Question

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 = 3 + 1 2 t, where x and y are measured in centimeters. The temperature function satisfies Tx(2, 4) = 3 and Ty(2, 4) = 5. How fast is the temperature rising on the bug's path after 2 seconds? (Round your answer to two decimal places.)

Answer 1

At t = 2

`(dT)/(dt) = 3*(1/4) + (1/2)*5 = 3.25`

Explanation:

Using the chain rule we get that

`(dT)/(dt) = (dT)/(dx)(dx)/(dt) + (dT)/(dy)(dy)/(dt)`

From the information we know that

`(dT)/(dx)(2,4) = 3\\\\(dT)/(dy)(2,4) = 5`

and using one variable calculus we know that

`(dx)/(dt) = (1)/(2√(2+t))\\\\\\(dy)/(dt) = (1)/(2)`

therefore at t = 2

`(dT)/(dt) = 3*(1/4) + (1/2)*5 = 3.25`