/³√V)
This is the rate of change between the volume and the mantle.
But ³√V, as we said earlier, is just the side length, so we can write it as M' = 4/a.
This means however, that the rate of expansion of the Area is dependent on the original side length of the cube.
Unfortunately this is what makes this task so ridiculous: Since we are given a change rate over time, this rate is literally only valid for less than a femto second, as the moment the cube grows, the side length of the cube changes and the entire calculation would have to be done again.
In other words, it is completely useless to calculate the rate of change of the mantle to the cube over time, as it changes constantly. It is not a fixed value!
You Can Also Solve It Like This If I'm Wrong Then Sorry
1. Assume H & B In Cube
2. In volume get The value of x
3. Put This x value in Ar. Equation (If Getting Simplified The Simplify)
4. Now Differentiate It
5. Put Differentiate equal to 0
6. Now Solve It U Shall Get Answer.
Thank you. This may be even hard for a math graduate who hasn't studied math for many years.
How would a grad in any field not have studied for many years?
What are you talking about? I know many people like this. Are you a troll?@@kristianbojinov6715
Good stuff 👌
Thank you!!!
What about your series of algebra videos
I dont know what should be next (I am not a teacher)
but I would like to see video about similar matrices , characteristic polynomial ,eigenvalues and eigenvectors , Cayley-Hamilton theorem , some matrix decomposition like diagonalization , Jordan form and maybe SVD
Good job 😍
but there is a way too much easier than yours
since that V=L³ , L=3 & dV/dt =4 cm³/min
Therefore : dv/dt = 3L² dl/dt =4
Therefore dl/dt = 4 /27
And since Area of cube = 6L²
Therefore dA/dt = 12 L dl/dt
= (12)(3)(4/27)
dA/dt=16/3 cm²/min
Easiest question for Egyptian high school students 😅
That works too! You noticed I only differentiated once. That was my goal.
I used the method of our Egyptian friend. For me the key was finding dL/dt first. That said, the professor's method was quite creative.
Impressive
Bro I am very weak in calculatas. Make a video on this topic.
A(x)=6x^2
so:
dA(x)/dt= 12*x*dx=4*(3*x*dx/dt) (1)
V(x)=x^3
so:
dV/dt=3x^2*dx/dt
(we will need 3*x*dx/dt)
3*x*dx/dt=(1/(x))*dv/dt (2)
after substitution (2) to (1) we get:
dA(x)/dt= 4*(1/x)*dv/dx
for x=3 and dV/dt= 4
dA/dt=16/3
I was thinking about a similar problem in calc today 🤔
you can also go straight dV/dt=3x2.dx/dt=4 so dx/dt=4/(3x2) and dA/dt=12x.dx/dt=16/x=16/3 when x=3
Yep, was looking for this comment
Good
dA/dt = 4/s* dV/dt, but the value of s is a function of time. How do you replace it with a constant value?
Let V = s^3, dV/dt = 3s^2 * ds/dt.
Then let A = 6s^2, dA/dt = 12s * ds/dt
ds/dt = 1/3s^2 * dV/dt
Then, dA/dt = 12s * 1/3s^2 * dV/dt = 4/s * dV/dt
So, dA/dt = 4/3 * 4 = 16/3
are you a Zimbabwean?
We can first find dx/dt
V=x^3
dv/dt = 3 x^2 dx/dt
4= 3 (3)^2 dx/d
dx/dt = 4/27
A = 6 x^2
dA/dt = 12x dx/dt
= 12 × 3 × 4/27
=16/3
Merci !
Thank you! Much appreciated 👏
typical mathematical approach. more physically with V(t) , O(t) would be dO/dt (27/4) =