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 = 1 + t , y = 9 + 1 3 t, where x and y are measured in centimeters. The temperature function satisfies Tx(2, 10) = 7 and Ty(2, 10) = 4. How fast is the temperature rising on the bug's path after 3 seconds? (Round your answer to two decimal places.)

Answer 1

Answer:

$\dfrac{dT}{dt}=59\ ^{\circ}C/s$

Explanation:

Given that

x = 1 + t , y = 9 + 1 3 t

Tx(2, 10) = 7 ,Ty(2, 10) = 4

T=T(x,y)

$\dfrac{dT}{dt}=\dfrac{dT}{dx}\dfrac{dx}{dt}+\dfrac{dT}{dy}\dfrac{d}{dt}$

$T_x=\dfrac{dT}{dx}\ ,T_y=\dfrac{dT}{dy}$

$\dfrac{dx}{dt}=1$

$\dfrac{dy}{dt}=13$

Tx(2, 10) = 7 ,Ty(2, 10) = 4

Now by putting the values

$\dfrac{dT}{dt}=\dfrac{dT}{dx}\dfrac{dx}{dt}+\dfrac{dT}{dy}\dfrac{d}{dt}$

$\dfrac{dT}{dt}=7\times 1+4\times 13\ ^{\circ}C/s$

$\dfrac{dT}{dt}=59\ ^{\circ}C/s$