Refer to the diagram shown.
Given:
[tex]h(n,e) = 1500 + 75n + 50e[/tex]
Define
[tex]\hat{r} = unit \, vector \, along \, \vec{v} \\ \hat{i} = unit \, vector \, east \\ \hat{j} = unit \, vector \, north \\ \nabla \equiv \hat{i} \frac{\partial}{\partial e} + \hat{j} \frac{\partial}{\partial n} [/tex]
[tex]\hat{r} = \frac{1}{ \sqrt{2} } (-\hat{i}+\hat{j} )[/tex]
Then the rate of change of h with respect to the vector v is
[tex]\nabla h . \hat{r} = \frac{1}{\sqrt{2}}(50\hat{i} + 75\hat{j}).(-\hat{i}+\hat{j}) = \frac{1}{\sqrt{2}} (-50+75) =17.68[/tex]
Answer: 17.7 ft per mile
