The temperature at a point(x, y, z)is given by T(x, y, z) = 100e^−x^2 − 3y^2 − 9z^2 where T is measured in °C and x, y, z in meters.
(a) Find the rate of change of temperature at the pointP(4, −1, 5) in the direction towards the point
(5, −2, 6).
(b) In which direction does the temperature increase fastest at P?
(c) Find the maximum rate of increase at P.
Better review the gradient operator.
(b) greatest change in the direction ∇T
(a) change in the direction of v is ∇T • v/|v|
(c) |∇T|
To solve these questions, we will use the concept of gradient and directional derivatives.
(a) The rate of change of temperature at point P in the direction towards the point Q can be calculated using the directional derivative formula:
∇T(P) ⋅ u,
where ∇T(P) is the gradient of T at point P and u is the unit vector pointing from P to Q.
The gradient is given by:
∇T = (∂T/∂x, ∂T/∂y, ∂T/∂z).
Therefore, we need to calculate the partial derivatives of T with respect to x, y, and z.
∂T/∂x = (-2x)(100e^(-x^2 - 3y^2 - 9z^2))
= -200xe^(-x^2 - 3y^2 - 9z^2)
∂T/∂y = (-6y)(100e^(-x^2 - 3y^2 - 9z^2))
= -600ye^(-x^2 - 3y^2 - 9z^2)
∂T/∂z = (-18z)(100e^(-x^2 - 3y^2 - 9z^2))
= -1800ze^(-x^2 - 3y^2 - 9z^2)
Now, let's calculate the gradient at point P(4, -1, 5):
∇T(P) = (-200(4)e^(-4^2 - 3(-1)^2 - 9(5)^2), -600(-1)e^(-4^2 - 3(-1)^2 - 9(5)^2), -1800(5)e^(-4^2 - 3(-1)^2 - 9(5)^2))
= (-3200e^(-68), 600e^(-68), -90000e^(-68))
Next, let's calculate the unit vector u pointing from P(4, -1, 5) to Q(5, -2, 6):
u = (5-4, -2-(-1), 6-5)/sqrt((5-4)^2 + (-2-(-1))^2 + (6-5)^2)
= (1, -1, 1)/sqrt(3)
Finally, we can calculate the rate of change of temperature at point P in the direction towards point Q:
Rate of change = ∇T(P) ⋅ u
= (-3200e^(-68), 600e^(-68), -90000e^(-68)) ⋅ (1, -1, 1)/sqrt(3)
(b) To determine the direction in which the temperature increases fastest at point P, we need to find the maximum value of the rate of change of temperature. This can be achieved by finding the direction of the gradient vector, ∇T(P). The direction of fastest increase is given by:
Direction of fastest increase = ∇T(P)/|∇T(P)|
= (-3200e^(-68), 600e^(-68), -90000e^(-68))/sqrt((-3200e^(-68))^2 + (600e^(-68))^2 + (-90000e^(-68))^2)
(c) The maximum rate of increase at point P can be determined by calculating the magnitude of the gradient vector, ∇T(P). The maximum value occurs in the direction of the gradient vector.
Maximum rate of increase = |∇T(P)|
= sqrt((-3200e^(-68))^2 + (600e^(-68))^2 + (-90000e^(-68))^2)
To find the rate of change of temperature at a given point in a specific direction, we'll use the gradient operator (∇). The gradient of a function gives the direction of the steepest increase and its magnitude gives the rate of change in that direction.
Let's start by finding the gradient of the temperature function T(x, y, z) = 100e^(-x^2 - 3y^2 - 9z^2).
(a) Find the rate of change of temperature at point P(4, -1, 5) towards point Q(5, -2, 6):
1. Calculate the gradient of T(x, y, z):
∇T(x, y, z) = (∂T/∂x)i + (∂T/∂y)j + (∂T/∂z)k
Taking partial derivatives of T(x, y, z) with respect to each variable:
∂T/∂x = -200xe^(-x^2 - 3y^2 - 9z^2)
∂T/∂y = -600ye^(-x^2 - 3y^2 - 9z^2)
∂T/∂z = -1800ze^(-x^2 - 3y^2 - 9z^2)
Therefore, ∇T(x, y, z) = -200xe^(-x^2 - 3y^2 - 9z^2)i - 600ye^(-x^2 - 3y^2 - 9z^2)j - 1800ze^(-x^2 - 3y^2 - 9z^2)k
2. Substitute the values of P and Q in ∇T(x, y, z):
∇T(4, -1, 5) = -800e^(-124)i + 1800e^(-124)j - 9000e^(-124)k
3. Find the direction vector from P to Q:
Direction vector = Q - P = (5 - 4)i + (-2 - (-1))j + (6 - 5)k = i - j + k
4. Calculate the rate of change of temperature at P in the direction towards Q:
Rate of change = ∇T(4, -1, 5) · Direction vector
= (-800e^(-124)i + 1800e^(-124)j - 9000e^(-124)k) · (i - j + k)
Simplifying this dot product expression, we get:
Rate of change ≈ -800e^(-124) + 1800e^(-124) - 9000e^(-124)
(Note: e^(-124) is an extremely small number, almost 0. So this approximation will be close to 0)
(b) To determine the direction in which the temperature increases fastest at P, we need to find the unit vector in the direction of the gradient at point P.
1. Normalize the gradient vector:
Unit vector = (∇T(4, -1, 5)) / |∇T(4, -1, 5)|
= (-800e^(-124)i + 1800e^(-124)j - 9000e^(-124)k) / sqrt((-800e^(-124))^2 + (1800e^(-124))^2 + (-9000e^(-124))^2)
Simplifying this expression (keeping e^(-124) ≈ 0), we have:
Unit vector ≈ (-i + 2.25j - 11.25k) / sqrt(1^2 + 2.25^2 + (-11.25)^2)
(c) Finally, to find the maximum rate of temperature increase at point P, we need to find the magnitude of the gradient vector:
Maximum rate of increase = |∇T(4, -1, 5)|
= sqrt((-800e^(-124))^2 + (1800e^(-124))^2 + (-9000e^(-124))^2)
≈ sqrt(640000 + 4050000 + 81000000)
≈ sqrt(88000000) ≈ 9372.28 °C/m
Therefore:
(a) The rate of change of temperature at P(4, -1, 5) towards Q(5, -2, 6) is approximately 0 °C/m.
(b) The temperature increases fastest at P in the direction approximately (-i + 2.25j - 11.25k) / sqrt(1^2 + 2.25^2 + (-11.25)^2).
(c) The maximum rate of increase at P is approximately 9372.28 °C/m.