Question

The temperature at a point (x,y,z) is given by T(x,y,z)=300e −x² −3y² −9z² where T is measured in ∘C and x,y,z in meters.

(a) Find the rate of change of temperature at the point P(2,−1,2) in the direction towards the point (3,−4,5).

Answer 1

To find the rate of change of temperature at the point P (2, -1, 2) in the direction towards the point (3, -4, 5), we first need to find the gradient vector of the temperature function. Then, we find the unit vector in the direction towards the point (3, -4, 5). Finally, we take the dot product of the gradient vector and the unit vector to obtain the rate of change of temperature at point P.

Step-by-step explanation:

To find the rate of change of temperature at the point P (2, -1, 2) in the direction towards the point (3, -4, 5), we first need to find the gradient vector of the temperature function. The gradient vector is given by ∇T = (∂T/∂x, ∂T/∂y, ∂T/∂z). So, we need to compute the partial derivatives of T(x, y, z) with respect to x, y, and z.

∂T/∂x = -2xT, ∂T/∂y = -6yT, ∂T/∂z = -18zT

Substituting the values of x, y, and z for point P, we get ∇T(P) = (-4(2)(300e^(-4)), -6(-1)(300e^(-6)), -18(2)(300e^(-36))) = (-2400e^(-4), 1800e^(-6), -10800e^(-36)).

Next, we need to find the unit vector in the direction towards the point (3, -4, 5). The magnitude of this vector is sqrt((3-2)^2 + (-4-(-1))^2 + (5-2)^2) = sqrt(1+9+9) = sqrt(19).

The unit vector is (1/sqrt(19))(3-2, -4-(-1), 5-2) = (1/sqrt(19))(1, -3, 3).

Finally, we can find the rate of change of temperature at point P in the direction towards the point (3, -4, 5) by taking the dot product of the gradient vector and the unit vector:

Rate of change of temperature = ∇T(P) · u = (-2400e^(-4), 1800e^(-6), -10800e^(-36)) · (1/sqrt(19))(1, -3, 3).