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 the following function, where x and y are measured in centimeters. x = sqrt(2 + t), y = 5 + 1/14 t The temperature function satisfies Tx(4, 6) = 8 and Ty(4, 6) = 4. How fast is the temperature rising on the bug's path after 14 seconds?

Answer:The rate of change of temperature is 1.29 degree Celsius per second.   Step-by-step explanation:We are given the following information in the question:The temperature at a point (x, y) is T(x, y), measured in degrees Celsius where x and y are measured in centimeters.[tex]x = \sqrt{2+t}\\\\y = 5 + \displaystyle\frac{1}{14}t[/tex][tex]T_x(4,6) = 8, T_y(4,6) = 4[/tex]We have to find the rate at which the temperature is rising on the bug's path after 14 seconds.At t = 14 seconds, we have,[tex]x = \sqrt{2+14} = 4\\\\y = 5 + \displaystyle\frac{1}{14}(14) = 5+1 = 6[/tex]To find rate of change of temperature, we differentiate,[tex]\displaystyle\frac{dT}{dt} = \frac{dT}{dx}\frac{dx}{dt} + \frac{dT}{dy}\frac{dy}{dt}\\\\\displaystyle\frac{dT}{dt} = T_x(x,y)(\frac{1}{2\sqrt{2+t}}) + T_y(x,y)\frac{1}{14}\\\\At~ t = 14, x = 4, y = 6\\\\\frac{dT}{dt} = T_x(4,6)(\frac{1}{2\sqrt{2+t}}) + T_y(4,6)\frac{1}{14}\\\\\frac{dT}{dt} = 8\times \frac{1}{8} +4\times \frac{1}{14} = 1 + 0.2857 = 1.2857[/tex]Thus, the rate of change of temperature is 1.29 degree Celsius per second.