The total differential | Calculus in a Nutshell | LetThereBeMath |
ฝัง
- เผยแพร่เมื่อ 19 ก.พ. 2017
- An extension of the chain rule allows us to find the total change of a function with respect to changes in all of its variables at once. This is what we call the total differential.
Thank you Let there be Math. I am a chem major and with the pandemic, I am scrambling to learn enough calculus for Physical Chem so I can finish my degree and graduate. Your video is very concise and much clearer than my professor
Thank you so much for these videos. I wish you were my math professor. 😄
Great video, thanks a lot!
you are the gem
Great lesson, I would also say that this method is used extensively in fluid mechanics where it is known as the substantial derivative or material derivative which defines any property of a fluid element moving with the flow of the fluid.
In the last example, your method doesn't account for the small squares at the corners. Try:
d=0.05
S(10+d, 12+d, 20+d) - S(10, 12, 20) = 8.415 # but your result is 8.4
The difference is 0.015 = 6*d**2
There are 6 squares of side length d you didn't account for. If you visualize in 3d - 3 squares are lonely, and 3 are at same corner.
Now it is crystal clear for me, thank you.
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Finding the total differential looks a lot like finding the gradient. Are they related in any significant way?
I believe so because the gradient is simply the partial in the x and the partial in the y which are both in the equation for total differential
Thank you!
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Thanks Man!!
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In the example with the calculation of difference in Watts, why cant you just solve the equation with the first scenario parameters and subtract the equation with the 2nd scenario parameters? Wouldnt that be equal to the difference in watt from the two scenarios? So P=V^2/R, and what i dont understand is why the following equation doesnt lead to the same result as the total differential: (200^2/8)-(195^2/7.8)?? Any help would be great. I have my exam in Calculus monday.
Never mind, it does. My calculus proffessor just told me that you couldnt do it that way, and i never really understood why.
Why is the partial derivative of P with respect to R -(V^2/R^2)? I thought it would just be V^2 because the R becomes a 1
You are differentiating with respect to R so you treat V as a constant and R as a variable. In that case you need to use the power rule for R. As you see, the original function is P = V^2/R or P = V^2 * R^(-1) which is the same as 1/R. When you use the power rule, you take the exponent ( which in this case is -1) and bring it in the fraction (where it is just a minus sign) and subtract -1 from the exponent of R^(-1) to get R^(-2). Thus, you get -V^2/R^2 or -V^2 * R^(-2).
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Wondering how it relates to the gradient too...
Why the approximation ?..why not integrate and get an 'exact' solution....doesn't make sense, unless it is impossible to find the integral.
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