The thermal sensation can be modeled by $$W =35.74 + 0.6215T +(0.4275T - 35.75)v^{0.16}$$ where $W$ (thermal sensation) and the temperature $T$ are given in Fahrenheit and v (wind speed) in miles per hour. Evaluate the partial derivative of $W$ with respect to $v$ in $(T,v) = (25,10)$ and interpret that partial derivative as a rate of change.
My answer
I evaluated the partial as follows:
\begin{align*} \frac{\partial W}{\partial v} &= 0 + 0 + (0.4275T - 35.75) \cdot (0.16) \cdot v^{- 0.84}\\ \frac{\partial W}{\partial v} \bigg\rvert_{(25,10)} &= -0.5796 \end{align*}
My question
I am struggling with answering the second part of the question and also, on finding the unit for $ \frac{\partial W}{\partial v}$. Does the partial mean that, keeping the temperature as 25 Fahrenheit, the thermal sensation will decrease $0.5796$ degrees for a $1$ mile per hour increment in the wind speed? Also, is the unit for $ \frac{\partial W}{\partial v}$ $\frac{F \cdot h}{mi}$?
Any help is highly appreciated.
You are correct. $\dfrac{\partial W}{\partial v}=-0.5796\;\;$ means that the thermal sensation will decrease wrt the velocity at a rate of $0.5796\;\;$ while temperature is constant. the units of $\dfrac{\partial W}{\partial v}$ are the same as $\dfrac{W}{v}$