Question

A bird is flying in a room with a velocity field of . Calculate the temperature change that the bird feels after 9 seconds of flight, as it flies through x

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The temperature change is $\frac{dT}{dt} = 1.016 ^oC/m$

Explanation:

From the question we are told that

   The velocity field with which the bird is flying is  $\vec V =  (u, v, w)= 0.6x + 0.2t - 1.4 \ m/s$

   The temperature of the room is  $T(x, y, u) =  400 -0.4y -0.6z-0.2(5 - x)^2 \  ^o C$

    The time considered is  t =  10 \  seconds

    The  distance that the bird flew is  x  =  1 m

 Given that the bird is inside the room then the temperature of the room is equal to the temperature of the bird

Generally the change in the bird temperature with time is mathematically represented as

      $\frac{dT}{dt} = -0.4 \frac{dy}{dt} -0.6\frac{dz}{dt} -0.2[2 *  (5-x)] [-\frac{dx}{dt} ]$

Here the negative sign in $\frac{dx}{dt}$ is because of the negative sign that is attached to x in the equation

 So

       $\frac{dT}{dt} = -0.4v_y  -0.6v_z -0.2[2 *  (5-x)][ -v_x]$

From the given equation of velocity field

    $v_x  =  0.6x$

    $v_y  =  0.2t$

     $v_z  =  -1.4$

So

$\frac{dT}{dt} = -0.4[0.2t]  -0.6[-1.4] -0.2[2 *  (5-x)][ -[0.6x]]$    

substituting the given values of x and t

$\frac{dT}{dt} = -0.4[0.2(10)]  -0.6[-1.4] -0.2[2 *  (5-1)][ -[0.61]]$      

$\frac{dT}{dt} = -0.8 +0.84 + 0.976$  

$\frac{dT}{dt} = 1.016 ^oC/m$