Question

The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperature is T (in degrees Fahrenheit). An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is I (T,H) = 45.33 + 0.6845T + 5.758H − 0.00365T 2 − 0.1565HT + 0.001HT2 (a) Calculate I at (T ,H) = (95, 50). (b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

Answer 1

Answer:

a) $I(95,50) = 73.19$ degrees

b) $I_{T}(95,50) = -7.73$

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

$I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}$

a) Calculate I at (T ,H) = (95, 50).

$I(95,50) = 45.33 + 0.6845*(95) + 5.758*(50) - 0.00365*(95)^{2} - 0.1565*95*50 + 0.001*50*95^{2} = 73.19$ degrees

(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is $I_{T}(T,H)$. So

$I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}$

$I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H$

$I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73$