Respuesta :
Answer: 107.67°C
Step-by-step explanation:
I guess that we could do a Taylor expansion around the point (2, 1)
Remember that a Taylor expansion around the point (a, b) is:
[tex]T(x,y) = T(a, b) + \frac{dT(a,b)}{dx}(x - a) + \frac{dT(a,b)}{dy}(y - b) + .....[/tex]
Where the latter terms need higher orders of the derivates of T, so we can not find them, regardless of that, this expansion will be accurate near the point (a, b),
Then, using this, we can write our expanssion as:
T(x,y) = 107 + 9*(x - 2) - 8*(y - 1)
Now we evaluate this in x = 2.03 and y = 0.95
T(2.03, 0.95) = 107 + 9(2.03 - 2) - 8*(0.95 - 1) = 107.67
Then a good estimation of the temperature at the point (2.03,0.95) is 107.67°C
The estimate of the temperature of the unevenly heated metal plate at temperature at the point (2.03,0.95) is;
T(2.03, 0.95) = 107.67 °C
Formula for taylor expansion around a point (a, b) is given as;
T(x, y) = T(a,b) + [tex]\frac{dT(a,b)}{dx}[/tex](x - a) + [tex]\frac{dT(a,b)}{dy}[/tex](y - b) + ....
We are given;
T(2,1) = 107
T'x(2,1) = [tex]\frac{dT(a,b)}{dx}[/tex] = 9
T'y(2,1) = [tex]\frac{dT(a,b)}{dy}[/tex] = −8
Plugging in the relevant values into our taylor expansion above gives;
T(x, y) = 107 + 9(x - 2) - 8(y - 1)
Now, we want to estimate the temperature at the point (2.03, 0.95).
Thus, we will plug in x = 2.03 and y = 0.95 to get;
T(2.03, 0.95) = 107 + 9(2.03 - 2) - 8(0.95 - 1)
T(2.03, 0.95) = 107 + 0.27 + 0.4
T(2.03, 0.95) = 107.67 °C
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