Part I Given that the temperature distribution in our classroom can be described
ID: 2989551 • Letter: P
Question
Part I
Given that the temperature distribution in our classroom can be described by
Part II
You are being chased by a mountain Yeti and in order to call for help you need to get to the top of the
mountain fast. The mountain
Explanation / Answer
part 1
dT/dx= 50 e-4x^2 -8y^3 -3z^2 * (-8x-0-0) = 50 e-4x^2 -8y^3 -3z^2 * (-8x)
dT/dy = 50 e-4x^2 -8y^3 -3z^2 * (0-24y-0) = 50 e-4x^2 -8y^3 -3z^2 * (-24y)
dT/dz = 50 e-4x^2 -8y^3 -3z^2 * (0-0-6z) = 50 e-4x^2 -8y^3 -3z^2 * (-6z)
At (1, 0.5, 1)
dT/dx = 50 e-4-1-3 * -8 = -400e-8
dT/dy = 50e-8 * -12 = -600 e-8
dT/dz = 50e-8 * -6 = -300e-8
maximum rate of change is in y direction
temperature decreases fastest in y direction
maximum rate of change is -600e^-8
part 2
dz/dx = 6x2 + 2x -5y
dz/dy = -5x
dz/dz =1
At (10,10,3600)
dz/dx= 6*100 + 2*10 -5*10 = 600+20-50 = 570
dz/dy = -50
dz/dz=1
climbing should be in x direction to reach fastest
part 3
dF/dx= 10yz
dF/dy = 2y -15z + 10 xz
dF/dz = -15y + 10xy
At (1,1,1)
dF/dx =10
dF/dy = 2-15+10 = -3
dF/dz= -15+10 = -5
The reduction in stench is maximum in z direction (-ve means reduction in stench)
pls rate :)
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