obtain the heat flow equation(1.3) as follows: the quantity of heat Q flowing ac

Question

obtain the heat flow equation(1.3) as follows: the quantity of heat Q flowing across a surface is proportional to the normal component of
the (negative) temperature gradient, (-[Del]T).n
compare chapter 6 equation (10.4) and apply the discussion of flow
of water given there to the flow of heat, thus show that the rate of
gain of heat per unit volume per unit time is proportional to [Del].
[Del]T . But
[PartialD]T/[PartialD]t is proportional to this gain in heat ; thus
show that T satisfies (1.3

Explanation / Answer

Linear: y=mx+q 8=3*m+q and 22=10*m+q m=(8-q)/3) 22=10((8-q)/3)+q --> q=2 m=2 y=2x+2 __________________________ Exponential: y=e^(x+a)+c --> y=(e^x)*(e^a)+c 8=(e^3)*(e^a)+c 22=(e^10)*(e^a)+c c=8-(e^3)*(e^a) 22=(e^10)*(e^a)+8-(e^3)*(e^a) c=8-(e^3)(e^a) e^a=14/(e^10-e^3)=0.000636179 a=ln(^a)=-7.360030372 c=8-0.012777996=7.987222003 y=e^(x-736..)+7.98.. ___________________________ Power: y=x^(a)+b 8=3^a+b 22=10^a+b y=x^1.254+4.034389522 Huh!!!! This last has been really hard for me! I resolved it by iteration...in limit of 0.001 is correct. If you want more precision, start to iterate from 1.2541 and 1.2549 as extremes. I suggest you (it's not a judgement, just a suggestion, forgive me if i say wrong) to apply yourself: i would dislike to make someone else's work, leaving him not to apprend. In the exponential i left ".." instead of complete result, but the numbers are written just before.

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