the formula for wind chill c (in degrees fahrenheit) is given by c = 35.74+0.6215t - 35.75v^0.16+0.4275tv^0.1…

the formula for wind chill c (in degrees fahrenheit) is given by c = 35.74+0.6215t - 35.75v^0.16+0.4275tv^0.16 where v is the wind speed in miles per hour and t is the temperature in degrees fahrenheit. the wind speed is 22 ± 3 miles per hour and the temperature is 8 ± 2°. use dc to estimate the maximum possible propagated error and relative error in calculating the wind - chill. step 1 first find the partial derivative with respect to t. ∂c/∂t = + 0.4275v^0.16 substitute for the median value v = 22 and simplify. (round your answer to four decimal places.)

the formula for wind chill c (in degrees fahrenheit) is given by c = 35.74+0.6215t - 35.75v^0.16+0.4275tv^0.16 where v is the wind speed in miles per hour and t is the temperature in degrees fahrenheit. the wind speed is 22 ± 3 miles per hour and the temperature is 8 ± 2°. use dc to estimate the maximum possible propagated error and relative error in calculating the wind - chill. step 1 first find the partial derivative with respect to t. ∂c/∂t = + 0.4275v^0.16 substitute for the median value v = 22 and simplify. (round your answer to four decimal places.)

Answer

Explanation:

Step1: Find partial - derivative of C with respect to T

Given $C = 35.74+0.6215T - 35.75v^{0.16}+0.4275Tv^{0.16}$. Using the sum - rule of differentiation $\frac{\partial}{\partial T}(u + v+w + z)=\frac{\partial u}{\partial T}+\frac{\partial v}{\partial T}+\frac{\partial w}{\partial T}+\frac{\partial z}{\partial T}$, where $u = 35.74$, $v = 0.6215T$, $w=-35.75v^{0.16}$, $z = 0.4275Tv^{0.16}$. Since $\frac{\partial(35.74)}{\partial T}=0$, $\frac{\partial(0.6215T)}{\partial T}=0.6215$, $\frac{\partial(-35.75v^{0.16})}{\partial T}=0$ and $\frac{\partial(0.4275Tv^{0.16})}{\partial T}=0.4275v^{0.16}$. So, $\frac{\partial C}{\partial T}=0.6215 + 0.4275v^{0.16}$.

Step2: Substitute $v = 22$

Substitute $v = 22$ into $\frac{\partial C}{\partial T}$: $\frac{\partial C}{\partial T}=0.6215+0.4275\times22^{0.16}$. First, calculate $22^{0.16}$. Let $x = 22^{0.16}$, then $\ln x=0.16\ln22$. $\ln22\approx3.09104$, so $0.16\ln22\approx0.16\times3.09104 = 0.4945664$. $x = e^{0.4945664}\approx1.6393$. Then $0.4275\times1.6393\approx0.7008$. $\frac{\partial C}{\partial T}=0.6215 + 0.7008=1.3223$.

Answer:

$1.3223$