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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.)

Respuesta :

Answer:

At  t = 2

[tex]\frac{dT}{dt} = 3*(1/4) + (1/2)*5 = 3.25[/tex]

Step-by-step explanation:

Using the chain rule we get that

[tex]\frac{dT}{dt} = \frac{dT}{dx}\frac{dx}{dt} + \frac{dT}{dy}\frac{dy}{dt}[/tex]

From the information we know that

[tex]\frac{dT}{dx}(2,4) = 3\\\\\frac{dT}{dy}(2,4) = 5[/tex]

and using one variable calculus we know that

[tex]\frac{dx}{dt} = \frac{1}{2\sqrt{2+t}}\\\\\\\frac{dy}{dt} = \frac{1}{2}[/tex]

therefore at  t = 2

[tex]\frac{dT}{dt} = 3*(1/4) + (1/2)*5 = 3.25[/tex]

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