The temperature at a point (x, y) in the plane is T(x, y) ◦C. If a bug crawls on the plane so that its position in the plane after t minutes is given by (x(t), y(t)) where x = √ 4 + t , y = 4 + 2 5 t , determine how fast the temperature is rising on the bug’s path at t = 5 when Tx(3, 6) = 24 , Ty(3, 6) = 15 .

rate = 34°C/min

Step-by-step explanation:

By the Chain Rule for partial differentiation, the rate of change of temperature on thebug’s path at (x(t), y(t)) is given by

dT/dt = dT(x(t),y(t))/dt = (∂T/∂x)(dx/dt)+(∂T/∂y)(dy/dt)

Now

dx/dt = 1/(2√(4 + t))

dy/dt = (2/5)*t

But when t= 5, the bug is at the point (3,6), while

dx/dt (t=5) = 1/(2√(4 + 5)) = 1/6

dy/dt (t=5) = (2/5)*5 = 2

and we are told that

Tx (3, 6) = 24,      Ty (3, 6) = 15

Consequently, after 5 minutes the temperature on the bug’s path is changing at a

rate = 24*(1/6)+15*(2) = 34°C/min