temp is T(x,y,z)=1500e−x^2−2y^2−z^2 Find the rate of change of temp at P(2,−1,2) to Q(3,−3,3). DPQf(2,−1,2)=, what direction does temp increase fastest at P?, Find the max rate of increase at P

Answer 1

To find the rate of change of temperature from point P(2, -1, 2) to Q(3, -3, 3) and the direction of the fastest temperature increase at P, you can use the gradient vector of the temperature function T(x, y, z). The gradient vector points in the direction of the steepest increase in temperature, and its magnitude gives the rate of change of temperature in that direction.

The gradient of T(x, y, z) is given as:

∇T = (∂T/∂x)i + (∂T/∂y)j + (∂T/∂z)k

Where:

∇T is the gradient vector.

(∂T/∂x) is the partial derivative of T with respect to x.

(∂T/∂y) is the partial derivative of T with respect to y.

(∂T/∂z) is the partial derivative of T with respect to z.

i, j, k are the unit vectors in the x, y, and z directions, respectively.

Calculate the partial derivatives:

∂T/∂x = -3000xe^(-x^2-2y^2-z^2)

∂T/∂y = -6000ye^(-x^2-2y^2-z^2)

∂T/∂z = -2000ze^(-x^2-2y^2-z^2)

Evaluate these partial derivatives at point P(2, -1, 2) to find the gradient vector at P:

∇T(P) = (-3000*2*e^(-2-2-4))i + (-6000*(-1)*e(-2-2-4))j + (-2000*2*e^(-2-2-4))k

∇T(P) = (-6000e^(-8))i + 12000e^(-8)j + (-4000e^(-8))k

Now, to find the rate of change of temperature from P to Q, calculate the displacement vector D from P to Q, which is given by:

D = Q - P = (3 - 2)i + (-3 - (-1))j + (3 - 2)k

D = i - 2j + k

Now, calculate the dot product of the gradient vector and the displacement vector:

DPQ = ∇T(P) · D

DPQ = [(-6000e^(-8))i + 12000e^(-8)j + (-4000e^(-8))k] · (i - 2j + k)

DPQ = (-6000e^(-8)) + (-24000e^(-8)) + (-4000e^(-8))

DPQ = -34000e^(-8)

So, DPQ at point P(2, -1, 2) is -34000e^(-8).

The direction in which temperature increases fastest at P is in the direction of the gradient vector ∇T(P). This direction is given by:

Direction = ∇T(P) / |∇T(P)|

Calculate the unit vector in this direction:

Direction = [(-6000e^(-8))i + 12000e^(-8)j + (-4000e^(-8))k] / |-34000e^(-8)|

Direction = [(-6000i + 12000j - 4000k) / 34000]

The max rate of temperature increase at P is |-34000e^(-8)|.