Find the rate at which the temperature is changing

The temperature (in degrees Celsius) at a point (x, y) on a metal plate in the
xy-plane is given by

T(x, y) =135xy/(1 + x^2 + y^2)
where x and y are measured in meters.

(a) Find the rate at which the temperature is changing at the point (1, 1) in the
direction of the vector a = 2i - j. Is the temperature increasing or decreasing
in that direction?

(b) An ant is located at the point (1, 1) where the temperature is a hot 45C. Understandably,
the ant wants to walk in the direction in which the temperature
drops most rapidly. Find a unit vector in that direction.

T(x,y) = 135xy / (1 + x² + y²)

∂T/∂x = (135y(1+x²+y²) − 135xy(2x)) / (1 + x² + y²)²
. . . . = 135y (1 + x² + y² − 2x²) / (1 + x² + y²)²
. . . . = 135y (1 − x² + y²) / (1 + x² + y²)²

∂T/∂y = (135x(1+x²+y²) − 135xy(2y)) / (1 + x² + y²)²
. . . . = 135x (1 + x² + y² − 2y²) / (1 + x² + y²)²
. . . . = 135x (1 + x² − y²) / (1 + x² + y²)²

------------------------------

(a)

∇T(x,y) = < ∂T/∂x, ∂T/∂y >
∇T(x,y) = < 135y(1−x²+y²)/(1+x²+y²)², 135x(1+x²−y²)/(1+x²+y²)² >

∇T(1,1) = < 135(1−1+1)/(1+1+1)², 135(1+1−1)/(1+1+1)² >
. . . . . . = < 135/9, 135/9 >
. . . . . . = < 15, 15 >

Unit vector in direction of vector < 2, −1 > is u = 1/√5 < 2, −1 > = < −2/√5, 1/√5 >

∇T(1,1) · u = < 15, 15 > · < −2/√5, 1/√5 > = −30/√5 + 15/√5 = −15/√5 = −3√5

Temperature is decreasing (since −3√5 < 0)

------------------------------

(b)

Temperature is increasing most rapidly in direction of gradient at point (1,1)
and temperature is decreasing most rapidly in opposite direction,
i.e. < −15, −15 >

Unit vector in that direction is < −1/√2, −1/√2 >

Rate at which temperature is changing in this direction is
< 15, 15 > · < −1/√2, −1/√2 > = −15√2

Mαthmφm