The temperature at a point (x, y) is T(x, y), measured in degrees Celsius. A bug crawls so that its position after t seconds is given by x = 2 + t , y = 1 + 1 2 t, where x and y are measured in centimeters. The temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3. How fast is the temperature rising on the bug's path after 2 seconds? (Round your answer to two decimal places.)

10.50°C

Step-by-step explanation:

Given x = 2 + t , y = 1 + 1/2t where x and y are measured in centimeters. Also, the temperature function satisfies Tx(2, 2) = 9 and Ty(2, 2) = 3

The rate of change in temperature of the bug path can be expressed using the composite formula:

dT/dt = Tx(dx/dt) + Ty(dy/dt)

If x = 2+t; dx/dt = 1

If y = 1+12t; dy/dt = 1/2

Substituting the parameters gotten into dT/dt we will have;

dT/dt = 9(1)+3(1/2)

dT/dt = 9+1.5

dT/dt = 10.50°C/s

Hence the rate at which the temperature is rising along the bug's path is 10.50°C/s

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