Related rates of volume and area of a cube

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  • Опубліковано 10 гру 2024

КОМЕНТАРІ • 42

  • @albertobuj4021
    @albertobuj4021 Рік тому +26

    I'm sure that if this guy were a professor at every high school, 90% of students would choose mathematics in college.

  • @suomynona898
    @suomynona898 10 місяців тому +6

    The way everything came together at the end made me smile. Maths is beautiful

  • @casar68
    @casar68 10 місяців тому +2

    55 years Oldman, love maths, and love when I discover a teacher like you seem to be !
    Thanks

  • @jdlescott4493
    @jdlescott4493 10 місяців тому +3

    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!

  • @girlthatyoufear6
    @girlthatyoufear6 Рік тому +11

    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 ☺️

  • @arbenkellici3808
    @arbenkellici3808 Рік тому +1

    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!

  • @SYAgencies0379
    @SYAgencies0379 Рік тому +2

    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.

  • @samueleczek
    @samueleczek Місяць тому

    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!

  • @VirKap99
    @VirKap99 Рік тому +1

    Helps a LOT. This is so well explained

  • @ramachandrakandikere245
    @ramachandrakandikere245 Рік тому +1

    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.😊❤

  • @punditgi
    @punditgi Рік тому +5

    Brilliant! 🎉😊

  • @jaikrishnay3569
    @jaikrishnay3569 Рік тому +2

    Bro l like your teaching

  • @mayanksethia4209
    @mayanksethia4209 7 місяців тому

    excellent expalnantion sir. Thank you so much

  • @thetajinos7790
    @thetajinos7790 4 місяці тому

    Great video 😮😮🎉

  • @m.h.6470
    @m.h.6470 11 місяців тому +1

    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!

  • @dreamofkinggaming
    @dreamofkinggaming Рік тому +2

    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.

  • @AubreyForever
    @AubreyForever Рік тому +1

    Thank you. This may be even hard for a math graduate who hasn't studied math for many years.

    • @kristianbojinov6715
      @kristianbojinov6715 11 місяців тому +1

      How would a grad in any field not have studied for many years?

    • @AubreyForever
      @AubreyForever 11 місяців тому

      What are you talking about? I know many people like this. Are you a troll?@@kristianbojinov6715

  • @slavinojunepri7648
    @slavinojunepri7648 3 місяці тому

    Good stuff 👌

  • @apriljohnson6191
    @apriljohnson6191 Рік тому +1

    Thank you!!!

  • @holyshit922
    @holyshit922 Рік тому +1

    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

  • @wnslogisticsservices
    @wnslogisticsservices Рік тому +5

    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

    • @wnslogisticsservices
      @wnslogisticsservices Рік тому +1

      Easiest question for Egyptian high school students 😅

    • @PrimeNewtons
      @PrimeNewtons  Рік тому +3

      That works too! You noticed I only differentiated once. That was my goal.

    • @jim2376
      @jim2376 11 місяців тому

      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.

  • @vee985
    @vee985 4 місяці тому

    Impressive

  • @FahimMohtasim
    @FahimMohtasim Рік тому +1

    Bro I am very weak in calculatas. Make a video on this topic.

  • @boguslawszostak1784
    @boguslawszostak1784 11 місяців тому +1

    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

  • @Stender_
    @Stender_ Рік тому +1

    I was thinking about a similar problem in calc today 🤔

  • @florianbasier
    @florianbasier 9 місяців тому

    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

    • @clayton97330
      @clayton97330 8 місяців тому

      Yep, was looking for this comment

  • @VanNguyen-kx6gx
    @VanNguyen-kx6gx 11 місяців тому +1

    Good

  • @giancarlolosciale1
    @giancarlolosciale1 4 місяці тому

    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?

  • @jaypee2558
    @jaypee2558 11 місяців тому +1

    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

  • @PatronelaHove
    @PatronelaHove Місяць тому

    are you a Zimbabwean?

  • @skwbusaidi
    @skwbusaidi 7 місяців тому

    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

  • @casar68
    @casar68 10 місяців тому

    Merci !

    • @PrimeNewtons
      @PrimeNewtons  10 місяців тому

      Thank you! Much appreciated 👏

  • @berndmayer3984
    @berndmayer3984 8 місяців тому

    typical mathematical approach. more physically with V(t) , O(t) would be dO/dt (27/4) =