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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 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?

Respuesta :

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.

oladayoademosu

The rate at which the temperature rising on the bug's path after 14 seconds is 1.286 °C/s

Given that the temperature at a point (x, y) is T(x, y), measured in degrees Celsius.

Also, given that

  • x = √(2 + t),
  • y = 5 + t/14,
  • Tx(4,6) = 8 and
  • Ty(4, 6) = 4.

The rate of temperature rise

The rate of change of temperature with time is

dT/dt = dT/dx × dx/dt + dT/dy × dy/dt

dT/dt = Tx × dx/dt + Ty × dy/dt

Now dx/dt = 1/[2√(2 + t)] and dy/dt = 1/14

Substituting the values of the variables into the equation, we have

So, dT/dt = Tx × dx/dt + Ty × dy/dt

dT/dt = Tx(4,6) × 1/[2√(2 + t)]  + Ty(4,6) × 1/14

dT/dt = 8 × 1/[2√(2 + t)]  + 4 × 1/14

dT/dt = 4/[√(2 + t)]  + 2/7

The rate of temperature rise when t = 14 s

So, when t = 14 s, we have

dT/dt = 4/[√(2 + t)] + 2/7

dT/dt = 4/[√(2 + 14)] + 2/7

dT/dt = 4/[√16] + 2/7

dT/dt = 4/4 + 2/7

dT/dt = 1 + 2/7

dT/dt = 9/7

dT/dt = 1.286 °C/s

So, the rate at which the temperature rising on the bug's path after 14 seconds is 1.286 °C/s

Learn more about rate of temperature rise here:

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