Question

A square plate RequalsStartSet (x comma y ): 0 less than or equals x less than or equals 1 comma 0 less than or equals y less than or equals 1 EndSet has a temperature distribution Upper T (x comma y )equals100 minus 25 x minus 40 y.
a. Sketch two level curves of the temperature in the plate.
b. Find the gradient of the temperature gradient Upper T (x comma y ).

Answer 1

Answer:

Step-by-step explanation:

We are given the following:

$R = \{(x,y)|0\leq x \leq 1, 0 \leq y \leq 1 \}$

and $T(x,y) = 100-25x - 40 y$

a). Recall that a level curve of a function f(x,y) is given by $R_c = \{(x,y) | f(x,y) = c\}$ where c is a constant. That is, all the points in the set of interest to which the function applied to the points is exactly the value c.

Consider c = 80. So we get

$100-25x-40y = 80$

which implies that $y = \frac{-25}{40}x+0.5$(Graph 1).

We can also consider c=60, which gives us

$100-25x-40y = 60$

which implies that $y = \frac{-25}{40}x+1$. (Graph 2)

b)Recall that the gradient of a function f(x,y) is given by

$\nabla f = (\frac{df}{dx}, \frac{df}{dy})$

In this case,

$\frac{dT}{dx} = -25, \frac{dT}{dy} = -40$

Thus, the gradient of T is given by

$\nabla T =(-25,-40)$